\section{\textbf{Introduction and Preliminary Results}}

In this section we recall some well-known results that will partly be needed in this paper and to some for the sake of their beauty and insurmountable importance in the field. We find some results concerning the distribution of some sequences in arithmetic progression useful in the current paper. First we state the celebrated Szemeredi theorem concerning arithmetic progression. The theorem has both infinite and finite version, but we have considered appropriate to state the finite version.

\begin{theorem}[Szemeredi]\label{szemeredi}
$\forall$ $\epsilon>0$ and $\forall k\in \mathbb{N}$ there exists an $n\in \mathbb{N}$ such that if $A\subset \mathbb{N}_n$
\footnote{see the notation in section 3.}
satisfies $|A|\geq \epsilon n$, then $A$ contains an arithmetic progression of length $k$.
\end{theorem}
The well-known Green-Tao theorem \cite{green2008primes} provides an extension in this direction as

\begin{theorem}[Green-Tao]\label{Green-Tao}
Let $\pi(n)$ denotes the number of primes no more than $n$. If $A\subset \mathbb{P}$ the set of all prime numbers such that 
\begin{align}
    \limsup \limits_{n\longrightarrow \infty}\frac{|A\cap \mathbb{N}_n|}{\pi(n)}>0\nonumber
\end{align}then $A$ contains infinitely many arithmetic progressions of length $k$ for any $k>0$.
\end{theorem}
\bigskip

In this paper, motivated in part by the binary Goldbach conjecture and its variants, we develop a method which we feel might be a valuable resource and a recipe for studying problems concerning partition of numbers in specified subsets of $\mathbb{N}$. The method is very elementary in nature and has parallels with configurations of points on the geometric circle.\\ Let us suppose that for any $n\in \mathbb{N}$ we can write $n=u+v$ where $u,v\in \mathbb{M}\subset \mathbb{N}$ then the circle embedding method associate each of this summands to points on the circle generated in a certain manner by $n>2$ and a line joining any such associated points on the circle. This geometric correspondence turns out to useful in our development, as the results obtained in this setting are then transformed back to results concerning the partition of integers. We study various features and statistics of a Combinatorial structure in this development, which we choose to call the \textbf{circle of partition}.
\bigskip


\section{\textbf{The Circle of Partition}}
In this section we introduce the notion of the circle of partition. We study this notion in-depth and explore some potential applications in the following sequel.

\begin{definition}\label{major}
Let $n\in \mathbb{N}$ and $\mathbb{M}\subset \mathbb{N}$. We denote with 
\begin{align}
\mathcal{C}(n,\mathbb{M})=\left \{[x]\mid x,y\in \mathbb{M},n=x+y\right \}\nonumber
\end{align}
the Circle of Partition generated by $n$ with respect to the subset $\mathbb{M}$. We will abbreviate this in the further text as CoP.  We call members of  $\mathcal{C}(n,\mathbb{M})$  as points and denote them by $[x]$. For the special case $\mathbb{M}=\mathbb{N}$ we denote the CoP shortly as $\mathcal{C}(n)$. 
\end{definition}


\begin{definition}\label{axis}
We denote the line $\mathbb{L}_{[x],[y]}$ joining the point $[x]$ and $[y]$ as an axis of the CoP $\mathcal{C}(n,\mathbb{M})$ if and only if $x+y=n$. 
We say the axis point $[y]$ is an axis partner of the axis point $[x]$ and vice versa. 
We do not distinguish between $\mathbb{L}_{[x],[y]}$ and $\mathbb{L}_{[y],[x]}$, since  it is essentially the  the same axis.
The point $[x]\in \mathcal{C}(n,\mathbb{M})$ such that $2x=n$ is the \textbf{center} of the CoP.
If it exists then we call it as a \textbf{degenerated axis} $\mathbb{L}_{[x]}$ in comparison to the \textbf{real axes} $\mathbb{L}_{[x],[y]}$. 
The line joining any two arbitrary point which are not axes partners on the CoP will be referred to as a \textbf{chord} of the CoP. The length of the chord $\mathcal{L}_{[x],[y]}$ joining the points $[x],[y]\in \mathcal{C}(n,\mathbb{M})$, denoted as $\Gamma([x],[y])$ is given by 
\begin{align}
\Gamma([x],[y])=|x-y|.\nonumber
\end{align}
\end{definition}
\bigskip

It is important to point out that the \textbf{median} of the weights of each co-axis point coincides with the center of the underlying CoP if it exists. That is to say, given all the real axes of the CoP $\mathcal{C}(n,\mathbb{M})$ as 
\begin{align}
\mathbb{L}_{[u_1],[v_1]},\mathbb{L}_{[u_2],[v_2]},\ldots, \mathbb{L}_{[u_k],[v_k]}\nonumber
\end{align}then the following relations hold
\begin{align}
\frac{u_1+v_1}{2}=\frac{u_2+v_2}{2}=\cdots=\frac{u_k+v_k}{2}=\frac{n}{2}\nonumber
\end{align}which is equivalent to the conditions for any of the pair of real axes $\mathbb{L}_{[u_i],[v_i]},\mathbb{L}_{[u_j],[v_j]}$ for $1\leq i,j\leq k$
\begin{align}
\Gamma([u_i],[u_j])=\Gamma([v_i],[v_j])\nonumber
\end{align}and 
\begin{align}
\Gamma([v_j],[u_i])=\Gamma([u_j],[v_i]).\nonumber
\end{align}
\bigskip

\begin{definition}\label{aligned_embedding}
Let $\mathbb{M}\subseteq\mathbb{N}$ and $\mathcal{C}(n,\mathbb{M})$ and $\mathcal{C}(m,\mathbb{M})$ be two distinct CoPs for which holds
\begin{align}
\mathcal{C}(n,\mathbb{M})&\subset\mathcal{C}(m,\mathbb{M})\label{E-embedding1}\\
&\mbox{or}\nonumber\\
\mathcal{C}(n,\mathbb{M})&\supset\mathcal{C}(m,\mathbb{M})\label{E-embedding2}.
\end{align}
Then we say the CoPs \textit{admit embedding}. We say the CoPs \textit{admit aligned embedding} if and only if with (\ref{E-embedding1}) holds $n<m$ and with (\ref{E-embedding2}) $n>m$ and $\mathcal{C}(n,\mathbb{M})=\mathcal{C}(m,\mathbb{M})$ holds if and only if $n=m$.
We say the CoPs \textit{admit reverse aligned embedding} if and only if with (\ref{E-embedding1}) holds $n>m$ and with (\ref{E-embedding2}) $n<m$.
\end{definition}
\bigskip

\paragraph*{\textbf{Notations}}
We let 
\begin{align}
\mathbb{N}_n=\left \{m\in \mathbb{N}\mid ~m\leq n\right \}\nonumber
\end{align}
be the \textbf{sequence} of the first $n$ natural numbers. Further we will denote 
\begin{align}
\Vert[x]\Vert:=x\nonumber
\end{align}
as the \textbf{weight} of the point $[x]$ and correspondingly the weight set of points in the CoP $\mathcal{C}(n,\mathbb{M})$ as $||\mathcal{C}(n,\mathbb{M})||$.
\bigskip

The above language in many ways could be seen as a criterion determining the plausibility of carrying out a partition in a specified set. Indeed this feasibility is trivial if we take the set $\mathbb{M}$ to be the set of natural numbers $\mathbb{N}$. The situation becomes harder if we take the set $\mathbb{M}$ to be a special subset of natural numbers $\mathbb{N}$, as the corresponding CoP $\mathcal{C}(n,\mathbb{M})$ may not always be non-empty for all $n\in \mathbb{N}$. One archetype of problems of this flavour is the binary Goldbach conjecture, when we take the base set $\mathbb{M}$ to be the set of all prime numbers $\mathbb{P}$. One could imagine the same sort of difficulty if we extend our base set to other special subsets of the natural numbers. As such we start by developing the theory assuming the base set of natural numbers $\mathbb{N}$ and latter extend it to other base sets  $\mathbb{M}$ equipped with certain important and subtle properties. 

\begin{remark}
It is important to notice that a typical CoP need not have a center. In the case of an absence of a center then we say the circle has a deleted center. However all CoPs $\mathcal{C}(n)$ with even generators have a center. It is easy to see that the CoP $\mathcal{C}(n)$ contains all points whose weights are positive integers from $1$ to $n-1$ inclusive: 
\[
\mathcal{C}(n)=\lbrace[x]\mid~x\in \mathbb{N},x<n\rbrace.
\] 
Therefore the CoP $\mathcal{C}(n)$ has $\left \lfloor \frac{n-1}{2}\right \rfloor$ different real axes.
\end{remark}

\begin{proposition}\label{unique}
Each axis is uniquely determined by points $[x]\in \mathcal{C}(n,\mathbb{M})$. 
\end{proposition}

\begin{proof}
A degenerated axis is determined by the center of the CoP. And this is unique if it exists.

Let $\mathbb{L}_{[x],[y]}$ be a real axis of the CoP $\mathcal{C}(n,\mathbb{M})$. Suppose as well that $\mathbb{L}_{[x],[z]}$ is also a real axis with $z\neq y$. Then it follows by Definition \ref{axis} that we must have $n=x+y=x+z$ and therefore $y=z$. This cannot be and the claim follows immediately.
\end{proof}

\begin{corollary}\label{partner}
Each point of a CoP $\mathcal{C}(n,\mathbb{M})$ excluding an existing center has exactly one real axis partner.
\end{corollary}
\begin{proof}
Let $[x]\in \mathcal{C}(n,\mathbb{M})$ be a point without a real axis partner. Then holds for every point $[y]\neq [x]$
\[
\Vert[x]\Vert+\Vert[y]\Vert\neq n.
\]
This contradiction to the Definition \ref{major}. Due to Proposition \ref{unique} the case of more than one axis partners is impossible. This completes the proof.
\end{proof}

\begin{corollary}\label{ordered_weights}
The weights of the points of 
\[
\mathcal{C}(n,\mathbb{M})=\lbrace[x_1],[x_2].\ldots,[x_k]\rbrace
\]
are \textbf{strictly totally ordered}.
\end{corollary}

\begin{proof}
W.l.o.g. we assume that 
\begin{align}
x_1&=\min\left(x\mid[x]\in\mathcal{C}(n,\mathbb{M})\right)\mbox{ and}\label{E-min} \\
x_k&=\max\left(x\mid[x]\in\mathcal{C}(n,\mathbb{M})\right).\label{E-max}
\end{align}
At first we assume that $x_1+x_k<n$. Then there is a weight $x_i$ with
\begin{align*}
&x_1<x_i<x_k\mbox{ and }n=x_1+x_i.\\
&\mbox{Because }x_i<x_k\mbox{ we get}\\
&n=x_1+x_i<x_1+x_k.
\end{align*}
This contradicts the assumption. Now we assume that $x_1+x_k>n$. Then there is a weight $x_i$ with
\begin{align*}
&x_1<x_i<x_k\mbox{ and }n=x_i+x_k.\\
&\mbox{Because }x_i>x_1\mbox{ we get}\\
&n=x_i+x_k>x_1+x_k.
\end{align*}
This also contradicts the assumption. Therefore remains $x_1+x_k=n$. Because of (\ref{E-min}) and (\ref{E-max}) holds
\[
x_1<x_2<x_{k-1}<x_k.
\]
Now we remove $x_1$ and $x_k$ out of the consideration and repeat the procedure above with $x_2$ and $x_{k-1}$ and obtain
$x_2+x_{k-1}=n$ and
\[
x_1<x_2<x_3<x_{k-2}<x_{k-1}<x_k.
\]
By repeating this procedure for $x_i$ and $x_{k+1-i}$ for $3\leq i \leq\left\lfloor\frac{k}{2}\right\rfloor$ we get finally
\[
x_1<x_2<x_3<x_4<\ldots<x_{k-3}<x_{k-2}<x_{k-1}<x_k.
\]
\end{proof}

\begin{proposition}\label{cup-subset}
Let $\mathcal{C}(n,\mathbb{M})$ and $\mathcal{C}(m,\mathbb{M})$ be two distinct CoPs admitting aligned embedding. Then holds
\[
\mathcal{C}(n,\mathbb{M})\cup\mathcal{C}(m,\mathbb{M})
\subset\mathcal{C}(n+m,\mathbb{M}).
\]
\end{proposition}
\begin{proof}
W.l.o.g. we assume $\mathcal{C}(n,\mathbb{M})\subset\mathcal{C}(m,\mathbb{M})$. Then holds
\begin{align*}
\mathcal{C}(n,\mathbb{M})\cup\mathcal{C}(m,\mathbb{M})
&=\mathcal{C}(m,\mathbb{M})\\
&\mbox{ and because of \textit{admitting aligned embedding}}\\
&\subset\mathcal{C}(n+m,\mathbb{M}) \mbox{ due to }m<n+m.
\end{align*} 
\end{proof}


\begin{theorem}\label{properties}
Let $n\in\mathbb{N}$ and $\mathcal{C}(n)$ be a CoP generated by $n$. Then $\mathcal{C}(n)$ admits aligned embedding.
\end{theorem}
\begin{proof}
W.l.o.g. we have to prove for two distinct CoPs
\[
\mathcal{C}(n)\subset\mathcal{C}(m)
\mbox{ if and only if }n<m\mid n,m\in\mathbb{N}.
\]
First let $n<m$. Then follows that
\begin{align*}
\mathcal{C}(n)&=\lbrace [x]\mid x\in\mathbb{N}, x<n\rbrace\\
&\subset\lbrace [x]\mid x\in\mathbb{N}, x<m\rbrace\\
&=\mathcal{C}(m).
\end{align*}
Conversely we suppose $\mathcal{C}(n)\subset\mathcal{C}(m)$. Then it follows that 
\[
\lbrace [x]\mid x\in\mathbb{N}, x<n\rbrace\subset
\lbrace [x]\mid x\in\mathbb{N}, x<m\rbrace
\]
and it holds $n<m$.
\end{proof}

\bigskip

Now we will see that Theorem \ref{properties} is always valid for some special subsets $\mathbb{M}$ instead of $\mathbb{N}$, the subsets containing arithmetic progressions.
Let be $\mathbb{M}_{a,d} \subset \mathbb{N}$ with
\begin{align}\label{Mad}
\mathbb{M}_{a,d}:= \lbrace x \in \mathbb{N} \mid x \equiv a \mood{d}, d \in \mathbb{N}\rbrace
\end{align}
and
\begin{align*}
\mathcal{C}(n,\mathbb{M}_{a,d}) &= \lbrace [x]\mid x+y=n\wedge x,y \in \mathbb{M}_{a,d}\rbrace, n \in \mathbb{M}_{2a,d}\\
&= \lbrace [x]\mid x \in \mathbb{M}_{a,d}\wedge  x \leq n-a \rbrace.
\end{align*}
For $x<y \in \mathbb{M}_{a,d}$ holds $y - x \equiv 0 \mood{d}$. On the other hand holds $x + y \equiv 2a \mood{d}$, so that $\mathcal{C}(n,\mathbb{M}_{a,d}) = \emptyset$ for $n \not\in \mathbb{M}_{2a,d}$.

\begin{theorem}\label{properties2}
Let $n \in \mathbb{M}_{2a,d}$ and $\mathcal{C}(n,\mathbb{M}_{a,d})$ be a CoP generated by $n$. Then the CoP admits aligned embedding an increment $d$.
\end{theorem}
\begin{proof}
W.l.o.g. we have to prove
\[
\mathcal{C}(n,\mathbb{M}_{a,d}) \subset \mathcal{C}(m,\mathbb{M}_{a,d}) 
 \mbox{ if and only if }n < m.
\]
At first let be $n < m$. Since $n,m\in\mathbb{M}_{2a,d}$ holds $m-n = k\cdot d$ has an increment $d$.
Further holds
\begin{align*}
\Vert\mathcal{C}(n,\mathbb{M}_{a,d})\Vert &=
\lbrace k \in \mathbb{M}_{a,d} \mid k \leq n-a \rbrace\\
&\mbox{and because of }n < m\\
&\subset \lbrace k \in \mathbb{M}_{a,d} \mid k \leq m-a \rbrace\\ 
&=\Vert\mathcal{C}(m,\mathbb{M}_{a,d})\Vert. 
\end{align*}

On the other hand
let be $\mathcal{C}(n,\mathbb{M}_{a,d})\subset \mathcal{C}(m,\mathbb{M}_{a,d})$. Then holds
\begin{align*}
\Vert\mathcal{C}(n,\mathbb{M}_{a,d})\Vert &=
\lbrace k \in \mathbb{M}_{a,d} \mid k \leq n-a \rbrace\\
&\subset \Vert\mathcal{C}(m,\mathbb{M}_{a,d})\Vert\\
&=\lbrace k \in \mathbb{M}_{a,d} \mid k \leq m-a \rbrace\\ 
&\mbox{and therefore must be}\\
&n < m.
\end{align*}
\end{proof}

\begin{corollary}\label{col-prop2}
Let $\mathcal{C}(n,\mathbb{M})$ and $\mathcal{C}(m,\mathbb{M})$ be two distinct CoPs admit align embedding. Then holds
\begin{align*}
\mathcal{C}(n,\mathbb{M})\supset\mathcal{C}(m,\mathbb{M})
\mbox{ if and only if }
n&>m.
\end{align*}
\end{corollary}

\begin{corollary}
Because of Proposition \ref{cup-subset} and Theorem \ref{properties2} holds for two distinct CoPs $\mathcal{C}(n,\mathbb{M}_{a,d})$ and $\mathcal{C}(m,\mathbb{M}_{a,d})$
\footnote{
$n+m-2a$ on the right side in order to get $n+m-2a\in\mathbb{M}_{2a,d}$ by $n,m\in\mathbb{M}_{2a,d}$.
}
\[
\mathcal{C}(n,\mathbb{M}_{a,d}) \cup \mathcal{C}(m,\mathbb{M}_{a,d})
\subset\mathcal{C}(n+m-2a,\mathbb{M}_{a,d}).
\]
\end{corollary}

\begin{remark}
CoPs $\mathcal{C}(n,\mathbb{P})$ with the set of all prime numbers as base set are important examples for CoPs not admitting embedding. The following example demonstrates this scenario.
\begin{align*}
\mathcal{C}(20,\mathbb{P})&=\lbrace [3],[7],[13],[17]\rbrace\mbox{ but}\\
\mathcal{C}(22,\mathbb{P})&=\lbrace [3],[5],[11],[17],[19]\rbrace.
\end{align*} 
\end{remark}

\begin{proposition}\label{commonfirst}
Let $\mathbb{M}\subseteq\mathbb{N}$ and $\mathcal{C}(n,\mathbb{M})\subset\mathcal{C}(m,\mathbb{M})$ two CoPs with a common base set $\mathbb{M}$ and $w_o$ and $z_o$ the weights of the median points of $\mathcal{C}(n,\mathbb{M})$ resp. $\mathcal{C}(m,\mathbb{M})$.
If $w_o<z_o$ then the CoPs admit aligned embedding, if $w_o>z_o$ the CoPs admit reverse aligned embedding.
\end{proposition}

\begin{proof}
Let
\begin{align*}
u_o:&=\min(u\in\Vert\mathcal{C}(n,\mathbb{M})\Vert)\mbox{ and}\\
x_o:&=\min(x\in\Vert\mathcal{C}(m,\mathbb{M})\Vert)
\mbox{ be the least weights of the CoPs and}\\
v_o:&=\max(v\in\Vert\mathcal{C}(n,\mathbb{M})\Vert)\mbox{ and}\\
y_o:&=\max(y\in\Vert\mathcal{C}(m,\mathbb{M})\Vert)
\mbox{ the greatest weights of the CoPs.}
\end{align*}
Because the CoPs are strictly totally ordered the minimal and the maximal points are unique.
Then 
\[
w_o:=\frac{u_o+v_o}{2}=\frac{n}{2}\mbox{ and }
z_o:=\frac{x_o+y_o}{2}=\frac{m}{2}
\]
are the weights of the median points of the CoPs and we can distinguish three cases
\begin{itemize}
\item[A.)]
$w_o < z_o$,
\item[B.)]
$w_o = z_o$,
\item[C.)]
$w_o > z_o$.
\end{itemize}
Because of $\mathcal{C}(n,\mathbb{M})\subset\mathcal{C}(m,\mathbb{M})$ all points of $\mathcal{C}(n,\mathbb{M})$ must be also points of $\mathcal{C}(m,\mathbb{M})$. Therefore must be
\begin{align*}
x_o\leq u_o<v_o\leq y_o.
\end{align*}
Now we consider the case A.):
\par From $w_o < z_o$ follows immediately $n<m$. That means $\mathcal{C}(n,\mathbb{M})$ admits aligned embedding. This includes the case of a common first point ($x_o=u_o$) of both CoPs.\\
The opposite we get in case C.):
\par From $w_o > z_o$ follows immediately $n>m$. That means $\mathcal{C}(n,\mathbb{M})$ admits reverse aligned embedding.. This includes the case of a common last point ($v_o=y_o$) of both CoPs.\\
In case B.) we would obtain $n=m$. Because of $\mathcal{C}(n,\mathbb{M})\subset\mathcal{C}(m,\mathbb{M})$ there must be at least one real axis $\mathbb{L}_{[s],[t]}~ \hat{\in}~ \mathcal{C}(m,\mathbb{M})$ which is not a real axis of $\mathcal{C}(n,\mathbb{M})$. But this is because of $n=m$ impossible. Therefore case B.) don't occur.
\end{proof}

\begin{example}
As an example for reverse aligned embedding
we consider the following CoPs
\begin{align*}
\mathcal{C}(36,\mathbb{P})&=\lbrace [5],\mathbf{[7]},[13],[17],\mathbf{[19]},[23],[29],\mathbf{[31]}\rbrace\mbox{ and }\\
\mathcal{C}(38,\mathbb{P})&=\lbrace [7],[19],[31]\rbrace.
\end{align*}
We see that $\mathcal{C}(38,\mathbb{P})\subset\mathcal{C}(36,\mathbb{P})$ but $38>36$.
\end{example}

\paragraph*{\textbf{Notation}}
Let us denote the assignment of an axis $\mathbb{L}_{[x],[y]}$ resp. $\mathbb{L}_{[x]}$ to a CoP $\mathcal{C}(n,\mathbb{M})$ as
\begin{align*}
&\mathbb{L}_{[x],[y]}~ \hat{\in}~ \mathcal{C}(n,\mathbb{M})
\mbox{ which means }
[x],[y] \in \mathcal{C}(n,\mathbb{M}) \mbox{ and } x+y=n\mbox{ resp.}\\
&\mathbb{L}_{[x]}~ \hat{\in}~ \mathcal{C}(n,\mathbb{M})
\mbox{ which means }
[x]\in \mathcal{C}(n,\mathbb{M}) \mbox{ and } 2x=n
\end{align*}
and the number of real axes of a CoP as
\[
\nu(n,\mathbb{M}):=\#\lbrace\mathbb{L}_{[x],[y]}~ \hat{\in}~ \mathcal{C}(n,\mathbb{M})\mid x<y\rbrace.
\]
Obviously holds
\[
\nu(n,\mathbb{M})=\left\lfloor\frac{k}{2}\right\rfloor ,\mbox{ if }
\vert\mathcal{C}(n,\mathbb{M})\vert =k.
\]

\begin{proposition}\label{increaarithmetic}
Let $\mathbb{M}\subset\mathbb{N}$ and $\mathcal{C}(n,\mathbb{M})$ be a CoP admitting aligned embedding.
Then $\nu(n,\mathbb{M})$ is a non-decreasing function for all $n$ such that $\mathcal{C}(n,\mathbb{M})$ is not empty.
\end{proposition}

\begin{proof}
Since the CoP $\mathcal{C}(n,\mathbb{M})$ admits aligned embedding it holds w.l.o.g.
\begin{align*}
\mathcal{C}(n,\mathbb{M})&\subset\mathcal{C}(m,\mathbb{M})
\mbox{ for }n<m \mbox{ and hence}\\
\vert\mathcal{C}(n,\mathbb{M})\vert &< \vert\mathcal{C}(m,\mathbb{M})\vert
\mbox{ and therefore}\\
\nu(n,\mathbb{M})&<\nu(m,\mathbb{M}).
\end{align*} 
\end{proof}

Let be
\begin{equation}\label{NStern}
\mathbb{N}^* = \lbrace n\in\mathbb{N}\mid n\equiv\pm 1\mood{6}\rbrace.
\end{equation}
Then holds that the set $\mathbb{P}^*$ of all primes $\geq 5$ is covered by $\mathbb{N}^*$.\\

\begin{proposition}\label{properties3}
The CoP $\mathcal{C}(n,\mathbb{N}^*)$ admits aligned embedding with an increment $6$ for all $n$ such that $n\equiv\pm 2\mood{6}$ or $n\equiv 0\mood{6}$.
\end{proposition}
\begin{proof}
If $n\equiv -2\mood{6}$ then must hold for the weights of all points $[x]\in\mathcal{C}(n,\mathbb{N}^*)$ $x\equiv -1\mood{6}$. Then all points of $\mathcal{C}(n,\mathbb{N}^*)$ are points of $\mathcal{C}(n,\mathbb{M}_{5,6})$.
In the other case $n\equiv +2\mood{6}$ must be $x\equiv +1\mood{6}$.
Hence holds
\begin{align*}
\mathcal{C}(n,\mathbb{N}^*)=
\begin{cases}
\mathcal{C}(n,\mathbb{M}_{5,6})&\mbox{ if }n\equiv -2\mood{6}\\
\mathcal{C}(n,\mathbb{M}'_{1,6})&\mbox{ if }n\equiv +2\mood{6}
\end{cases}
\end{align*}
where $\mathbb{M}'_{1,6}:=\mathbb{M}_{1,6}\setminus \lbrace 1\rbrace$. Because of Theorem \ref{properties2} follows the claim for $n\equiv\pm 2\mood{6}$.

If $n\equiv 0\mood{6}$  then it must be for every real axis $\mathbb{L}_{[x],[y]}$
\[
x \opn{Mod} 6 =-y \opn{Mod} 6.
\]
This means that if $x\in\mathbb{M}_{5,6}$ then must be $y\in\mathbb{M}'_{1,6}$ and reverse.
W.l.o.g. we assume $x\in\mathbb{M}_{5,6}$ and $y\in\mathbb{M}'_{1,6}$ with $x,y\in\mathcal{C}(n,\mathbb{N}^*)$ and $x<y$. Then is 
\begin{align*}
&x+2\in\mathbb{M}'_{1,6}\mbox{ and }y-2\in\mathbb{M}_{5,6}\mbox{ and due to }x+2+y-2=n\\
&\mbox{holds }x+2,y-2\in\mathcal{C}(n,\mathbb{N}^*)
\mbox{ with }\mathbb{L}_{[x+2],[y-2]}~ \hat{\in}~ \mathcal{C}(n,\mathbb{N}^*)\\
&\mbox{and we have a chain of weights of }\mathcal{C}(n,\mathbb{N}^*)\\
&x<x+2<y-2<y.
\end{align*}
Also holds 
\begin{align*}
&\mathbb{L}_{[x],[y-2]}~ \hat{\in}~ \mathcal{C}(n-2,\mathbb{M}_{5,6})\mbox{ because of }x+y-2=n-2\mbox{ and}\\
&\mathbb{L}_{[x+2],[y]}~ \hat{\in}~ \mathcal{C}(n+2,\mathbb{M}'_{1,6})\mbox{ due to }x+2+y=n+2.
\end{align*}
Therefore to each real axis $\mathbb{L}_{[x],[y]}~ \hat{\in}~ \mathcal{C}(n,\mathbb{N}^*)$ belong  
\begin{align*}
&\mbox{a second real axis }\mathbb{L}_{[x+2],[y-2]}~ \hat{\in}~ \mathcal{C}(n,\mathbb{N}^*)
\mbox{ and}\\
&\mbox{a real axis }\mathbb{L}_{[x],[y-2]}~ \hat{\in}~ \mathcal{C}(n-2,\mathbb{M}_{5,6})\mbox{ and}\\
&\mbox{a real axis }\mathbb{L}_{[x+2],[y]}~ \hat{\in}~ \mathcal{C}(n+2,\mathbb{M}'_{1,6})
\end{align*}
and each point of $\mathcal{C}(n,\mathbb{N}^*)$ is a point of either $\mathcal{C}(n-2,\mathbb{M}_{5,6})$ or $\mathcal{C}(n+2,\mathbb{M}'_{1,6})$.

Now let us consider a real axis $\mathbb{L}_{[u],[v]}~ \hat{\in}~ \mathcal{C}(n-2,\mathbb{M}_{5,6})$ with $u<v$. Then is because of $u+v=n-2$
\begin{align*}
&\mathbb{L}_{[u+2],[v]}~ \hat{\in}~ \mathcal{C}(n,\mathbb{N}^*)\mbox{ because of }u+2+v=n\mbox{ and}\\
&\mathbb{L}_{[u],[v+2]}~ \hat{\in}~ \mathcal{C}(n,\mathbb{N}^*)\mbox{ due to }u+v+2=n\\
&\mbox{and we have a chain of weights of }\mathcal{C}(n,\mathbb{N}^*)\\
&u<u+2<v<v+2.
\end{align*}
And for a real axis $\mathbb{L}_{[w],[z]}~ \hat{\in}~ \mathcal{C}(n,\mathbb{M}'_{1,6})$ with $w<z$ we have because of $w+z=n+2$
\begin{align*}
&\mathbb{L}_{[w-2],[z]}~ \hat{\in}~ \mathcal{C}(n,\mathbb{N}^*)\mbox{ because of }w-2+z=n\mbox{ and}\\
&\mathbb{L}_{[w],[z-2]}~ \hat{\in}~ \mathcal{C}(n,\mathbb{N}^*)\mbox{ due to }w+z-2=n\\
&\mbox{and we have a chain of weights of }\mathcal{C}(n,\mathbb{N}^*)\\
&w-2<w<z-2<z.
\end{align*}
If we assume w.l.o.g. $u<w$ then we have a chain of weights of $\mathcal{C}(n,\mathbb{N}^*)$
\[
u<u+2<w-2<w<z-2<z<v<v+2.
\]
Therefore all points of $\mathcal{C}(n-2,\mathbb{M}_{5,6})$ and $\mathcal{C}(n+2,\mathbb{M}'_{1,6})$ belong to $\mathcal{C}(n,\mathbb{N}^*)$ too and there is no point of $\mathcal{C}(n,\mathbb{N}^*)$ which is not a point either of $\mathcal{C}(n-2,\mathbb{M}_{5,6})$ or of $\mathcal{C}(n+2,\mathbb{M}'_{1,6})$.

Since additional the CoPs $\mathcal{C}(n-2,\mathbb{M}_{5,6})$ and $\mathcal{C}(n+2,\mathbb{M}'_{1,6})$ are disjunct because $\mathbb{M}_{5,6}$ and $\mathbb{M}'_{1,6}$ are disjunct holds finally
\[
\mathcal{C}(n,\mathbb{N}^*)=\mathcal{C}(n-2,\mathbb{M}_{5,6})\cup\mathcal{C}(n+2,\mathbb{M}'_{1,6}).
\]
Since $\mathcal{C}(n-2,\mathbb{M}_{5,6})$ and $\mathcal{C}(n+2,\mathbb{M}'_{1,6})$ due to Theorem \ref{properties2} admit aligned embedding with an increment $6$ and they are disjunct the CoP $\mathcal{C}(n,\mathbb{N}^*)$ admits aligned embedding with an increment $6$ too.
\end{proof}

\begin{corollary}\label{C-centerStern}
If $n\equiv \pm 2\mood{6}$ then the CoP $\mathcal{C}(n,\mathbb{N}^*)$ has a center if and only if $\frac{n}{2}\equiv \pm 1\mood{6}$.
In the case $n\equiv 0\mood{6}$ the CoP $\mathcal{C}(n,\mathbb{N}^*)$ has no center because all weights $x$ of it are $\equiv\pm 1\mood{6}$ and therefore 
\[
2x\equiv\pm 2\mood{6}\not\equiv 0\mood{6}.
\]
\end{corollary}

\begin{corollary}\label{C-centerP}
Because of $\mathbb{P}^*\subset\mathbb{N}^*$ the CoP $\mathcal{C}(n,\mathbb{P}^*)$ has a center if and only if $\frac{n}{2}$ is a prime. In the case $n\equiv 0\mood{6}$ there is no center in $\mathcal{C}(n,\mathbb{P}^*)$ with the same justification like in Corollary \ref{C-centerStern}.
\end{corollary}

\section{\textbf{A Fundamental Theorem and its Conclusion}}

\begin{theorem}[Fundamental]\label{central}
Let $n,r \in \mathbb{N}$, $\mathbb{M}\subseteq\mathbb{N}$ and $\mathcal{C}(n,\mathbb{M})$ be a nonempty CoP with an axis $\mathbb{L}_{[x],[n-x]}~ \hat{\in}~ \mathcal{C}(n,\mathbb{M})$
\footnote{
The axis can also be a degenerated axis with $x=n-x=\frac{n}{2}$ if it exists.}
.
If holds $x+r\in\mathbb{M}$ then $\mathcal{C}(n+r,\mathbb{M})$ is a nonempty CoP too. 
\end{theorem}
\begin{proof}
Since $\mathbb{L}_{[x],[n-x]}~ \hat{\in}~ \mathcal{C}(n,\mathbb{M})$, $x$ and $n-x$ are members of $\mathbb{M}$. And due to the premise also $x+r\in\mathbb{M}$. Then holds
\[
n+r-(x+r)=n-x\in\mathbb{M}.
\]
Ergo there is an axis $\mathbb{L}_{[x+r],[n+r-(x+r)]}~ \hat{\in}~ \mathcal{C}(n+r,\mathbb{M})$ and $\mathcal{C}(n+r,\mathbb{M})$ is nonempty.
\end{proof}

\begin{corollary}\label{C-ext_inftyChain}
Let the requirements of Theorem \ref{central} be fulfilled. If the base set $\mathbb{M}$ is an infinite set and there exists a nonempty CoP $\mathcal{C}(n_o,\mathbb{M})$ then there exist infinitely many positive integers $n> n_o$ with nonempty CoPs $\mathcal{C}(n,\mathbb{M})$.
\end{corollary}
\begin{proof}
Let $\mathbb{L}_{[x],[n-x]}$ be an axis of $\mathcal{C}(n_o,\mathbb{M})$. Then is due to Theorem \ref{central} also $\mathcal{C}(n_o+r_1,\mathbb{M})$ with $r_1>0$ nonempty if $x+r_1\in\mathbb{M}$. From this CoP we can continue this process with $r_2>0$ to the nonempty CoP $\mathcal{C}(n_o+r_1+r_2,\mathbb{M})$. Since the base set is an infinite set this process can be repeated infinitely many. 
\end{proof}

\begin{theorem}[Almost Goldberg]\label{almostBGC}
Let the requirements of Theorem \ref{central} be fulfilled for the base set $\mathbb{P}$. Then there exist infinitely many positive integers $n\geq 6$ which have at least one representation as sum of two primes.
\end{theorem}
\begin{proof}
For $n=6$ holds $\Vert\mathcal{C}(6,\mathbb{P})\Vert=\lbrace 3\rbrace$. And because $3+r_1=3+2=5$ is also prime then holds $\mathbb{L}_{[5],[8-5]}$ is an axis of $\mathcal{C}(6+2,\mathbb{P})$. We continue with the last weight. It is $5$ and $7$ is the next prime greater than $5$. Hence we come to $\mathbb{L}_{[7],[10-7]}$ as an axis of $\mathcal{C}(8+2,\mathbb{P})$.
We repeat this again with last point of $\mathcal{C}(10,\mathbb{P})$. This is $7$ and $11$ is the next prime after $7$. Hence we come to $\mathbb{L}_{[11],[14-11]}$ as an axis of $\mathcal{C}(10+4,\mathbb{P})$. 

Because of Corollary \ref{C-ext_inftyChain} this process can be infinitely many often repeated always with the greatest weight. On this way we get infinitely many CoPs $\mathcal{C}(n,\mathbb{P})\mid n\geq 6$ such that the first point of all CoPs is $[3]$ since the first 
\[
n-x=6-3=3.
\]
Other chains of CoPs also can be constructed by another selection strategy.
\end{proof}
T. Estermann \cite{estermann1938goldbach} could prove in 1938 a similar result using analytical methods. But our proof uses only elemantary instruments and is more constructive.


\section{\textbf{Rotation and Dilation of Circles of Partition}}
In this section we introduce the notion of the \textbf{Rotation} and \textbf{Dilation} of CoPs produced by a given generator. We launch the following formal terminology.

\begin{definition}
Let $\mathbb{M}\subseteq \mathbb{N}$ with $n\in \mathbb{N}$ and $\mathcal{C}(n,\mathbb{M})$ be the  CoP generated by $n$. The map 
\begin{align}
\varpi_r:\mathcal{C}(n,\mathbb{M})\longrightarrow \mathcal{C}^r(n,\mathbb{M})\nonumber 
\end{align}
will be the $r^{th}$ level rotation of the CoP $\mathcal{C}(n,\mathbb{M})$ with 
\begin{align*}
\mathcal{C}^r(n,\mathbb{M}):=&\lbrace [k]\in \mathcal{C}(n,\mathbb{M})\mid [x]\in \mathcal{C}(n,\mathbb{M}),~x+r\equiv k\mood{n},~r\in \mathbb{Z},\\
&\mbox{ if }x+r\equiv 0\mood{n}\mbox{ then }k:=(n+r)\opn{Mod}n
\rbrace.
\end{align*}
If the sign is positive then we say the $r^{th}$ level rotation is clockwise.  Otherwise, it is an anti-clockwise $r^{th}$ level rotation for $r\neq 0$. However, if we take $r=0$, then the rotation is trivial and the real axes joining points on the CoP remains stable.
It is important to say that the result of a rotation must not be necessarily a CoP.
Due to the condition $[k]\in \mathcal{C}(n,\mathbb{M})$ it is even possible that the target set is empty. In this case we say that the $r^{th}$ level rotation fails to exist.
\end{definition}

\begin{theorem}\label{T_rotation-C(n)}
The CoP $\mathcal{C}(n)$ remains invariant under the $r^{th}$ level rotation $\varpi_r$. That is
\[
\varpi_r: \mathcal{C}(n)\longrightarrow\mathcal{C}(n).
\]
\end{theorem}
\begin{proof}
The set of weights of the images of $\mathcal{C}(n)$ is
\footnote{We denote by $\lbrace a,b,\ldots,z\rbrace_n$ the set
$\lbrace a\opn{Mod}n,b\opn{Mod}n,\ldots,z\opn{Mod}n\rbrace$.
}
\[
\Vert\mathcal{C}^r(n)\Vert=\lbrace r+1,r+2,\ldots,r+n-1\rbrace_n.
\]
The missing value is $(r+n-k)_n$ if $r+n-k\equiv 0\mood{n}$. Therefore holds
\[
k=(n+r)\opn{Mod}n.
\]
And this is the substituted value by virtue of the definition.
\end{proof}
If the inequality  $-n<r<n$ is valid then we get 
\[
k=
\begin{cases}
r &\mbox{ if } r>0\\
n-\vert r\vert &\mbox{ if } r<0.
\end{cases}
\]
\begin{example}
$n=8, r=+2$\\
$\Vert\mathcal{C}(8)\Vert=\lbrace 1,2,3,4,5,6,7\rbrace$.\\
The critical point is $[6]$ 
because $6+2\equiv 0\pmod{8}$.
The set of the weights of the images of all points except of $[6]$ is
$\lbrace 3,4,5,6,7,-,1\rbrace$. Absent is $2$. \\
As image of $[6]$ we set $[(8+2)\opn{Mod}8]=[2]$
and we get as target set\\
$\Vert\varpi_{3}\left(\mathcal{C}(8)\right)\Vert=\lbrace 3,4,5,6,7,2,1\rbrace\rightarrow$ $\lbrace 1,2,3,4,5,6,7\rbrace=\Vert\mathcal{C}(8)\Vert$.\\

\noindent
$n=8, r=-2$\\
The critical point is $[2]$ 
because $2-2\equiv 0\pmod{8}$.
The set of the weights of the images of all points except of $[2]$ is
$\lbrace 7,-,1,2,3,4,5\rbrace$. Absent is $6$. \\
As image of $[2]$ we set $[(8-2)\opn{Mod}8]=[6]$
and we get as target set\\
$\Vert\varpi_{3}\left(\mathcal{C}(8)\right)\Vert=\lbrace 7,6,1,2,3,4,5\rbrace\rightarrow$ $\lbrace 1,2,3,4,5,6,7\rbrace=\Vert\mathcal{C}(8)\Vert$.\\
\end{example}

\begin{proposition}\label{P_not-rotation-Mad}
Let $\mathcal{C}(n,\mathbb{M}_{a,d})$ be a CoP defined as in (\ref{Mad}). Then there exists not an $r^{th}$ level rotation for $r\equiv c\mood{d}$ with $0<c<d$ and $c\not\equiv 2a\mood{d}$.
\end{proposition}
\begin{proof}
W.l.o.g. we let $c\leq n$. \\
We observe $[n-a-kd]$ is a point of $\mathcal{C}(n,\mathbb{M}_{a,d})$ for $k=0(1)\frac{n-2a}{d}$
\footnote{Because of $n\in\mathbb{M}_{2a,d}$ is it a positive integer.}
. By applying the rotation $\varpi_r$ its weight will be transformed to
\begin{align*} 
(n-a-kd+c)\opn{Mod}n&=(c-a-kd)\opn{Mod}n \mbox{ and because of }c\leq n\\
&=c-a-kd \\
&\equiv (c-a)\mood{d}\mbox{ and because of }c\not\equiv 2a\mood{d}\\
&\not\equiv a\mood{d}.
\end{align*}
Hence all rotated points of $\mathcal{C}(n,\mathbb{M}_{a,d})$ are not points of $\mathcal{C}(n,\mathbb{M}_{a,d})$ and therefore the target set of the rotation is an \textbf{empty set}.
\end{proof}

\begin{proposition}\label{P_ext-rotation-Mad}
Let $\mathcal{C}(n,\mathbb{M}_{a,d})$ be a CoP defined as in (\ref{Mad}). Then $\mathcal{C}(n,\mathbb{M}_{a,d})$ remains invariant under the $r^{th}$ level rotation $\varpi_r$ provided $d=2a$ and $r\equiv 0\mood{d}$.
\end{proposition}
\begin{proof}
First we recall that $n\equiv 2a\mood{d}$. Under the assumption $d=2a$ it certainly follows that $n\equiv 0\mood{d}$. Now, let $(x+r)\opn{Mod}n=c$ be the weight of a rotated point $[x]$. Then it is easy to see that the following congruence condition is valid
\begin{align*}
x+r&\equiv c\mood{n}\mbox{ and because }n\equiv 0\mood{d}\\
&\equiv c\mood{d}.
\end{align*}
On the other hand the congruence conditions $x\equiv a\mood{d}$ and $r\equiv 0\mood{d}$ imply
\begin{align*}
x+r&\equiv a\mood{d}.
\end{align*}
Hence we have $a=c$ and $x+r\equiv a\mood{d}$. Therefore all image points $\mathcal{C}(n,\mathbb{M}_{a,d})$ are members of $\mathbb{M}_{a,d}$ and less than $n$. In principle all image points of the $r^{th}$ level rotation of the CoP $\mathcal{C}(n,\mathbb{M}_{a,d})$ are again points of the CoP $\mathcal{C}(n,\mathbb{M}_{a,d})$. This proves the claim that CoPs of the form $\mathcal{C}(n,\mathbb{M}_{a,d})$ remains invariant under some $r^{th}$ level rotation with special conditions.
\end{proof}

\begin{example}
$n=24$, $a=2$, $d=4$, $r=4$\\
$\Vert\mathcal{C}(24,\mathbb{M}_{2,4})\Vert=\lbrace 2,6,10,14,18,22\rbrace$.
Then is\\
$\Vert\varpi_4\left(\mathcal{C}(24,\mathbb{M}_{2,4})\right)\Vert=\lbrace 6,10,14,18,22,2\rbrace$
$\rightarrow\lbrace 2,6,10,14,18,22\rbrace$.
\end{example}

\begin{corollary}
For  conditions espoused in Proposition \ref{P_not-rotation-Mad} and of Proposition \ref{P_ext-rotation-Mad} the $r^{th}$ level rotation of a CoP $\mathcal{C}(n,\mathbb{M}_{a,d})$ results in a set which is a real subset of $\mathcal{C}(n,\mathbb{M}_{a,d})$. 
\end{corollary}

\begin{definition}
Let $\mathbb{M}\subseteq \mathbb{N}$ with $n\in \mathbb{N}$ and  $\mathcal{C}(n,\mathbb{M})$ be the CoP generated by $n$. The map 
\begin{align}
\delta_r:\mathcal{C}(n,\mathbb{M})\longrightarrow \mathcal{C}_r(n,\mathbb{M})\nonumber
\end{align}
will be the $r^{th}$ scale dilation of the CoP $\mathcal{C}(n,\mathbb{M})$ with 
\begin{align}
\mathcal{C}_r(n,\mathbb{M}):=\left \{[x]\in \mathcal{C}(n+r,\mathbb{M})\mid~r\in \mathbb{Z},~n+r>1\right \}.\nonumber
\end{align}
If the sign is positive then we say the $r^{th}$ scale dilation is an expansion. Otherwise, it is an $r^{th}$ scale compression for $r\neq 0$. However if we take $r=0$, then the dilation is a trivial dilation and the CoP remains invariant under the dilation.
\end{definition}
\bigskip

\begin{remark}\label{remC(n)}
It is important to note that if the base set is taken to be the set of natural numbers $\mathbb{N}$, then the image set of dilation collapses to the following 
\begin{align}\label{C(n)}
\delta_r\big(\mathcal{C}(n)\big)&:=\mathcal{C}_r(n)\nonumber\\
&=\left \{[x]\mid x\in\mathbb{N}_{n+r-1},~r\in \mathbb{Z},~n+r>1\right \}\nonumber \\
&=\mathcal{C}(n+r).
\end{align}
Additionally, it is important to point out that in case $r<0$ some points of $\mathcal{C}(n)$ have the same image where as in the case $r>0$ some points of $\mathcal{C}(n)$ have more than one image.
\end{remark}
\bigskip


As it happens, dilation at any scale between CoPs have the natural tendency of translating the generator of the source CoP by the size of the scale of the dilation. However it is somewhat difficult to define dilation on individual points in a given CoP. Any perceived dilation map could manifestly work on a typical CoP but it may proved handicapped for some other CoPs. In the sense that some points may poke outside the target CoP under this fixed dilation. In light of this anomaly, we ask the following questions

\begin{question}\label{question1}
Let $\mathbb{M}\subseteq \mathbb{N}$. Does there exists a  well-defined dilation 
\begin{align}\delta_r:\mathcal{C}(n,\mathbb{M})\longrightarrow \mathcal{C}(m,\mathbb{M})\nonumber
\end{align} 
on each $[x]\in \mathcal{C}(n,\mathbb{M})$ for all CoPs?
\end{question}
\bigskip

Put it differently, Question \ref{question1} asks if there exists a fixed map that assigns each points in a typical CoP to its target CoP in a sufficiently uniform way. That is to say, the map we seek should avoid the subtleties as espoused in our earlier discussion.

\begin{theorem}\label{existence}
Let $n,m\in \mathbb{N}$, $\mathbb{M}\subseteq\mathbb{N}$ and $\mathcal{C}(n,\mathbb{M})$ be a CoP admitting aligned embedding. Then there exists some dilation $\delta_r$ such that 
\begin{align*}
\delta_r:\mathcal{C}(n,\mathbb{M})\longrightarrow \mathcal{C}(m,\mathbb{M}).
\end{align*} 
\end{theorem}

\begin{proof}
It is evident that for $m=n$ the trivial dilation $\delta_0$ meets the claim. For the case $m\neq n$ we break the proof into several cases. The case $r$ is positive and the case it is negative. Let $\delta_r$ be any dilation for $r>0$ and suppose for any two CoP $\mathcal{C}(n,\mathbb{M})$ and $\mathcal{C}(m,\mathbb{M})$ with $\mathcal{C}(m,\mathbb{M})\subset \mathcal{C}(n,\mathbb{M})$ there exists no dilation associating them. 
By virtue of the property that the CoPs admitting embedding
exactly one of the following embedding holds
\begin{align*}
\delta_r\big(\mathcal{C}(m,\mathbb{M})\big)\subset \mathcal{C}(n,\mathbb{M})  \mbox{ or } \mathcal{C}(n,\mathbb{M})\subset \delta_r\big(\mathcal{C}(m,\mathbb{M})\big).
\end{align*}
We analyze each of these sub-cases. First let us assume that $\delta_r\big(\mathcal{C}(m,\mathbb{M})\big)\subset \mathcal{C}(n,\mathbb{M})$. It follows that there exists some CoP $\mathcal{C}(s,\mathbb{M})$ with  $\delta_r(\mathcal{C}(m,\mathbb{M}))\subseteq \mathcal{C}(s,\mathbb{M})$ such that $\mathcal{C}(s,\mathbb{M})\subset \mathcal{C}(n,\mathbb{M})$. Since there exists no dilation between CoPs the following proper embedding must necessarily hold 
\begin{align*}
\delta_r\big(\mathcal{C}(m,\mathbb{M})\big)\subset \mathcal{C}(s,\mathbb{M})\subset \mathcal{C}(n,\mathbb{M}).
\end{align*}
Again there exists some CoP $\mathcal{C}(t,\mathbb{M})$ with $\delta_r(\mathcal{C}(m,\mathbb{M}))\subseteq \mathcal{C}(t,\mathbb{M})$ such that $\mathcal{C}(t,\mathbb{M})\subset \mathcal{C}(s,\mathbb{M})$. Then under the underlying assumption that there exists no dilation between CoPs, we obtain the following proper embedding 
\begin{align*}
\delta_r(\mathcal{C}(m,\mathbb{M}))\subset \mathcal{C}(t,\mathbb{M})\subset \mathcal{C}(s,\mathbb{M})\subset \mathcal{C}(n,\mathbb{M}).
\end{align*}
By repeating the argument in this manner, we obtain the following infinite descending chains of covers of the smallest CoP
\begin{align*}
\mathcal{C}(m+r,\mathbb{M}):=\delta_r\big(\mathcal{C}(m,\mathbb{M})\big)\subset \cdots \subset \mathcal{C}(t,\mathbb{M}) \subset \mathcal{C}(s,\mathbb{M})\subset \mathcal{C}(n,\mathbb{M}).
\end{align*}
Because the CoPs admit aligned embedding
we obtain the infinite descending sequence of positive integers towards the generator $m+r$ of the last CoP  
\begin{align*}
n>s>t>\cdots>\cdots>m+r.
\end{align*}
This is absurd, thereby ending the proof of the first sub-case. We now turn to the case $\mathcal{C}(n,\mathbb{M})\subset \delta_r(\mathcal{C}(m,\mathbb{M}))$. Then in a similar fashion there must exist some CoP $\mathcal{C}(t,\mathbb{M})$ with $\mathcal{C}(t,\mathbb{M})\subseteq \delta_r(\mathcal{C}(m,\mathbb{M}))$ such that $\mathcal{C}(n,\mathbb{M})\subset \mathcal{C}(t,\mathbb{M})$. Then under the assumption that there exists no dilation between CoP, we have the following embedding
\begin{align*}
\mathcal{C}(n,\mathbb{M})\subset \mathcal{C}(t,\mathbb{M})\subset \delta_r(\mathcal{C}(m,\mathbb{M})).
\end{align*}
Again there exists some CoP $\mathcal{C}(s,\mathbb{M})$ with $\mathcal{C}(s,\mathbb{M})\subseteq \delta_r(\mathcal{C}(m,\mathbb{M}))$ such that $\mathcal{C}(t,\mathbb{M})\subset \mathcal{C}(s,\mathbb{M})$. Under the assumption that there exists no dilation between CoP, we have the following embedding 
\begin{align*}
\mathcal{C}(n,\mathbb{M})\subset \mathcal{C}(t,\mathbb{M})\subset \mathcal{C}(s,\mathbb{M})\subset \delta_r(\mathcal{C}(m,\mathbb{M})).
\end{align*}
By repeating this argument indefinitely we obtain the following infinite sequence of embedding 
\begin{align*}
\mathcal{C}(n,\mathbb{M})\subset \mathcal{C}(t,\mathbb{M})\subset \mathcal{C}(s,\mathbb{M})\cdots \subset \delta_r(\mathcal{C}(m,\mathbb{M})):=\mathcal{C}(m+r,\mathbb{M}).
\end{align*}
By virtue of the CoPs admitting aligned embedding, we obtain an infinite ascending sequence of positive integers towards the generator of the last CoP in the chain 
\begin{align*}
n<t<s<\cdots<m+r.
\end{align*}
This is absurdity, since we cannot have positive integers approaching a fixed positive integer for infinite amount of time. This completes the proof for the case $r>0$. We now turn to the case $r<0$ for any two CoP $\mathcal{C}(m,\mathbb{M}), \mathcal{C}(n,\mathbb{M})$ with $\mathcal{C}(n,\mathbb{M})\subset \mathcal{C}(m,\mathbb{M})$. Under the main assumption exactly one of the following embedding must hold 
\begin{align*}
\delta_r\big(\mathcal{C}(m,\mathbb{M})\big)\subset \mathcal{C}(n,\mathbb{M}) \mbox{ or } \mathcal{C}(n,\mathbb{M})\subset \delta_r\big(\mathcal{C}(m,\mathbb{M})\big).
\end{align*}
A similar analysis could be carried out for each of the above cases.
\end{proof}

\begin{corollary}
Because of Theorem \ref{properties} the CoP $\mathcal{C}(n)$ admits aligned embedding and there is
the dilation $\delta_1:\mathcal{C}(n)\longrightarrow \mathcal{C}(n+1)$ with 
\begin{align}\label{E_Cn-Cn+1}
\delta_1([x]):=
\begin{cases}
[x] \quad &\mbox{for }~1\leq x\leq n-1 \\
[n] \quad &\mbox{additional for}~x=1
\end{cases}
\end{align} 
that can produce an infinite ascending chain of CoPs 
\begin{align}
\mathcal{C}(n)\subset \mathcal{C}(n+1)\subset \mathcal{C}(n+2)\subset \cdots. \nonumber
\end{align}
\end{corollary}

It is easy to see that the assignment of $[n]$ as also an image of $[1]$ is not the only possibility. Also possible would be $[n]$ as the image of $[2]\ldots[n-1]$. In all cases we would have a correct point-to-point mapping.
Hence a subset of the cross set $\mathcal{C}(n)\times\mathcal{C}(n+1)$ for which holds:
\begin{itemize}
\item
for each point of $\mathcal{C}(n)$ there is at least one image point of $\mathcal{C}(n+1)$ and
\item
for each image point of $\mathcal{C}(n+1)$ there is only one preimage point of $\mathcal{C}(n)$
\end{itemize}
is not a well-defined pointwise definition of the map $\mathcal{C}(n)\longrightarrow\mathcal{C}(n+1)$ because there are several such subsets.

\begin{corollary}
In light of Theorem \ref{properties2} the CoP $\mathcal{C}(n,\mathbb{M}_{a,d})$ admits aligned embedding and there is
the dilation $\delta_d:\mathcal{C}(n,\mathbb{M}_{a,d})\longrightarrow \mathcal{C}(n+d,\mathbb{M}_{a,d})$ with 
\[
\delta_d([x]):=\begin{cases}
[x] \quad &\mbox{for } a\leq x\leq n-a \\
[n-a+d] \quad &\mbox{additional for}~x=a
\end{cases}
\] 
that can generate an infinite ascending chain of CoPs 
\begin{align}
\mathcal{C}(n,\mathbb{M}_{a,d})\subset \mathcal{C}(n+d,\mathbb{M}_{a,d})\subset \mathcal{C}(n+2d,\mathbb{M}_{a,d})\subset \cdots. \nonumber
\end{align}
\end{corollary}

\section{\textbf{Special Maps of Circles of Partition}}
In this section we introduce and study the notion of several special maps of circles of partition. We launch more formally the following languages.

\begin{definition}\label{flip}
Let $\mathbb{M}\subseteq \mathbb{N}$ and $\mathcal{C}(n,\mathbb{M})\neq \emptyset$ be a CoP containing the axis $\mathbb{L}_{[a],[b]}$. By the \textbf{flipping} of the CoP $\mathcal{C}(n,\mathbb{M})$ along the so called flipping axis $\mathbb{L}_{[a],[b]}$, we mean the map 
\begin{align*}
\vartheta_{[a],[b]}:\mathcal{C}(n,\mathbb{M})\longrightarrow \mathcal{C}(m,\mathbb{M})
\end{align*}
with $\vartheta_{[a],[b]}([a])=[a]$ and $\vartheta_{[a],[b]}([b])=[b]$ such that for any two $[x],[y]\in \mathcal{C}(n,\mathbb{M})$ with $[x],[y]\neq [a],[b]$ holds
\begin{align*}
\Vert\vartheta_{[a],[b]}([x])\Vert+\Vert\vartheta_{[a],[b]}([y])\Vert \neq n 
\end{align*}
A flipping axis can also be a degenerated axis $\mathbb{L}_{[a]}$.
We say the CoP $\mathcal{C}(n,\mathbb{M})$ is susceptible to flipping if there exists such a map.
\end{definition}
\bigskip

\begin{example}
Let be $\mathbb{M}=\mathbb{P}$ and $n=20$. The CoP $\mathcal{C}(20,\mathbb{P})$ is the set $\lbrace [3],[7],[13],[17]$ with two axes $\mathbb{L}_{[3],[17]}$ and $\mathbb{L}_{[7],[13]}$. Then the map
\[
\vartheta_{[3],[17]}: \mathcal{C}(20,\mathbb{P})\longrightarrow
\mathcal{C}(22,\mathbb{P})
\]
with $\mathcal{C}(22,\mathbb{P})=\lbrace [3],[5],[11],[17],[19]\rbrace$
is a flipping of $\mathcal{C}(20,\mathbb{P})$ along the axis $\mathbb{L}_{[3],[17]}$ if f.i.
\begin{align*}
\vartheta_{[3],[17]}([3])&=[3]\\
\vartheta_{[3],[17]}([7])&=[5]\\
\vartheta_{[3],[17]}([13])&=[11]\mbox{ and }[19]\\
\vartheta_{[3],[17]}([17])&=[17].
\end{align*}
Hence we get $\Vert[5]\Vert+\Vert[11]\Vert=16\neq 20$ or  $\Vert[5]\Vert+\Vert[19]\Vert=24\neq 20$.

Vice versa there are
no axis points of $\mathcal{C}(22,\mathbb{P})$ that are also points of $\mathcal{C}(20,\mathbb{P})$. Hence there exists no flipping from $\mathcal{C}(22,\mathbb{P})$ to $\mathcal{C}(20,\mathbb{P})$ along an axis of $\mathcal{C}(22,\mathbb{P})$.
\end{example}

\begin{proposition}\label{flip_Mad}
Let $\mathbb{M}_{a,d}$ be defined as in (\ref{Mad}) with $0<a\leq d$. Then the CoP $\mathcal{C}(n,\mathbb{M}_{a,d})$ is susceptible to flipping if and only if $n>m$. 
\end{proposition}
\begin{proof}
We must regard that in order to get $\mathcal{C}(n,\mathbb{M}_{a,d})\neq\emptyset$ it must be $n\in \mathbb{M}_{2a,d}$. Then is $n-a\in \mathbb{M}_{a,d}$. The same is valid for $\mathcal{C}(m,\mathbb{M}_{a,d})$.\newline
We assume that $n>m$. Then holds with Corollary \ref{col-prop2}
\[
\mathcal{C}(n,\mathbb{M}_{a,d}) \supset \mathcal{C}(m,\mathbb{M}_{a,d}).
\]
Due to $n\in\mathbb{M}_{2a,d}$ holds $\frac{n-2a}{d}\in\mathbb{N}$. The weights of $\mathcal{C}(n,\mathbb{M}_{a,d})$ are
\begin{align*}
\Vert\mathcal{C}(n,\mathbb{M}_{a,d})\Vert
=\left\lbrace a+k\cdot d\mid k=0,1,2,\ldots,\frac{n-2a}{d}\right\rbrace.
\end{align*}
Hence $\mathcal{C}(n,\mathbb{M}_{a,d})$ has 
\[
l_n=\frac{n-2a}{d}+1 \mbox{ members.}
\] 
This is in accordance with the general counting function for CoPs:
\begin{align*}
\vert\mathcal{C}(n,\mathbb{M}_{a,d})\vert
&=1+\sum_{\indbox{1\leq x \leq n-a}{x\equiv a\mood{d}}}1\\
&=1+\frac{n-2a}{d}.
\end{align*}
The addition of $1$ is required because the counting starts with $0$.
Now we must distinguish two cases
\begin{itemize}
\item[rC:]
The CoP $\mathcal{C}(n,\mathbb{M}_{a,d})$ has a real center.
\item[dC:]
The CoP $\mathcal{C}(n,\mathbb{M}_{a,d})$ has a deleted center.
\end{itemize} 
In the case rC holds $l_n$ is odd and $l_n$ is even in the other case.
Now we choose in the case rC the degenerated axis $\mathbb{L}_{[u]}$ of the CoP $\mathcal{C}(n,\mathbb{M}_{a,d})$ as the flipping axis, 
in the case dC those which is the closest one to the center of the CoP. The weights of $[u],[v]$ are $u=v=\frac{n}{2}$ for the case rC and $u=\frac{n-d}{2}$, $v=\frac{n+d}{2}$ in the other case.
In order to satisfy the conditions of Definition \ref{flip}
\begin{align*}
\vartheta_{[u],[v]}([u])=[u]\mbox{ and }
\vartheta_{[u],[v]}([v])=[v]
\end{align*}
the last point of $\mathcal{C}(m,\mathbb{M}_{a,d})$ should be $[v]$. 
Due to Corollary \ref{partner} we get for $m$ as the sum of the weights of the first and the last member of CoP $\mathcal{C}(m,\mathbb{M}_{a,d})$ 
\begin{align}\label{E-m_rC}
m=
\begin{cases}
a+\dfrac{n}{2}&\mbox{ for rC}\\
a+\dfrac{n+d}{2}&\mbox{ for dC}.
\end{cases}
\end{align}
Analogously to $\mathcal{C}(n,\mathbb{M}_{a,d})$ holds for the number of members of $\mathcal{C}(m,\mathbb{M}_{a,d})$
\begin{align*}
l_m-1 &:= \sum_{\indbox{1\leq x \leq m-a}{x\equiv a\mood{d}}}1=\frac{m-2a}{d}\\
&=
\begin{cases}
\dfrac{a+\frac{n}{2}-2a}{d}=\dfrac{n-2a}{2d}=\dfrac{l_n-1}{2}&\mbox{ for rC}\\
\dfrac{a+\frac{n+d}{2}-2a}{d}=\dfrac{n-2a}{2d}+\dfrac{1}{2}=\dfrac{l_n}{2}&\mbox{ for dC}.
\end{cases}
\end{align*}
Hence we obtain for both cases
\begin{align*}
\l_m = \left\lfloor\frac{l_n}{2}\right\rfloor+1.
\end{align*}
All these fulfills the \textbf{following map}
\begin{align*}
\vartheta_{[u],[v]}(x)&=a+k(x)\cdot d \mbox{ with}\\
\frac{x-a}{d}&\equiv k(x)\mood{l_m}.
\end{align*}
This map assigns each point of $\mathcal{C}(n,\mathbb{M}_{a,d})$ to a point of $\mathcal{C}(m,\mathbb{M}_{a,d})$.
\bigskip

The heaviest point of CoP $\mathcal{C}(m,\mathbb{M}_{a,d})$ is $[m-a]$. In the case rC the flipping axis is $\mathbb{L}_{[u]}$ with $u=\frac{n}{2}$ and we get with (\ref{E-m_rC})
\begin{align*}
\left\Vert\vartheta_{[v],[v]}\left(\left[\frac{n}{2}\right]\right)\right\Vert=m-a&=\frac{n}{2}.
\end{align*}
Hence the requirement $\Vert\vartheta_{[v],[v]}([v])\Vert=u=v=\frac{n}{2}$ is fulfilled.
In the case dC we get with (\ref{E-m_rC})
\begin{align*}
\left\Vert\vartheta_{[u],[v]}\left(\left[\frac{n+d}{2}\right]\right)\right\Vert=m-a&=\frac{n+d}{2}=v.
\end{align*}
Therefore holds $u=v-d=\frac{n-d}{2}$.
And for each two points $[x],[y]\in\mathcal{C}(n,\mathbb{M}_{a,d})$ with $[x],[y]\neq [u],[v]$ holds
\[
\Vert\vartheta_{[u],[v]}([x])\Vert + \Vert\vartheta_{[u],[v]}([y])\Vert < n
\]
because $\vartheta_{[u],[v]}([u])=[u]$ and $\vartheta_{[u],[v]}([v])=[v]$ are the two heaviest points of $\mathcal{C}(m,\mathbb{M}_{a,d})$ in case dC respectively is the heaviest point of $\mathcal{C}(m,\mathbb{M}_{a,d})$ in rC with the sum of weights of the two heaviest points $\leq n$. The weight sum of all others is lesser.
Thereby the first part of the claim is proven.

If on the other hand holds $n\leq m$ then the source CoP is a subset of the target CoP. All axes points of $\mathcal{C}(n,\mathbb{M}_{a,d})$ are identically mapped into $\mathcal{C}(m,\mathbb{M}_{a,d})$. And for all these 
$\vartheta_{[u],[v]}([x])$ and $\vartheta_{[u],[v]}([y])$ from any axis $\mathbb{L}_{[x],[y]}$ of $\mathcal{C}(n,\mathbb{M}_{a,d}$
holds
\[
\Vert\vartheta_{[u],[v]}([x])\Vert + \Vert\vartheta_{[u],[v]}([y])\Vert = n.
\]
This is a contradiction to the requirements of the claim.
\end{proof}

\begin{remark}
Note that due to $\mathbb{M}_{1,1}=\mathbb{N}$ this statement also holds for each CoP $\mathcal{C}(n)$.
\end{remark}

\begin{proposition}
The chosen axis closest to the center of the CoP $\mathcal{C}(n,\mathbb{M}_{a,d})$ resp. the degenerated axis in case of existing center
is the only one for flipping along an axis in the case of $\mathbb{M}=\mathbb{M}_{a,d}$.
\end{proposition}
\begin{proof}
For all axes $\mathbb{L}_{[x],[y]}~ \hat{\in}~ \mathcal{C}(n,\mathbb{M}_{a,d})$ holds
\footnote{W.l.o.g. we assume $x\leq y$ for all axes $\mathbb{L}_{[x],[y]}$. 
In case of existing center is $\mathbb{L}_{[x]}=\mathbb{L}_{[x],[x]}$.}
\[
x \leq \frac{n}{2} \leq y.
\]
Therefore there is no axis $\mathbb{L}_{[x],[y]}$ with $y<\frac{n}{2}$.
For the chosen axis $\mathbb{L}_{[u],[v]}$ closest to the center of $\mathcal{C}(n,\mathbb{M}_{a,d})$ holds
\[
\frac{n-d}{2}\leq\Vert[u]\Vert\leq\Vert[v]\Vert\leq\frac{n+d}{2}.
\]
The only opposite of this are axes $\mathbb{L}_{[w],[z]}$ with $\Vert[w]\Vert<\frac{n-d}{2}$
and its axis partner with $\Vert[z]\Vert>\frac{n+d}{2}$. 
Then between $[w]$ and $[z]$ there is at least one axis $\mathbb{L}_{[x],[y]}$ with $w<x\leq y<z$ and $x+y=n$. This is a contradiction to the requirements of flipping along the axes $\mathbb{L}_{[w],[z]}$. 
Hence only the axis $\mathbb{L}_{[u],[v]}$ resp. $\mathbb{L}_{[u]}$ with
\begin{itemize}
\item[rC:]
$\Vert[u]\Vert=\dfrac{n}{2}$
\item[dC:]
$\Vert[u]\Vert=\dfrac{n-d}{2},\Vert[v]\Vert=\dfrac{n+d}{2}$
\end{itemize} 
satisfies the requirements of a flipping axis.  
\end{proof}

It is quite suggestive from this proposition the notion of flipping of CoPs under $\mathbb{M}=\mathbb{M}_{a,d}$ can be thought of  as the process of slicing a circle into two equal half and overturning one half to lie perfectly on top of the other half, thereby forming a geometric structure akin to the semi-circle.\newline

\begin{example}
We choose $a=2, d=4$ and hence $\mathbb{M}=\mathbb{M}_{2,4}$. Then with $n=28$ is
\begin{align*}
\Vert\mathcal{C}(28,\mathbb{M}_{2,4})\Vert
&=\lbrace 2,6,10,14,18,22,26\rbrace,\\
l_n&=\frac{28-2\cdot 2}{4}+1=7,\\
l_m&=\left\lfloor\frac{7}{2}\right\rfloor+1=4\mbox{ and}\\
m&=2+\frac{28}{2}=16
\end{align*}
with the flipping axis $\mathbb{L}_{[14]}$.
Hence is
\begin{align*}
\Vert\vartheta_{[14],[14]}\left(\mathcal{C}(28,\mathbb{M}_{2,4})\right)\Vert
&=\Vert\mathcal{C}(16,\mathbb{M}_{2,4})\Vert\\
&=\lbrace 2,6,10,14\rbrace.
\end{align*}
All weight sums of any two members of $\lbrace [2],[6],[10],[14]\rbrace\setminus\lbrace 14\rbrace$ are less than $28$. If we would take $\mathbb{L}_{[6],[22]}$ as flipping axis we would obtain as target set
\[
\mathcal{C}(24,\mathbb{M}_{2,4})=\lbrace [2],[6],[10],[14],[18],[22]\rbrace.
\]
And here would be possible out of $\lbrace[6],[22]\rbrace$ one weight sum contradicting to the requirements:
\[
10+18=28.
\]
\end{example}
\bigskip

Now we introduce and study the concept of filtration of the CoPs. At first we deal with the filtration along an axis.

\begin{definition}\label{filtration}
Let $\mathbb{M}\subseteq \mathbb{N}$ with the corresponding CoP $\mathcal{C}(n,\mathbb{M})$ containing the axis $\mathbb{L}_{[x],[y]}$. By the \textbf{filtration} of the CoP $\mathcal{C}(n,\mathbb{M})$ along the filtration axis $\mathbb{L}_{[x],[y]}$ we mean the  map
\begin{align*}
\Phi_{[x],[y]}:\mathcal{C}(n,\mathbb{M})\longrightarrow \mathcal{C}(m,\mathbb{M})
\end{align*}
such that  $[x],[y]\not\in\mathcal{C}(m,\mathbb{M})$ for some $m\in \mathbb{N}\setminus \{1\}$ and there exists the so called co-axis $\mathbb{L}_{[u],[v]}$ of $\mathcal{C}(n,\mathbb{M})$ so that  $\mathbb{L}_{[u],[a]}$ and $\mathbb{L}_{[v],[b]}$ are axes of $\mathcal{C}(m,\mathbb{M})$ for some $[a],[b]\in \mathbb{M}$. 
We say the CoP $\mathcal{C}(n,\mathbb{M})$ is susceptible to filtration if there exists such a map.
The filtration axis can also be a degenereated axis.
\end{definition}
Also here an example will demonstrate this special map.

\begin{example}
Let be again $\mathbb{M}=\mathbb{P}$ and $n=20$. Then the map 
\[
\Phi_{[7],[13]}: \mathcal{C}(20,\mathbb{P})\longrightarrow\mathcal{C}(22,\mathbb{P})
\]
is a filtration of $\mathcal{C}(20,\mathbb{P})$ along the filtration axis $\mathbb{L}_{[7],[13]}$ due to the target CoP 
\[
\mathcal{C}(22,\mathbb{P})=\lbrace [3],[5],[11],[17],[19]\rbrace
\]
contains the axes $\mathbb{L}_{[3],[19]}$ and $\mathbb{L}_{[17],[5]}$
where $\mathbb{L}_{[3],[17]}$ is the co-axis of $\mathcal{C}(20,\mathbb{P})$.  
\end{example}

\begin{example}
Again we take $\mathbb{M}=\mathbb{P}$ but $n=46$. It is
\begin{align*}
\Vert\mathcal{C}(46,\mathbb{P})\Vert&=\lbrace 3,5,17,23,29,41,43\rbrace
\mbox{ and}\\
\Vert\mathcal{C}(50,\mathbb{P})\Vert&=\lbrace 3,7,13,19,31,37,43,47\rbrace.
\end{align*}
Then the map
\[
\Phi_{[23]}:\mathcal{C}(46,\mathbb{P})\longrightarrow\mathcal{C}(50,\mathbb{P})
\]
is a filtration of $\mathcal{C}(46,\mathbb{P})$ along the degenerated axis $\mathbb{L}_{[23]}$ due to the target CoP contains $\mathbb{L}_{[3],[47]}$ and $\mathbb{L}_{[7],[43]}$ where $\mathbb{L}_{[3],[43]}$ is the co-axis of $\mathcal{C}(46,\mathbb{P})$.
\end{example}

\begin{proposition}\label{nex_filt_Mad}
The CoP $\mathcal{C}(n,\mathbb{M})$ admits aligned embedding is \textbf{not} susceptible to filtration along an axis.
\end{proposition}
\begin{proof}
The claim is true if one of the following statements holds
\begin{enumerate}
\item[(A)]
The CoP $\mathcal{C}(n,\mathbb{M})$ has no filtration axis.
\item[(B)]
The CoP $\mathcal{C}(n,\mathbb{M})$ has no co-axis
\end{enumerate}
We suppose at first $n\leq m$. Then holds with Theorem \ref{properties2}
\begin{align*}
\mathcal{C}(n,\mathbb{M})\subseteq\mathcal{C}(m,\mathbb{M}).
\end{align*}
Then the images of all axis points of the source CoP are points of the target CoP. Hence there is no filtration axis (A).\newline
Now we look for $m<n$. In this case holds with Corollary \ref{col-prop2}
\begin{align*}
\mathcal{C}(n,\mathbb{M})\supset\mathcal{C}(m,\mathbb{M}).
\end{align*}
At first let be $m<\frac{n}{2}$. In this case the images of the end points of all axes of $\mathcal{C}(n,\mathbb{M})$ do not exist in $\mathcal{C}(m,\mathbb{M})$. Hence there is no co-axis (B).\newline
At last we look for $\frac{n}{2}\leq m <n$. In this case the images of the begin points of all axes of $\mathcal{C}(n,\mathbb{M})$ are points of $\mathcal{C}(m,\mathbb{M})$. Hence there is no filtration axis (A). 
\end{proof}


\begin{definition}\label{reduct}
Let $\mathbb{M}\subseteq \mathbb{N}$ with the corresponding CoP $\mathcal{C}(n,\mathbb{M})$ containing the axis $\mathbb{L}_{[x],[y]}$. By the \textbf{reduction} of the CoP $\mathcal{C}(n,\mathbb{M})$ in the base set $\mathbb{M}$ we mean the  map
\begin{align*}
\phi_{[x],[y]}:\mathcal{C}(n,\mathbb{M})\longrightarrow \mathcal{C}(n,\mathbb{M^*})
\end{align*}
with $\mathbb{M}^* := \mathbb{M}\setminus\lbrace x,y\rbrace$.
We say the CoP $\mathcal{C}(n,\mathbb{M})$ is susceptible to reduction if there exists such a map.
\end{definition}

\begin{proposition}
Let $\mathbb{M}_{a,d}$ be defined as in (\ref{Mad}). Then the CoP $\mathcal{C}(n,\mathbb{M}_{a,d})$ is susceptible to reduction.
\end{proposition}
\begin{proof}
W.l.o.g. we suppose $x<y$
and take
\begin{align*}
\phi_{[x],[y]}([u])=\left\lbrace
\begin{array}{ll}
~[u] &\mbox{ if }u\neq x \mbox{ and }u\neq y \\
~[u+d] &\mbox{ if }u=x\\
~[u-d] &\mbox{ if }u=y
\end{array}
\right.
\end{align*}
for all points $[u]\in\mathcal{C}(n,\mathbb{M}_{a,d})$.
Due to all members of $\mathbb{M}_{a,d}$ have the same distance $d$ it holds that if $u\in\mathbb{M}_{a,d}$ then is also $u\pm d\in\mathbb{M}_{a,d}$ and 
\begin{align*}
\Vert\phi_{[x],[y]}([x])\Vert+\Vert\phi_{[x],[y]}([y])\Vert &= x+d+y-d=n
\end{align*}
because $\mathbb{L}_{[x],[y]}$ is an axis of $\mathcal{C}(n,\mathbb{M}_{a,d})$.
\end{proof}
Due to $\mathbb{M}_{1,1}=\mathbb{N}$ this proposition holds for $\mathcal{C}(n)$ too.

\section{\textbf{Stable and Unstable Points on the Circle of Partition}}
In this section we launch the notion of stability of a sequence under a given dilation.

\begin{definition}
Let $\Theta(n)$ be a subsequence of $\mathbb{N}_n$ and suppose the CoP $\mathcal{C}(n,\mathbb{M})\neq \emptyset$.  Let $\mathbb{L}_{[x],[y]}$ be a real axis of the CoP $\mathcal{C}(n,\mathbb{M})$ with $x,y\in \Theta(n)$. Then we say the point $[x]\in \mathcal{C}(n,\mathbb{M})$ is \textbf{stable} relative to the subsequence $\Theta(n)$ under the $r^{th}$ level rotation $\varpi_r:\mathcal{C}(n,\mathbb{M})\longrightarrow \mathcal{C}(n,\mathbb{M})$ if $||\varpi_r([x])||\in \Theta(n)$ and $\exists z\in \Theta(n)$ such that $\mathbb{L}_{[\varpi_r([x])],[z]}$ is also a real axis of the CoP $\mathcal{C}(n,\mathbb{M})$. We say the subsequence $\Theta(n)$ is \textbf{stable} under the $r^{th}$ level rotation $\varpi_r$ if all points in $[x]\in \mathcal{C}(n,\mathbb{M})$ with $x\in \Theta(n)$ are stable.
\end{definition}



\begin{definition}
Let $\Theta(n)$ be a subsequence of $\mathbb{N}_n$ and suppose the CoP $\mathcal{C}(n,\mathbb{M})\neq \emptyset$.  Let $\mathbb{L}_{[x],[y]}$ be a real axis of the CoP $\mathcal{C}(n,\mathbb{M})$ with $x,y\in \Theta(n)$. Then we say the point $[x]\in \mathcal{C}(n,\mathbb{M})$ is \textbf{stable} relative to the subsequence $\Theta(n)$ under the $r^{th}$ scale dilation $\delta_r:\mathcal{C}(n,\mathbb{M})\longrightarrow \mathcal{C}(s,\mathbb{M})$ if $||\delta_r([x])||\in \Theta(n)$ and $\exists z\in \Theta(n)$ such that $\mathbb{L}_{[\delta_r([x])],[z]}$ is also a real axis of the CoP $\mathcal{C}(s,\mathbb{M})$. We say the subsequence $\Theta(n)$ is \textbf{stable} under the $r^{th}$ scale dilation $\delta_r$ if all points in $[x]\in \mathcal{C}(n,\mathbb{M})$ with $x\in \Theta(n)$ are stable.
\end{definition}

Next we establish an important result in the special case where the base set is the set $\mathbb{N}$ of natural numbers.

\begin{proposition}
Let $\Theta(n)=\mathbb{N}_{n-1}$ and let $\delta_r:\mathcal{C}(n)\longrightarrow \mathcal{C}(m)$ be a dilation. Then the subsequence $\Theta(n)$ is stable if and only if $n\geq m$.
\end{proposition}

\begin{proof}
In the case $m=n$ then the dilation is trivial and the claim is trivially true. Suppose the sequence $\Theta(n)$ is stable under the dilation \begin{align}\delta_r:\mathcal{C}(n)\longrightarrow \mathcal{C}(m)\nonumber
\end{align}and assume to the contrary that $n<m$. Then the dilation is an expansion. It follows that for all $[x]\in \mathcal{C}(n)$  with $x\in \Theta(n)$ there exists $z\in \Theta(n)$ such that $z+||\delta_r([x])|| =m$. Under the assumption $n<m$ and by virtue of Theorem \ref{properties} we have the embedding $\mathcal{C}(n)\subset \mathcal{C}(m)$ and for all $x\in \Theta(n)$ holds $[x]\in \mathcal{C}(n)$ and $1+x\leq n<m$. There exist some $[y]\in \mathcal{C}(n)$ such that $\delta_r([y])=[1]$ but there exists no $z\in \Theta(n)$ such that $1+z=m$. It follows that the point $[y]$ is not a stable point under $\delta_r$. This contradicts the claim that $\Theta(n)$ is stable and so $n<m$ is impossible. Conversely let us suppose that $m<n$ and consider the dilation 
\begin{align}
\delta_r:\mathcal{C}(n)\longrightarrow \mathcal{C}(m).\nonumber
\end{align}
We note that for any point $[x]\in \mathcal{C}(n)$ there exist some $k<m<n$ such that $||\delta_r([x])||+k=m$. 
Because $k \in \mathbb{N}_{n-1}=\Theta(n)$ it follows that the subsequence $\Theta(n)$ is stable under any dilation $\delta_r$.
\end{proof}
\bigskip

Next we show that any consecutive subsequence of $\mathbb{N}_n$ containing none of its degenerate terms must be stable under the simple dilation. We formalize this assertion in the following results.

\begin{proposition}
Let  $\Theta(n):=\left \{x,x+1,\ldots,n-x,n-x+1\right \}$ be a subsequence of $\mathbb{N}_n$ for any $1<x<\frac{n}{2}$  and  $\delta_r:\mathcal{C}(n)\longrightarrow \mathcal{C}(n+1)$ be an expansion. Then $\Theta(n)$ is stable under the expansion $\delta_r$.
\end{proposition}

\begin{proof}
For any point $[x]\in \mathcal{C}(n)$ we see that $\mathbb{L}_{[x],[n-x]}$ is a real axis of the CoP. By enforcing $1<x<\frac{n}{2}$, then we observe that the dilation $\delta_1:\mathcal{C}(n)\longrightarrow \mathcal{C}(n+1)$ with 
\begin{align}
\delta_1([x]):=
\begin{cases}
[x] \quad &\mbox{for }~1\leq x\leq n-1 \\
[n] \quad &\mbox{additional for}~x=1
\end{cases}
\end{align}
is achievable. 
It follows that for each $1<x<\frac{n}{2}$ the line $\mathbb{L}_{[x],[n-x+1]}$ is also a real axis of the CoP $\mathcal{C}(n+1)$. This proves that $\Theta(n)$ is stable under the dilation $\delta_r$.
\end{proof}

\section{\textbf{The Density of Points on the Circle of Partition}}
In this section we introduce the notion of density of points on CoP $\mathcal{C}(n,\mathbb{M})$ for $\mathbb{M}\subseteq \mathbb{N}$. We launch the following language in that regard. We consider in this section only real axes. Therefore we don't use the attribute \textit{real} in this section.

\begin{definition}
Let be $\mathbb{H}\subset\mathbb{N}$. Then the quantity 
\[
\mathcal{D}\left(\mathbb{H}\right)=\lim_{n\rightarrow\infty}
\frac{\vert\mathbb{H}\cap \mathbb{N}_n\vert}{n}
\]
denotes the density of $\mathbb{H}$.
\end{definition}

\begin{definition}\label{pointdensity}
Let $\mathcal{C}(n,\mathbb{M})$ be CoP with $\mathbb{M}\subset \mathbb{N}$ and $n\in \mathbb{N}$. Suppose $\mathbb{H}\subset \mathbb{M}$ then by the density of points $[x]\in \mathcal{C}(n,\mathbb{M})$ such that $x\in \mathbb{H}$, denoted $\mathcal{D}(\mathbb{H}_{\mathcal{C}(\infty,\mathbb{M})})$, we mean the quantity 
\begin{align}
\mathcal{D}\left(\mathbb{H}_{\mathcal{C}(\infty,\mathbb{M})}\right)=\lim \limits_{n\longrightarrow \infty}\frac{\# \lbrace\mathbb{L}_{[x],[y]} ~ \hat{\in}~  \mathcal{C}(n,\mathbb{M})|~ \{x,y\} \cap \mathbb{H}\neq \emptyset \rbrace}{ \nu(n,\mathbb{M})}.\nonumber
\end{align}
\end{definition}
\bigskip

\begin{remark}
The notion of the density of points as espoused in Definition \ref{pointdensity} provides  a passage between the density of the corresponding weight set of points. This possibility renders this type of density as a black box in studying problems concerning partition of numbers into specialized sequences taking into consideration their density.
\end{remark}

\begin{proposition}\label{inequality}
Let $\mathcal{C}(n)$ with $n\in \mathbb{N}$ be a CoP and $\mathbb{H}\subset \mathbb{N}$. Then the following inequality holds 
\begin{align}
\mathcal{D}(\mathbb{H})=\lim \limits_{n\longrightarrow \infty}\frac{\left \lfloor \frac{|\mathbb{H}\cap \mathbb{N}_n|}{2}\right \rfloor}{\left \lfloor \frac{n-1}{2}\right \rfloor}\leq \mathcal{D}(\mathbb{H}_{\mathcal{C}(\infty)})\leq \lim \limits_{n\longrightarrow \infty}\frac{|\mathbb{H}\cap \mathbb{N}_n|}{\left \lfloor \frac{n-1}{2}\right \rfloor}=2\mathcal{D}(\mathbb{H}).\nonumber
\end{align}
\end{proposition}

\begin{proof}
The upper bound is obtained from a configuration where no two points $[x],[y]\in \mathcal{C}(n)$ such that $x,y\in \mathbb{H}$ lie on the same axis of the CoP. That is, by the uniqueness of the axes of CoPs with $\nu(n,\mathbb{H})=0$, we can write
\begin{align}
   \# \left \{\mathbb{L}_{[x],[y]}\in \mathcal{C}(n)|~\{x,y\}\cap \mathbb{H}\neq \emptyset \right \}&=\nu(n,\mathbb{H})+\# \left \{\mathbb{L}_{[x],[y]}\in \mathcal{C}(n)|~x\in \mathbb{H},~y\in \mathbb{N}\setminus \mathbb{H}\right \} \nonumber \\&=\# \left \{\mathbb{L}_{[x],[y]}\in \mathcal{C}(n)|~x\in \mathbb{H},~y\in \mathbb{N}\setminus \mathbb{H}\right \} \nonumber \\&=|\mathbb{H}\cap \mathbb{N}_n|.\nonumber
\end{align}
The lower bound however follows from a configuration where any two points $[x],[y]\in \mathcal{C}(n)$ with $x,y\in \mathbb{H}$ are joined by a axis of the CoP. That is, by the uniqueness of the axis of CoPs with $\# \left \{\mathbb{L}_{[x],[y]}\in \mathcal{C}(n)|~x\in \mathbb{H},~y\in \mathbb{N}\setminus \mathbb{H}\right \} =0$, then we can write 
\begin{align}
    \# \left \{\mathbb{L}_{[x],[y]}\in \mathcal{C}(n)|~\{x,y\}\cap \mathbb{H}\neq \emptyset \right \}&=\nu(n,\mathbb{H})\nonumber \\&=\left \lfloor \frac{|\mathbb{H}\cap \mathbb{N}_n|}{2}\right \rfloor \nonumber
\end{align}
\end{proof}
\bigskip

\begin{proposition}\label{propertydensity}
Let $\mathbb{H}\subset \mathbb{N}$ and suppose $\mathcal{D}(\mathbb{H}_{\mathcal{C}(\infty)})$ exists. Then the following properties hold:
\begin{enumerate}
    \item [(i)] $\mathcal{D}(\mathbb{N}_{\mathcal{C}(\infty)})=1$ and $\mathcal{D}(\mathbb{H}_{\mathcal{C}(\infty)})\leq 1$ and additionally that $\mathcal{D}(\mathbb{H}_{\mathcal{C}(\infty)})<1$ provided $\mathcal{D}(\mathbb{N}\setminus \mathbb{H})>0$.
    
    \item [(ii)] $1-\lim \limits_{n\longrightarrow \infty}\dfrac{\nu(n,\mathbb{N}\setminus\mathbb{H})}{\nu(n,\mathbb{N})}=\mathcal{D}(\mathbb{H}_{\mathcal{C}(\infty)})$.
    
    \item [(iii)] If $|\mathbb{H}|<\infty$ then $\mathcal{D}(\mathbb{H}_{\mathcal{C}(\infty)})=0$.
\end{enumerate}
\end{proposition}

\begin{proof}
It is easy to see that the first part of \textbf{Property} $(i)$ and $(iii)$ are both easy consequences of the definition of density of points on the CoP $\mathcal{C}(n)$ and Proposition \ref{inequality}. We establish the second part of property $(i)$ and \textbf{Property} $(ii)$, which is the less obvious case. We observe by the uniqueness of the axes of CoPs that we can write 
\begin{align*}
    1&=\lim \limits_{n\longrightarrow \infty}\frac{\nu(n,\mathbb{N})}{\nu(n,\mathbb{N})}\\
    &=\lim \limits_{n\longrightarrow \infty}\frac{\# \lbrace\mathbb{L}_{[x],[y]} ~ \hat{\in}~  \mathcal{C}(n)|~ x\in \mathbb{H}~,y\in \mathbb{N}\setminus \mathbb{H}\rbrace}{\nu(n,\mathbb{N})}\\
    &+\lim \limits_{n\longrightarrow \infty}\frac{\nu(n,\mathbb{H})}{\nu(n,\mathbb{N})}
    +\lim \limits_{n\longrightarrow \infty}\frac{\nu(n,\mathbb{N}\setminus\mathbb{H})}{\nu(n,\mathbb{N})}\\
    &=\mathcal{D}(\mathbb{H}_{\mathcal{C}(\infty)})
    +\lim \limits_{n\longrightarrow \infty}\frac{\nu(n,\mathbb{N}\setminus\mathbb{H})}{\nu(n,\mathbb{N})}
\end{align*}
and $(ii)$ follows immediately. The second part of $(i)$ follows from the above expression and exploiting the inequality
\begin{align}
   \lim \limits_{n\longrightarrow \infty}\frac{\nu(n,\mathbb{N}\setminus\mathbb{H})}{\nu(n,\mathbb{N})}&\leq \lim \limits_{n\longrightarrow \infty}\frac{\left \lfloor \frac{|\mathbb{N}\setminus \mathbb{H}\cap \mathbb{N}_n|}{2}\right \rfloor}{\left \lfloor \frac{n-1}{2}\right \rfloor}=\mathcal{D}(\mathbb{N}\setminus \mathbb{H})\nonumber
\end{align}
\end{proof}
\bigskip

It is important to notice that the same result may not hold if we replace the set of natural numbers $\mathbb{N}$ with a special subset $\mathbb{M}$. Next we transfer the notion of the density of a sequence to the density of corresponding points on the CoP $\mathcal{C}(n)$. This notion will play a crucial role in our latter developments.

\begin{proposition}
Let $\epsilon\in(0,1]$ and $\mathbb{H}$ be a sequence with $\mathbb{H}\subset \mathbb{N}$ and $\mathcal{C}(n)$ be a CoP. If $\mathcal{D}\left(\mathbb{H}\right)\geq \epsilon$ then $\mathcal{D}\left(\mathbb{H}_{\mathcal{C}(\infty)}\right)\geq \epsilon$.
\end{proposition}

\begin{proof}
The result follows by exploiting the inequality in Proposition \ref{inequality}
\end{proof}

\begin{proposition}
Let $\mathbb{H}$ be a sequence with $\mathbb{H}\subset \mathbb{N}$. For $\epsilon\in(0,1]$ and any $k\in \mathbb{N}$ if 
\begin{align}|\mathbb{H}\cap \mathbb{N}_n|\geq n\epsilon \nonumber
\end{align}
and the common difference of arithmetic progressions in $\bigg(\mathbb{N}\setminus \mathbb{H}\bigg)\cap \mathbb{N}_n$ are different from those in $\mathbb{H}\cap \mathbb{N}_n$, then there exists some rotation $\varpi_r$ such that the CoP $\mathcal{C}(n)$ contains at least $(k-1)$ stable points $[x]$ for $x\in \mathbb{H}\cap \mathbb{N}_n$.
\end{proposition}

\begin{proof}
Suppose $\mathbb{H}\subset \mathbb{N}$ with the underlying conditions, then by Theorem \ref{szemeredi} the sequence $\mathbb{H}$ contains fairly long arithmetic progressions of length $k$. We enumerate them as follows 
\begin{align}
x,x+s,x+2s,\ldots,x+(k-1)s\nonumber
\end{align}
for $s\in \mathbb{N}$. It follows that the corresponding points on the CoP $\mathcal{C}(n)$, namely 
\begin{align}
[x],[x+s],[x+2s],\ldots,[x+(k-1)s]\in \mathcal{C}(n)\nonumber
\end{align}
are equally spaced and the chord joining two of these adjacent points are of equal distance. Similarly points on the other end of the axis are equally spaced and the chords joining any of these two adjacent points are of equal distance $s$. Let us enumerate them as follows 
\begin{align}
[n-x],[n-x-s],[n-x-2s],\ldots, [n-x-(k-1)s]\in \mathcal{C}(n).\nonumber
\end{align}
Apply the rotation $\varpi_r$ by choosing $r=s$ then we have 
\begin{align}
\varpi_s([x]),\varpi_s([x+s]),\ldots,\varpi_s([x+(k-1)s]).\nonumber
\end{align}
The image of these points under the rotation is given by 
\begin{align}
[x+s],[x+2s],\ldots,[x+(k-1)s], [x+ks].\nonumber
\end{align}
Since the point $[x+ks]$ a priori was not on any of the axes considered at least $(k-1)$ points on these axes will be transferred to their immediate next point on an axis containing all points $[x]$ with $x\in \mathbb{H}\cap \mathbb{N}_n$. Similarly under the rotation the corresponding images of the points on the other half of the CoP lying on the same axis with these points have the images 
\begin{align}
\varpi_s([n-x]),\varpi_s([n-x-s]),\ldots,\varpi_s([n-x-(k-1)s])\nonumber
\end{align}
which we can recast as 
\begin{align}
[n-x-s],[n-x-2s],\ldots,[n-x-(k-1)s],[n-x-ks].\nonumber
\end{align}
At least $(k-1)$ of these points are points on the previous axis and they lying on the same axis with the points on the other half of the CoP. Since the sequence 
\begin{align}
n-x-s,n-x-2s,\ldots,n-x-ks\nonumber
\end{align}
are in arithmetic progression, it follows by the assumption
\begin{align}
n-x-s,n-x-2s,\ldots,n-x-ks\in \mathbb{H}\cap \mathbb{N}_n.\nonumber
\end{align}
This completes the proof.
\end{proof}

\begin{proposition}
Let $\mathbb{H}\subset \mathbb{N}$ such that $\mathbb{H}=\mathbb{J}\cup \mathbb{T}$ with $\mathbb{J}\cap \mathbb{T}=\emptyset$ and $\mathcal{D}(\mathbb{T})=0$. Then the following inequalities hold for density of CoPs
\begin{align}
   \mathcal{D}(\mathbb{H}_{\mathcal{C}(\infty)})=\mathcal{D}(\mathbb{J}_{\mathcal{C}(\infty)})\nonumber
\end{align}and 
\begin{align}
    \mathcal{D}((\mathbb{N}\setminus \mathbb{H})_{\mathcal{C}(\infty)})\leq \mathcal{D}((\mathbb{N}\setminus \mathbb{J})_{\mathcal{C}(\infty)}).\nonumber
\end{align}
\end{proposition}

\begin{proof}
Appealing to Proposition \ref{propertydensity}, it follows by the uniqueness of the axes of CoPs the following decomposition
\begin{align}
   \mathcal{D}(\mathbb{H}_{\mathcal{C}(\infty)})&=\lim \limits_{n\longrightarrow \infty}\frac{\# \left \{\mathbb{L}_{[x],[y]}\in \mathcal{C}(n)|~\{x,y\}\cap (\mathbb{J}\cup \mathbb{T})\neq \emptyset \right\}}{\left \lfloor \frac{n-1}{2}\right \rfloor}\nonumber \\&= \lim \limits_{n\longrightarrow \infty}\frac{\# \left \{\mathbb{L}_{[x],[y]}\in \mathcal{C}(n,\mathbb{J})\right\}}{\left \lfloor \frac{n-1}{2}\right \rfloor}\nonumber \\&+\lim \limits_{n\longrightarrow \infty}\frac{\# \left \{\mathbb{L}_{[x],[y]}\in \mathcal{C}(n,\mathbb{T})\right\}}{\left \lfloor \frac{n-1}{2}\right \rfloor}\nonumber \\&+\lim \limits_{n\longrightarrow \infty}\frac{\# \left \{\mathbb{L}_{[x],[y]}\in \mathcal{C}(n)|~x\in \mathbb{J},~y\in \mathbb{T}\right\}}{\left \lfloor \frac{n-1}{2}\right \rfloor}\nonumber \\&+ \lim \limits_{n\longrightarrow \infty}\frac{\# \left \{\mathbb{L}_{[x],[y]}\in \mathcal{C}(n)|~x\in \mathbb{J},~y\in \mathbb{N}\setminus \mathbb{J}\right\}}{\left \lfloor \frac{n-1}{2}\right \rfloor}\nonumber \\&+\lim \limits_{n\longrightarrow \infty}\frac{\# \left \{\mathbb{L}_{[x],[y]}\in \mathcal{C}(n)|~x\in \mathbb{T},~y\in \mathbb{N}\setminus \mathbb{T}\right\}}{\left \lfloor \frac{n-1}{2}\right \rfloor}\nonumber
\end{align}Under the inequalities 
\begin{align}
    \lim \limits_{n\longrightarrow \infty}\frac{\# \left \{\mathbb{L}_{[x],[y]}\in \mathcal{C}(n,\mathbb{T})\right\}}{\left \lfloor \frac{n-1}{2}\right \rfloor}\leq \lim \limits_{n\longrightarrow}\frac{\lfloor \frac{|\mathbb{T}\cap \mathbb{N}_n|}{2}\rfloor}{\left \lfloor \frac{n-1}{2}\right \rfloor}=0\nonumber
\end{align}and 
\begin{align}
    \lim \limits_{n\longrightarrow \infty}\frac{\# \left \{\mathbb{L}_{[x],[y]}\in \mathcal{C}(n)|~x\in \mathbb{J},~y\in \mathbb{T}\right\}}{\left \lfloor \frac{n-1}{2}\right \rfloor}\leq \lim \limits_{n\longrightarrow \infty}\frac{|\mathbb{T}\cap \mathbb{N}_n|}{\left \lfloor \frac{n-1}{2}\right \rfloor}=0\nonumber
\end{align}and 
\begin{align}
    \lim \limits_{n\longrightarrow \infty}\frac{\# \left \{\mathbb{L}_{[x],[y]}\in \mathcal{C}(n)|~x\in \mathbb{T},~y\in \mathbb{N}\setminus \mathbb{T}\right\}}{\left \lfloor \frac{n-1}{2}\right \rfloor}&\leq \lim \limits_{n\longrightarrow \infty}\frac{|\mathbb{T}\cap \mathbb{N}_n|}{\left \lfloor \frac{n-1}{2}\right \rfloor}=0\nonumber
\end{align}we have 
\begin{align}
    \mathcal{D}(\mathbb{H}_{\mathcal{C}(\infty)})&=\lim \limits_{n\longrightarrow \infty}\frac{\# \left \{\mathbb{L}_{[x],[y]}\in \mathcal{C}(n)|~\{x,y\}\cap (\mathbb{J}\cup \mathbb{T})\neq \emptyset \right\}}{\left \lfloor \frac{n-1}{2}\right \rfloor}\nonumber \\&=\lim \limits_{n\longrightarrow \infty}\frac{\# \left \{\mathbb{L}_{[x],[y]}\in \mathcal{C}(n,\mathbb{J})\right\}}{\left \lfloor \frac{n-1}{2}\right \rfloor}\nonumber \\&+\lim \limits_{n\longrightarrow \infty}\frac{\# \left \{\mathbb{L}_{[x],[y]}\in \mathcal{C}(n)|~x\in \mathbb{J},~y\in \mathbb{N}\setminus \mathbb{J}\right\}}{\left \lfloor \frac{n-1}{2}\right \rfloor}\nonumber \\&=\mathcal{D}(\mathbb{J}_{\mathcal{C}(\infty)}).\nonumber
\end{align}Again we have the following decomposition by virtue of $\mathbb{J}\cap \mathbb{T}=\emptyset$
\begin{align}
    \mathcal{D}((\mathbb{N}\setminus \mathbb{H})_{\mathcal{C}(\infty)})&= \lim \limits_{n\longrightarrow \infty}\frac{\# \left \{\mathbb{L}_{[x],[y]}\in \mathcal{C}(n)|~\{x,y\}\cap (\mathbb{N}\setminus \mathbb{J}\cup \mathbb{T})\neq \emptyset \right\}}{\left \lfloor \frac{n-1}{2}\right \rfloor}\nonumber \\&= \lim \limits_{n\longrightarrow \infty}\frac{\# \left \{\mathbb{L}_{[x],[y]}\in \mathcal{C}(n,\mathbb{N}\setminus \mathbb{J}\cap \mathbb{N}\setminus \mathbb{T})\right\}}{\left \lfloor \frac{n-1}{2}\right \rfloor}\nonumber \\&+\lim \limits_{n\longrightarrow \infty}\frac{\# \left \{\mathbb{L}_{[x],[y]}\in \mathcal{C}(n)|~x\in \mathbb{N}\setminus \mathbb{J}\cap \mathbb{N}\setminus \mathbb{T},~y\in \mathbb{J}\cup \mathbb{T}\right\}}{\left \lfloor \frac{n-1}{2}\right \rfloor}\nonumber \\&\leq \lim \limits_{n\longrightarrow \infty}\frac{\# \left \{\mathbb{L}_{[x],[y]}\in \mathcal{C}(n,\mathbb{N}\setminus \mathbb{J})\right\}}{\left \lfloor \frac{n-1}{2}\right \rfloor}\nonumber  \\&+ \lim \limits_{n\longrightarrow \infty}\frac{\# \left \{\mathbb{L}_{[x],[y]}\in \mathcal{C}(n)|~x\in \mathbb{N}\setminus \mathbb{J},~y\in \mathbb{J}\right\}}{\left \lfloor \frac{n-1}{2}\right \rfloor}\nonumber \\&+\lim \limits_{n\longrightarrow \infty}\frac{\# \left \{\mathbb{L}_{[x],[y]}\in \mathcal{C}(n)|~x\in \mathbb{N}\setminus \mathbb{J},~y\in \mathbb{T}\right\}}{\left \lfloor \frac{n-1}{2}\right \rfloor}\nonumber
\end{align}since $\mathbb{N}\setminus \mathbb{J}\cap \mathbb{N}\setminus \mathbb{T}\subset \mathbb{N}\setminus \mathbb{J}$. By exploiting the inequality
\begin{align}
    \lim \limits_{n\longrightarrow \infty}\frac{\# \left \{\mathbb{L}_{[x],[y]}\in \mathcal{C}(n)|~x\in \mathbb{N}\setminus \mathbb{J},~y\in \mathbb{T}\right\}}{\left \lfloor \frac{n-1}{2}\right \rfloor}&\leq \lim \limits_{n\longrightarrow \infty}\frac{|\mathbb{T}\cap \mathbb{N}_n|}{\left \lfloor \frac{n-1}{2}\right \rfloor}=0\nonumber
\end{align}it follows that 
\begin{align}
    \mathcal{D}((\mathbb{N}\setminus \mathbb{H})_{\mathcal{C}(\infty)})&\leq \lim \limits_{n\longrightarrow \infty}\frac{\# \left \{\mathbb{L}_{[x],[y]}\in \mathcal{C}(n,\mathbb{N}\setminus \mathbb{J})\right\}}{\left \lfloor \frac{n-1}{2}\right \rfloor}\nonumber  \\&+ \lim \limits_{n\longrightarrow \infty}\frac{\# \left \{\mathbb{L}_{[x],[y]}\in \mathcal{C}(n)|~x\in \mathbb{N}\setminus \mathbb{J},~y\in \mathbb{J}\right\}}{\left \lfloor \frac{n-1}{2}\right \rfloor}\nonumber \\&=\mathcal{D}((\mathbb{N}\setminus \mathbb{J})_{\mathcal{C}(\infty)}).\nonumber
\end{align}It follows by the above analysis the inequalities
\begin{align}
\mathcal{D}\left(\mathbb{H}_{\mathcal{C}(\infty)}\right)=\mathcal{D}(\mathbb{J}_{\mathcal{C}(\infty)})\nonumber
\end{align}and 
\begin{align}
\mathcal{D}\left((\mathbb{N}\setminus \mathbb{H})_{\mathcal{C}(\infty)}\right)\leq \mathcal{D}\left((\mathbb{N}\setminus \mathbb{J})_{\mathcal{C}(\infty)}\right).\nonumber
\end{align}
\end{proof}
\bigskip

In the accompanying proof we will make use of degenerate and non-degenerate points of a given set of points on a CoP. However intricate the proof might seem to be, it can be pinned down to just a simple principle. The highly dense nature of the sequence allows us to break their components into several boxes. The closest components in each of these boxes are equidistant from each other. The residue which are not dense will be thrown away into another box whose components are very sparse. We then translate a component by their gap if it ever happens to be in some dense box at the same time live on the same axis with other component. This forces the second component to also belong to some dense box. If the component on the same axis with another component does not belong to the dense box, then the components and the associated components must live in the sparse box. We can then move them into the dense box and repeat the arguments.  We make these terminologies more precise in the following definitions and then present our argument.

\begin{definition}
Let $\mathcal{P}\subseteq \mathcal{C}(n,\mathbb{M})$ with $\mathbb{M}\subseteq \mathbb{N}$. Then a point $[x]\in \mathcal{P}$ is a degenerate point if the line joining the point $[x]$ to the centre (resp. deleted centre) of the CoP $\mathcal{C}(n,\mathbb{M})$ is a boundary of the largest sector induced by the points in $\mathcal{P}$. Otherwise, we say it is a non-degenerate point in $\mathcal{P}$.
\end{definition}
\bigskip

\begin{theorem}\label{master}
Let $\mathbb{H}\subset \mathbb{N}$ and suppose that $\mathcal{C}(n,\mathbb{H})\neq \emptyset$. If for any $\epsilon\in (0,1]$ holds 
\begin{align}
|\mathbb{H}\cap \mathbb{N}_n|\geq n\epsilon\nonumber
\end{align}
with 
\begin{align}
\mathcal{D}(\mathbb{N}\setminus \mathbb{H})=\lim \limits_{n\longrightarrow \infty}\frac{|(\mathbb{N}\setminus \mathbb{H})\cap \mathbb{N}_n|}{n}<\mathcal{D}(\mathbb{H}) \nonumber
\end{align}
then there exists a dilation $\delta_r:\mathcal{C}(n,\mathbb{H})\longrightarrow \mathcal{C}(n+r,\mathbb{H})$ such that 
\begin{align} 
\mathcal{C}(n+r,\mathbb{H})\neq \emptyset. \nonumber
\end{align}
\end{theorem}

\begin{proof}
Under the assumption $|\mathbb{H}\cap \mathbb{N}_n|\geq n\epsilon$ for any $\epsilon\in(0,1]$, then $\mathbb{H}$ contains fairly long arithmetic progressions. Let us enumerate them as follows 
\begin{align}
\mathbb{G}_1=\left \{x_1+kd_1\in \mathbb{H}\right \}_{k=0}^{s_1;s_1\geq 1}.\nonumber
\end{align}
Let us consider the residual set 
\begin{align}
\mathbb{G}_2=\mathbb{H}\setminus \left \{x_1+kd_1\in \mathbb{H}\right \}_{k=0}^{s_1;s_1\geq 1}.\nonumber
\end{align}
Then we can partition the  sequence $\mathbb{H}$ in the following way
\begin{align}
\mathbb{H}=\mathbb{G}_1\cup \mathbb{G}_2.\nonumber
\end{align}
If $\mathbb{G}_2$ is still dense then we can repeat this process and obtain further a partition of $\mathbb{H}$ into three subsequence 
\begin{align}
\mathbb{H}=\mathbb{G}_1\cup \mathbb{G}_2\cup \mathbb{G}_3.\nonumber
\end{align}
By induction, we can partition the sequence $\mathbb{H}$ in the following way 
\begin{align}
\mathbb{H}=\bigcup_{i=1}^{m}\mathbb{G}_i \cup \mathbb{T}\label{decomposition}
\end{align}
where 
\begin{align}
\lim \limits_{n\longrightarrow \infty}\frac{|\mathbb{T}\cap \mathbb{N}_n|}{n}=0\nonumber
\end{align}and $\mathbb{J}=\bigcup_{i=1}^{m}\mathbb{G}_i $ with $\mathcal{D}(\mathbb{J})=\mathcal{D}(\mathbb{H})$, since $\mathbb{J}\cap \mathbb{T}=\emptyset$ and $\mathbb{G}_i=\left \{x_i+kd_i\in \mathbb{H}\right \}_{k=0}^{s_i;s_i\geq 1}$. Now it suffices to work with the corresponding points on the CoP $\mathcal{C}(n,\mathbb{H})$. Since by assumption $\mathcal{C}(n,\mathbb{H})\neq \emptyset$, It follows that there exist some axes $\mathbb{L}_{[a],[b]}~ \hat{\in}~  \mathcal{C}(n,\mathbb{H})$. Now let us suppose that 
\begin{align}
[b]\notin \bigcup_{i=1}^{m}\left \{[x_i+kd_i]\in \mathcal{C}(n,\mathbb{H})\right \}_{k=0}^{s_i;s_i\geq 1}\nonumber
\end{align}
for $b\in \mathbb{H}$, then it follows that no two adjacent chords of equal length joining points in 
\begin{align}\bigcup_{i=1}^{m}\left \{[x_i+kd_i]\in \mathcal{C}(n,\mathbb{H})\right \}_{k=0}^{s_i;s_i\geq 1}\nonumber
\end{align}
contains the point $[b]$. Let us suppose on the contrary that
\begin{align}[a]\in \bigcup_{i=1}^{m}\left \{[x_i+kd_i]\in \mathcal{C}(n,\mathbb{H})\right \}_{k=0}^{s_i;s_i\geq 1}\nonumber
\end{align}
then it follows that $[a]\in \left \{[x_i+kd_i]\in \mathcal{C}(n,\mathbb{H})\right \}_{k=0}^{s_i;s_i\geq 1}$ for some $1\leq i\leq m$. We consider two cases. The case $[a]$ is a degenerate point in the set and the case it is non-degenerate point in the set. If $[a]$ is a degenerate point in the set $\left \{[x_i+kd_i]\in \mathcal{C}(n,\mathbb{H})\right \}_{k=0}^{s_i;s_i\geq 1}$, in particular, $[a]$ is the first point in the set. Then it follows that the following points 
\begin{align}
[a], [x_i+d_i], [x_i+2d_i],\ldots [x_i+sd_i]\nonumber
\end{align}
are equally spaced with $b=n-x_i$. It follows that $b$ is contained in the arithmetic progression 
\begin{align}
n-x_i,n-x_i-d_i,\ldots,n-x_i-sd_i\nonumber
\end{align}
which contradicts the assumption that $[b]$ cannot lie on at least one of any two adjacent chords of equal length. Otherwise 
\begin{align}
n-x_i,n-x_i-d_i,\ldots,n-x_i-sd_i\in \bigg(\mathbb{N}\setminus \mathbb{J}\bigg)\cap \mathbb{N}_n\nonumber
\end{align}
and it follows each element in the weight set $\mathbb{J}\cap \mathbb{N}_n=\mathbb{K}$ of the corresponding point set $\mathbb{K}^{*}=\bigcup_{i=1}^{m}\left \{[x_i+kd_i]\in \mathcal{C}(n,\mathbb{H})\right \}_{k=0}^{s_i;s_i\geq 1}$ uniquely generates an element in the set $\bigg(\mathbb{N}\setminus \mathbb{J}\bigg)\cap \mathbb{N}_n$, so that 
\begin{align}
    |\mathbb{J}\cap \mathbb{N}_n|\leq |\bigg(\mathbb{N}\setminus \mathbb{J}\bigg)\cap \mathbb{N}_n|.\nonumber
\end{align}It follows that 
\begin{align}
  \mathcal{D}(\mathbb{N}\setminus \mathbb{H})&<\mathcal{D}(\mathbb{H})\nonumber \\&=\mathcal{D}(\mathbb{J})\nonumber \\&\leq \mathcal{D}(\mathbb{N}\setminus \mathbb{J})\nonumber
\end{align}
so that we have the inequality $\mathcal{D}(\mathbb{N}\setminus \mathbb{H})<\mathcal{D}(\mathbb{N}\setminus \mathbb{J})$. This contradicts the equality under equality $\mathcal{D}(\mathbb{H})=\mathcal{D}(\mathbb{J})$
\begin{align}
\mathcal{D}(\mathbb{N}\setminus \mathbb{H})&=1-\mathcal{D}(\mathbb{H})\nonumber \\&=1-\mathcal{D}(\mathbb{J})\nonumber \\&=\mathcal{D}(\mathbb{N}\setminus \mathbb{J}).\nonumber
\end{align}
For the case $[a]=[x_i+sd_i]$, then we obtain the a priori arithmetic progression with $b=n-x_i-sd_i$. The corresponding point $[b]$ also violates the required specification. If the point $[a]\in \left \{[x_i+kd_i]\in \mathcal{C}(n,\mathbb{H})\right \}_{k=0}^{s_i;s_i\geq 1}$ is a a non-degenerate point, then $a=x_i+jd_i$ for some $0<j<s$. The same analysis can be carried out to yield a contradiction. Now for the case 
\begin{align}
[a]\in \bigcup_{i=1}^{m}\left \{[x_i+kd_i]\in \mathcal{C}(n,\mathbb{H})\right \}_{k=0}^{s_i;s_i\geq 1}\nonumber
\end{align}
then we choose the dilation $\delta_r$ with $r=d_j$ such that $[b]\in \left \{[x_j+kd_j]\in \mathcal{C}(n,\mathbb{H})\right \}_{k=0}^{s_i;s_i\geq 1}$ for $r<0$ if $[b]$ is the last degenerate point in the set and $r>0$ if $[b]$ is the first degenerate point or a non-degenerate point in the set, so that we have 
\begin{align}
\mathbb{L}_{[a],[b+d_j]}~ \hat{\in}~  \mathcal{C}(n+d_j,\mathbb{H}).\nonumber
\end{align}
This completes the first part of the proof. For the second part let us assume that for the axis $\mathbb{L}_{[a],[b]}$ of $\mathcal{C}(n,\mathbb{H})$, then
\begin{align}
[a]\notin \bigcup_{i=1}^{m}\left \{[x_i+kd_i]\in \mathcal{C}(n,\mathbb{H})\right \}_{k=0}^{s_i;s_i\geq 1}\nonumber
\end{align}
then it must necessarily be that 
\begin{align}
[a]\in \mathbb{T}_n^{*}\nonumber
\end{align}
where $\mathbb{T}_n^{*}$ is the corresponding point set of elements in $\mathbb{T}\cap \mathbb{N}_n$. Since 
\begin{align}
|\mathbb{T}_n^{*}|\leq \bigg |\bigcup_{i=1}^{m}\left \{[x_i+kd_i]\in \mathcal{C}(n,\mathbb{H})\right \}_{k=0}^{s_i;s_i\geq 1}\bigg|,\nonumber
\end{align}
there exists some rotation $\varpi_t$ such that the point $\varpi_t([a])\in \bigcup_{i=1}^{m}\left \{[x_i+kd_i]\in \mathcal{C}(n,\mathbb{H})\right \}_{k=0}^{s_i;s_i\geq 1}$. In particular 
\begin{align}
\varpi_t([a])\in \left \{[x_j+kd_j]\in \mathcal{C}(n,\mathbb{H})\right \}_{k=0}^{s_i;s_i\geq 1}\nonumber
\end{align}
for some $1\leq j\leq m$. It follows there must exist a point
\begin{align}[v]\in \bigcup_{i=1}^{m}\left \{[x_i+kd_i]\in \mathcal{C}(n,\mathbb{H})\right \}_{k=0}^{s_i;s_i\geq 1}\nonumber
\end{align}
such that $\mathbb{L}_{[v],[\varpi_t([a])]}$ is an axis of the CoP $\mathcal{C}(n,\mathbb{H})$, by virtue of the previous arguments. Otherwise, we discard this choice of point and scout for a point with such property by varying the scale of the rotation $\varpi_t$. The proof is completed by choosing the dilation $\delta_r$ such that $r=d_j$ for $r<0$ if $\varpi_t([a])]$ is the last degenerate point in the set and $r>0$ if $\varpi_t([a])]$ is the first degenerate point or a non-degenerate point in the set, so that $\mathbb{L}_{[v],[||\varpi_t([a])||+d_j]}$ is an axis of the CoP
\begin{align}
\mathcal{C}(n+d_j,\mathbb{H}).\nonumber
\end{align}
\end{proof}

\begin{theorem}\label{squarefree}
There are infinitely many $n\in \mathbb{M}_{a,d}$ with fixed $a,d\in\mathbb{N}$ such that the representation 
\begin{align}
n=z_1+z_2\nonumber
\end{align}
where $\mu(z_1)=\mu(z_2)\neq 0$, $z_1,z_2\in\mathbb{N}$ and $\mu$ is the M\"obius function defined as 
\begin{align}
\mu(m)=\begin{cases}1 \quad \mathrm{if}\quad m=1\\0 \quad \mathrm{if}\quad p^k|m,~k\in \mathbb{N}\setminus \{1\}\\ (-1)^r \quad \mathrm{if}\quad m=p_1p_2\cdots p_r\end{cases}\nonumber
\end{align}
is valid.
\end{theorem}

\begin{proof}
The set of square-free integers 
\begin{align}
\mathcal{Q}:=\left \{m\in \mathbb{N}:\mu(m)\neq 0\right \}\nonumber
\end{align}
has natural density $\frac{6}{\pi^2}$ \cite{montgomery2007multiplicative, hildebrand2006introduction}. For $n$ large enough there exists some fixed $N_0>n$ such that the representation is valid 
\begin{align}
N_o=z_1+z_2\nonumber
\end{align}
with $\mu(z_1),\mu(z_2)\neq 0$.
Invoking Theorem \ref{master} there exist some $t\in \mathbb{N}$ such that the representation is valid 
\begin{align}
N_t:=N_o+t=v_1+v_2\nonumber
\end{align}
with $\mu(v_1)=\mu(v_2)\neq 0$. The result follows by an upwards induction in this manner.
\end{proof}
\bigskip

\begin{corollary}
There are infinitely many $n\in \mathbb{M}_{a,d}$ with fixed $a,d\in\mathbb{N}$ such that the representation  \begin{align}
n=z_1+z_2\nonumber
\end{align}
with $\gcd(z_1,z_2)=1$ and $z_1,z_2\in\mathbb{N}$ is valid.
\end{corollary}

\begin{proof}
The set 
\begin{align}\mathcal{R}:=\left \{(m,n):\gcd(m,k)=1,~1\leq m<k\right\} \nonumber
\end{align}
has natural density $\mathcal{D}(\mathcal{R})=\frac{6}{\pi^2}$  with relatively small density for the residual set \cite{hildebrand2006introduction}. The result follows by adapting a similar reasoning in Theorem \ref{squarefree}.
\end{proof}
\bigskip 

Let be
\begin{align}\label{E-Q}
&\mathbb{Q}_p:=\lbrace q\in\mathbb{N}\mid \left(q,P(p)\right)=1\rbrace\\
&\mbox{with }p\in\mathbb{P}\mbox{ and }P(p):= \prod_{i=1}^{\pi(p)} p_i.\nonumber
\end{align}


\begin{theorem}\label{T-Q_density}
The set $\mathbb{Q}_p$ has for every $p\in\mathbb{P}$ a positive density
\[
\mathcal{D}(\mathbb{Q}_p)=\lim_{n\rightarrow\infty}\frac{\vert\mathbb{Q}_p\cap\mathbb{N}_n\vert}{n}>0.
\]
\end{theorem}
\begin{proof}
Let us consider the set $\mathbb{Q}_p$ as result of the sieve of Eratosthenes. Each prime $p_i\mid i=1,\ldots,p_{\pi(p)}=p$ sieves in each interval with a length of $p_i$ exactly one number. Then remain unsieved from $\mathbb{N}_{P(p)}$ exactly 
\[
P(p)\cdot\prod_{i=1}^{\pi(p)}(p_i-1)
\]
numbers. Therefore $\mathbb{Q}_p$ has in $\mathbb{N}_{P(p)}$ a density
\[
\alpha(p):= \frac{\prod_{i=1}^{\pi(p)}(p_i-1)}{P(p)}=
\prod_{i=1}^{\pi(p)}\frac{p_i-1}{p_i}>0.
\]
Then is (for $n$ large enough) 
\[
\vert\mathbb{Q}_p\cap\mathbb{N}_n\vert \sim n\cdot\alpha(p)
\]
which means
\[
\mathcal{D}(\mathbb{Q}_p)=\lim_{n\rightarrow\infty}\frac{n\cdot\alpha(p)}{n}
=\alpha(p)>0.
\]
\end{proof}
\bigskip

\subsection{\textbf{Application of density of points to partitions}}
In this section we explore  the connection between the notion of density of points in a typical CoP to the possibility of partitioning number into certain sequences. This method tends to work very efficiently for sets of integers having a positive density. 

\begin{theorem}\label{density to partition}
Let $\mathbb{H}\subset \mathbb{N}$ such that $\mathcal{D}(\mathbb{H})>\frac{1}{2}$. Then every sufficiently large $n\in\mathbb{N}$ has representation of the form 
\begin{align}
  n=z_1+z_2 \nonumber
\end{align}where $z_1,z_2\in \mathbb{H}$.
\end{theorem}

\begin{proof}
Appealing to Proposition \ref{inequality} we can write 
\begin{align}
    \lim \limits_{n\longrightarrow \infty}\frac{\left \lfloor \frac{|\mathbb{H}\cap \mathbb{N}_n|}{2}\right \rfloor}{\left \lfloor \frac{n-1}{2}\right \rfloor}\leq \mathcal{D}(\mathbb{H}_{\mathcal{C}(\infty)})\leq \lim \limits_{n\longrightarrow \infty}\frac{|\mathbb{H}\cap \mathbb{N}_n|}{\left \lfloor \frac{n-1}{2}\right \rfloor}.\nonumber
\end{align}By the uniqueness of the axes of CoPs we can write
\begin{align}
    \# \left \{\mathbb{L}_{[x],[y]}~ \hat{\in}~  \mathcal{C}(n)|~\{x,y\}\cap \mathbb{H}\neq \emptyset \right \}&=\nu(n,\mathbb{H})+\# \left \{\mathbb{L}_{[x],[y]}~ \hat{\in}~  \mathcal{C}(n)|~x\in \mathbb{H},~y\in \mathbb{N}\setminus \mathbb{H}\right \}.\nonumber
\end{align}Let us assume $\nu(n,\mathbb{H})=0$ then it follows by appealing to Definition \ref{pointdensity}
\begin{align}
    \mathcal{D}(\mathbb{H}_{\mathcal{C}(\infty)})&=2\mathcal{D}(\mathbb{H})\nonumber \\&>2\times \frac{1}{2}=1.\nonumber
\end{align}This contradicts the inequality $\mathcal{D}(\mathbb{H}_{\mathcal{C}(\infty)})\leq 1$ in Proposition \ref{propertydensity}. This proves that $\nu(n,\mathbb{H})>0$ for all sufficiently large values of $n\in \mathbb{N}$.
\end{proof}

\begin{corollary}
Let $\mathbb{R}:=\left \{m\in \mathbb{N}|~\mu(m)\neq 0\right \}$. Then every sufficiently large $n\in \mathbb{N}$ can be written in the form 
\begin{align}
    n=z_1+z_2\nonumber
\end{align}where $\mu(z_1)=\mu(z_2)\neq 0$.
\end{corollary}

\begin{proof}
By the uniqueness of the axes of CoPs we can write
\begin{align}
    \# \left \{\mathbb{L}_{[x],[y]}~ \hat{\in}~  \mathcal{C}(n)|~\{x,y\}\cap \mathbb{R}\neq \emptyset \right \}&=\nu(n,\mathbb{R})+\# \left \{\mathbb{L}_{[x],[y]}~ \hat{\in}~  \mathcal{C}(n)|~x\in \mathbb{R},~y\in \mathbb{N}\setminus \mathbb{R}\right \}.\nonumber
\end{align}Let us assume $\nu(n,\mathbb{R})=0$ then it follows by appealing to Definition \ref{pointdensity} and Theorem \ref{density to partition}
\begin{align}
    \mathcal{D}(\mathbb{R}_{\mathcal{C}(\infty)})&=2\mathcal{D}(\mathbb{R})\nonumber \\&=\frac{12}{\pi^2}>1\nonumber
\end{align}since $\mathcal{D}(\mathbb{R})=\frac{6}{\pi^2}$. This contradicts the inequality $\mathcal{D}(\mathbb{R}_{\mathcal{C}(\infty)})\leq 1$ in Proposition \ref{propertydensity}. This proves that $\nu(n,\mathbb{R})>0$ for all sufficiently large values of $n\in \mathbb{N}$.
\end{proof}
\bigskip

One could ever hope and dream of this strategy to work when we replace the set $\mathbb{R}$ of square-free integers with the set of prime numbers. There we would certainly ran into complete deadlock, since the prime in accordance with the prime number theorem have density zero. Any success in this regard could conceivably work by introducing some exotic forms of the notion of density of points and carefully choosing a subset of the integers that is somewhat dense among the set of integers and covers that primes. We propose a strategy somewhat akin to the above method for possibly getting a handle on the binary Goldbach conjecture and it's variants. Before that we state and prove a conditional theorem concerning the binary Goldbach conjecture.

\begin{theorem}\label{conditional Goldbach}
Let $\mathbb{B}\subset \mathbb{N}$ such that $\mathbb{P}\subset \mathbb{B}$ with $|\mathcal{C}(n,\mathbb{B})|=|\mathbb{B}\cap \mathbb{N}_n|$ for all $n\in \mathbb{N}$ so that $$\lim \limits_{n\longrightarrow  \infty}\frac{|\mathbb{P}\cap \mathbb{N}_n|}{\eta(n)}>\frac{1}{2}$$ where $\eta(n)$ is the counting function of all integers belonging to the set $\mathbb{B}\cap \mathbb{N}_n$. Then $\nu(n,\mathbb{P})>0$ for all sufficiently large values of $n\in 2\mathbb{N}$.
\end{theorem}

\begin{proof}
First let us upper and lower bound the density of points in the CoP $\mathcal{C}(n,\mathbb{B})$ belonging to the set of the primes $\mathbb{P}$ so that under the condition $|\mathcal{C}(n,\mathbb{B})|=|\mathbb{B}\cap \mathbb{N}_n|$ for all $n\in \mathbb{N}$, we obtain the inequality
\begin{align}
\lim \limits_{n\longrightarrow \infty}\frac{\left \lfloor \frac{|\mathbb{P}\cap \mathbb{N}_n|}{2}\right \rfloor}{\left \lfloor \frac{\eta(n)-1}{2}\right \rfloor}\leq \mathcal{D}(\mathbb{P}_{\mathcal{C}(\infty,\mathbb{B})})\leq \lim \limits_{n\longrightarrow \infty}\frac{|\mathbb{P}\cap \mathbb{N}_n|}{\left \lfloor \frac{\eta(n)-1}{2}\right \rfloor}=2\lim \limits_{n\longrightarrow  \infty}\frac{|\mathbb{P}\cap \mathbb{N}_n|}{\eta(n)}.\nonumber
\end{align}Appealing to the uniqueness of the axes of CoPs, we can write
\begin{align}
\# \left \{\mathbb{L}_{[x],[y]}~ \hat{\in}~  \mathcal{C}(n,\mathbb{B})|~\{x,y\}\cap \mathbb{P}\neq \emptyset \right \}&=\nu(n,\mathbb{P})+\# \left \{\mathbb{L}_{[x],[y]}~ \hat{\in}~  \mathcal{C}(n,\mathbb{B})|~x\in \mathbb{P},~y\in \mathbb{B}\setminus \mathbb{P}\right \}.\nonumber
\end{align}Let us assume to the contrary $\nu(n,\mathbb{P})=0$, then it follows that no two points in the CoP $\mathcal{C}(n,\mathbb{B})$ with weight in the set $\mathbb{P}$ are axes partners, so that under the requirement $$\lim \limits_{n\longrightarrow  \infty}\frac{|\mathbb{P}\cap \mathbb{N}_n|}{\eta(n)}>\frac{1}{2}$$ where $\eta(n)$ is the counting function of all integers belonging to the set $\mathbb{B}\cap \mathbb{N}_n$, we obtain the inequality 
\begin{align}
    \mathcal{D}(\mathbb{P}_{\mathcal{C}(\infty,\mathbb{B})})&=2\lim \limits_{n\longrightarrow  \infty}\frac{|\mathbb{P}\cap \mathbb{N}_n|}{\eta(n)}\nonumber \\&>2\times \frac{1}{2}=1.\nonumber
\end{align}This contradicts the inequality $\mathcal{D}(\mathbb{P}_{\mathcal{C}(\infty,\mathbb{B})})\leq 1$ in Proposition \ref{propertydensity}. This proves that $\nu(n,\mathbb{P})>0$ for all sufficiently large values of $n\in 2\mathbb{N}$.
\end{proof}

\subsection{\textbf{Binary Goldbach conjecture proof technique via circles of partition}}
In this subsection we propose series of steps that could be taken to confirm the truth of the binary Goldbach conjecture. We enumerate the strategies chronologically as follows:

\begin{enumerate}
\item [$\bullet$] First construct a subset of the integers $\mathbb{B}$ that covers the primes with $|\mathcal{C}(n,\mathbb{B})|=|\mathbb{B}\cap \mathbb{N}_n|$ for all $n\in \mathbb{N}$ so that $$\lim \limits_{n\longrightarrow  \infty}\frac{|\mathbb{P}\cap \mathbb{N}_n|}{\eta(n)}>\frac{1}{2}$$ where $\eta(n)$ is the counting function of all integers belonging to the set $\mathbb{B}\cap \mathbb{N}_n$.
\bigskip

\item [$\bullet$] Next we remark that the following inequality also hold and this can be obtain by replacing the set $\mathbb{N}$  with the set $\mathbb{B}$
\begin{align}
\lim \limits_{n\longrightarrow \infty}\frac{\left \lfloor \frac{|\mathbb{P}\cap \mathbb{N}_n|}{2}\right \rfloor}{\left \lfloor \frac{\eta(n)-1}{2}\right \rfloor}\leq \mathcal{D}(\mathbb{P}_{\mathcal{C}(\infty,\mathbb{B})})\leq \lim \limits_{n\longrightarrow \infty}\frac{|\mathbb{P}\cap \mathbb{N}_n|}{\left \lfloor \frac{\eta(n)-1}{2}\right \rfloor}=2\lim \limits_{n\longrightarrow  \infty}\frac{|\mathbb{P}\cap \mathbb{N}_n|}{\eta(n)}.\nonumber
\end{align}
\bigskip

\item [$\bullet$] Appealing to the uniqueness of the axes of CoPs, we can write  
\begin{align*}
    &\# \left \{\mathbb{L}_{[x],[y]}~ \hat{\in}~  \mathcal{C}(n,\mathbb{B})|~\{x,y\}\cap \mathbb{P}\neq \emptyset \right \}\\
    &=\nu(n,\mathbb{P})+\# \left \{\mathbb{L}_{[x],[y]}~ \hat{\in}~  \mathcal{C}(n,\mathbb{B})|~x\in \mathbb{P},~y\in \mathbb{B}\setminus \mathbb{P}\right \}.\nonumber
\end{align*}
\bigskip

\item [$\bullet$] Let us assume $\nu(n,\mathbb{P})=0$ then it follows by appealing to Definition \ref{pointdensity}
\begin{align}
    \mathcal{D}(\mathbb{P}_{\mathcal{C}(\infty,\mathbb{B})})&=2\lim \limits_{n\longrightarrow  \infty}\frac{|\mathbb{P}\cap \mathbb{N}_n|}{\eta(n)}\nonumber \\&>2\times \frac{1}{2}=1.\nonumber
\end{align}This contradicts the inequality $\mathcal{D}(\mathbb{P}_{\mathcal{C}(\infty,\mathbb{B})})\leq 1$ in Proposition \ref{propertydensity}. This proves that $\nu(n,\mathbb{P})>0$ for all sufficiently large values of $n\in 2\mathbb{N}$.
\end{enumerate}

\section{\textbf{Open and Connected Circles of Partition}}
In this section we introduce the notion of open CoP. We first launch the notion of a path connecting CoP and examine in-depth the concept of connected CoPs and their interplay with other notions launched thus far. Also here and in the following sections only real axes are considered, the attribute \textit{real} is not used.

\begin{definition}
Let $\mathbb{M}\subseteq \mathbb{N}$ with $\mathcal{C}(n,\mathbb{M})\neq \emptyset$ and $\mathcal{C}(s,\mathbb{M})\neq \emptyset$ be any two distinct CoPs. 
Then by the path joining the CoP $\mathcal{C}(n,\mathbb{M})$ to the CoP $\mathcal{C}(s,\mathbb{M})$ we mean the line joining $[x]\in \mathcal{C}(n,\mathbb{M})$ to $[y]\in \mathcal{C}(s,\mathbb{M})$, denoted as $\mathcal{L}_{[x],[y]}$, such that $\mathcal{L}_{[x],[y]}$ is an axis of the CoP $\mathcal{C}(s,\mathbb{M})$
\[
\mathcal{L}_{[x],[y]}=\mathbb{L}_{[x],[y]}~ \hat{\in}~ \mathcal{C}(s,\mathbb{M}).
\]
We say the CoP $\mathcal{C}(n,\mathbb{M})$ is connected to the CoP $\mathcal{C}(s,\mathbb{M})$ if there exists such a path.\\ 
We say the CoP $\mathcal{C}(n,\mathbb{M})$ is \textbf{strongly connected} to some CoP $\mathcal{C}(m,\mathbb{M})$ if the connection exists for all possible dilations
\begin{align*}
\delta_r:\mathcal{C}(n,\mathbb{M})\longrightarrow \mathcal{C}(s,\mathbb{M})
\mbox{ by }s=n+r.
\end{align*}
with $\delta_r([x])=[y]$.
We say the CoP $\mathcal{C}(n,\mathbb{\mathbb{M}})$ is fully connected to the CoP $\mathcal{C}(s,\mathbb{M})$ if there exists such a path for each $[x]\in \mathcal{C}(n,\mathbb{M})$.
\end{definition}

\begin{proposition}\label{P-common_points}
Let $\mathbb{M}\subseteq \mathbb{N}$ with $\mathcal{C}(n,\mathbb{M})\neq \emptyset$ and $\mathcal{C}(s,\mathbb{M})\neq \emptyset$ be any two distinct CoPs with a common point $[x]$. Then and only then $\mathcal{C}(n,\mathbb{M})$ is connected to $\mathcal{C}(s,\mathbb{M})$.
\end{proposition}
\begin{proof}
Since $[x]\in\mathcal{C}(s,\mathbb{M})$ there must be an axis $\mathbb{L}_{[x],[s-x]}~ \hat{\in}~ \mathcal{C}(s,\mathbb{M})$. Since $[x]\in\mathcal{C}(n,\mathbb{M})$ there exists the path $\mathcal{L}_{[x],[s-x]}$. Hence $\mathcal{C}(n,\mathbb{M})$ is connected to $\mathcal{C}(s,\mathbb{M})$. \\
If otherwise there exists such a path $\mathcal{L}_{[x],[y]}$ with
a fixed $[x]\in\mathcal{C}(n,\mathbb{M})$ and
any $[y]\in\mathcal{C}(s,\mathbb{M})$ such that $\mathbb{L}_{[x],[y]}~ \hat{\in}~ \mathcal{C}(s,\mathbb{M})$ then it must certainly be that $[y]=[s-x]$ and $[x]$ is also a point of $\mathcal{C}(n,\mathbb{M})$. 
\end{proof}
\bigskip

\begin{proposition}
Let $\mathbb{M}\subseteq \mathbb{N}$ and $\mathcal{C}(n,\mathbb{M})$ be a CoP. If $\mathcal{C}(n,\mathbb{M})$ is fully connected to $\mathcal{C}(s,\mathbb{M})$ then 
\begin{align}
     \mathcal{C}(n,\mathbb{M})\subseteq \mathcal{C}(s,\mathbb{M}).\nonumber
\end{align}
\end{proposition}

\begin{proof}
Let $\mathbb{M}\subseteq \mathbb{N}$ and suppose the CoP $\mathcal{C}(n,\mathbb{M})$ is connected to the CoP $\mathcal{C}(s,\mathbb{M})$ then for each point $[x]\in \mathcal{C}(n,\mathbb{M})$ there exists an axis $\mathbb{L}_{[x],[y]}~ \hat{\in}~  \mathcal{C}(s,\mathbb{M})$ for some $[y]\in \mathcal{C}(s,\mathbb{M})$. It follows that $[x]\in \mathcal{C}(s,\mathbb{M})$, thereby ending the proof since the point $[x]$ is an arbitrary point in the CoP $\mathcal{C}(n,\mathbb{M})$.
\end{proof}
\bigskip

\begin{theorem}\label{connectexistence}
Let $\mathbb{M}\subseteq\mathbb{N}$ and $\mathcal{C}(n,\mathbb{M})$ be any CoP admits aligned embedding. Then $\mathcal{C}(n,\mathbb{M})$ is strongly connected to some CoP $\mathcal{C}(m,\mathbb{M})$ admits aligned embedding.
\end{theorem}


\begin{proof}
We assume that  $\mathcal{C}(n,\mathbb{M})$ is not strongly connected to any $\mathcal{C}(m,\mathbb{M})$, by virtue of the definition. Invoking the virtue the CoPs admit aligned embedding, we can assume $\mathcal{C}(n,\mathbb{M})\subset \mathcal{C}(s,\mathbb{M})$. The line $\mathbb{L}_{[x],[n-x]}$ is an axis of $\mathcal{C}(n,\mathbb{M})$ for any $[x]\in \mathcal{C}(n,\mathbb{M})$. 
It follows that $\mathbb{L}_{[x],[s-x]}$ is also an axis of the CoP $\mathcal{C}(s,\mathbb{M})$. 
Since no two CoPs are strongly connected and because of Theorem \ref{existence} there exists some dilation $\delta_{r_1}:\mathcal{C}(n,\mathbb{M})\longrightarrow \mathcal{C}(s,\mathbb{M})$ such that $[s-x]\neq \delta_{r_1}([x])$ for each $[x]\in \mathcal{C}(n,\mathbb{M})$. 
Let us produce a line $\mathcal{L}_{[x],[\delta_{r_1}([x])]}$ by joining $[x]$ to $\delta_{r_1}([x])$. Now, we can certainly partition these lines as axes of large and small CoPs relative to the CoP $\mathcal{C}(s)$ as below
\begin{align}
\lbrace\mathbb{L}_{[x],\delta_{r_1}([x])}~ \hat{\in}~  \mathcal{C}(v,\mathbb{M})|~n<v\leq s-1\rbrace\bigcup  \lbrace\mathbb{L}_{[x],\delta_{r_1}([x])}~ \hat{\in}~  \mathcal{C}(k,\mathbb{M})|~k>s\rbrace.\nonumber
\end{align}
Let us now pick arbitrarily a small CoP relative to the CoP $\mathcal{C}(s,\mathbb{M})$ and large relative to the CoP $\mathcal{C}(n,\mathbb{M})$. That is we pick a CoP $\mathcal{C}(v,\mathbb{M})$ from the first set arbitrarily. Then we obtain the strict embedding
\begin{align}
    \mathcal{C}(n,\mathbb{M})\subset \mathcal{C}(v,\mathbb{M})\subset \mathcal{C}(s,\mathbb{M}).\nonumber
\end{align}
Otherwise the CoP $\mathcal{C}(n,\mathbb{M})$ will have the axis $\mathbb{L}_{[x],[\delta_{r_1}([x])]}$, which will contradict our assumption. Under the assumption that no two CoPs are strongly connected, it follows that there exist some dilation
\begin{align}
    \delta_{r_2}:\mathcal{C}(n,\mathbb{M})\longrightarrow \mathcal{C}(v,\mathbb{M})\nonumber
\end{align}
such that for each $[x]\in \mathcal{C}(n,\mathbb{M})$ then $\delta_{r_2}([x])\neq [v-x]$. By repeating the argument in this manner under the assumption that no two CoPs are connected we obtain the following infinite embedding into the CoP $\mathcal{C}(n,\mathbb{M})$ as follows
\begin{align}
    \mathcal{C}(n,\mathbb{M})\subset \cdots \subset \mathcal{C}(t,\mathbb{M})\subset \mathcal{C}(v,\mathbb{M})\subset \mathcal{C}(s,\mathbb{M})\nonumber
\end{align}
and we have the following infinite descending sequence of generators toward the generator $n$
\begin{align}
    n<\cdots <t<v<s.\nonumber
\end{align}
This is absurd, thereby ending the proof of the claim.
\end{proof}
\bigskip

\begin{corollary}
Let $\mathcal{C}(n,\mathbb{M})$ and $\mathcal{C}(n,\mathbb{M})$  be two CoPs admit aligned embedding. If holds $n<m$
then $\mathcal{C}(n,\mathbb{M})$ is fully connected to the CoP $\mathcal{C}(m,\mathbb{M})$.
\end{corollary}

\begin{proof}
Due to Theorem \ref{properties2} holds
\[
\mathcal{C}(n,\mathbb{M})\subset\mathcal{C}(m,\mathbb{M}).
\]
Hence each point of $\mathcal{C}(n,\mathbb{M})$ is also a point of $\mathcal{C}(m,\mathbb{M})$. Because of Proposition \ref{P-common_points}, it follows that $\mathcal{C}(n,\mathbb{M})$ is connected to $\mathcal{C}(m,\mathbb{M})$ for each point $[x]\in\mathcal{C}(n,\mathbb{M})$. Hence $\mathcal{C}(n,\mathbb{M})$ is fully connected to $\mathcal{C}(m,\mathbb{M})$
\end{proof}

One could imagine an analogous results of fully connected CoPs if we take the base set $\mathbb{M}$ to be the set $\mathbb{P}$ of prime numbers. There things become a lot more complicated and will require a careful analysis of the situation. But this endeavour is not far from reach by examining additional concept that are somewhat on par with the subtleties and properties of these sequence. To that end, we make the following conjecture.

\begin{conjecture}\label{prime connectedness}
There are infinitely many pairs of fully connected CoPs of the form $\mathcal{C}(n,\mathbb{P})$.
\end{conjecture}

\begin{definition}
Let $\mathbb{M}\subseteq \mathbb{N}$ and $\mathcal{C}(n,\mathbb{M})$ be a CoP with $\mathbb{L}_{[x],[y]}~ \hat{\in}~  \mathcal{C}(n,\mathbb{M})$. 
Then by the open CoP induced by the point $[x],[y]$, we mean the exclusion $\mathcal{C}(n,\mathbb{M})\setminus [x],[y]$. 
We call the points $[x],[y]$ the gates to the interior of the open CoP. We denote the induced open CoP by $\widehat{\mathcal{C}(n,\mathbb{M})_{[x],[y]}}\subset \mathcal{C}(n,\mathbb{M})$. 
We say the CoPs $\mathcal{C}(s,\mathbb{M})$ and $\mathcal{C}(n,\mathbb{M})$ forms a two-member community if and only if there is a path joining the gate $[x],[y]$ of $\widehat{\mathcal{C}(n,\mathbb{M})_{[x],[y]}}$ to the CoP $\mathcal{C}(s,\mathbb{M})$.
\end{definition}


\section{\textbf{Children, Offspring and Family Induced by Circles of Partition}}
In this section we introduce the notion of children, the offspring and the family induced by a typical CoP. We relate this notion to the notion of connected CoPs.

\begin{definition}
Let $\mathbb{M}\subseteq \mathbb{N}$ and $\mathcal{C}(n,\mathbb{M})\neq \emptyset$ and let $\left \{\mathbb{L}_{[u_i],[v_i]}\right \}_{i=1}^{N;N\geq 2}$ for some $N\geq 2$ be the set of all the axes. Then we say the CoP $\mathcal{C}(s,\mathbb{M})$ is a \textbf{child} of the CoP $\mathcal{C}(n,\mathbb{M})$ if there exist some axes $\mathbb{L}_{[u_k],[v_k]},\mathbb{L}_{[u_j],[v_j]}\in \left \{\mathbb{L}_{[u_i],[v_i]}\right \}_{i=1}^{N;N\geq 2}$ such that at least one of  $\mathbb{L}_{[u_k],[u_j]},\mathbb{L}_{[u_k],[v_j]}, \mathbb{L}_{[v_k],[u_j]},\mathbb{L}_{[v_k],[v_j]}$ is an axis of the child CoP $\mathcal{C}(s,\mathbb{M})$. This axis forms the \textbf{principal axis} of the child CoP.
We call the collection of all CoPs generated in this manner the \textbf{offspring} of the \textbf{parent} CoP $\mathcal{C}(n,\mathbb{M})$. The parent CoP $\mathcal{C}(n,\mathbb{M})$ together with its offspring forms a \textbf{complete family} of CoPs. The size of the family of CoPs is the number of CoPs in the family. A subset of a family is said to be an \textbf{incomplete family} of CoPs.
\end{definition}

\begin{example}
Let us consider the CoP with $\Vert\mathcal{C}(20,\mathbb{P})\Vert=\left \{3,7,13,17\right \}$ with axes $\mathbb{L}_{[3],[17]}$ and $\mathbb{L}_{[7],[13]}$. We consider the following chords $\mathcal{L}_{[3],[7]}, \mathcal{L}_{[3],[13]},\mathcal{L}_{[7],[17]},\mathcal{L}_{[13],[17]}$. These chords correspond as principal axes to the following CoPs
\begin{align}
\mathcal{C}(10,\mathbb{P}),~\mathcal{C}(16,\mathbb{P}),~\mathcal{C}(24,\mathbb{P}),~\mathcal{C}(30,\mathbb{P}).\nonumber
\end{align}
Hence we obtain a complete family of CoPs of size $5$.
\end{example}

\begin{proposition}\label{P-child}
Let $\mathcal{C}(n,\mathbb{M})$ a non-empty CoP. Then each axis point $[x]$ together with a point $[u]$ of another axis of $\mathcal{C}(n,\mathbb{M})$ generates a child $\mathcal{C}(s,\mathbb{M})$ of the parent $\mathcal{C}(n,\mathbb{M})$ with $s=\Vert[x]\Vert+\Vert[u]\Vert$.
\end{proposition}
\begin{proof}
Let $\mathbb{L}_{[x],[y]}$ and $\mathbb{L}_{[u],[v]}$ be two axes of $\mathcal{C}(n,\mathbb{M})$. Appealing to Proposition \ref{unique}, we have 
\[
\Vert[x]\Vert+\Vert[u]\Vert=s\neq n.
\]
Hence $[x]$ and $[u]$ form the axis $\mathbb{L}_{[x],[v]}~ \hat{\in}~ \mathcal{C}(s,\mathbb{M})$ and $\mathcal{C}(s,\mathbb{M})$ is a child of $\mathcal{C}(n,\mathbb{M})$.
\end{proof}

\begin{proposition}\label{P-infinite_Mad}
Let $n\in \mathbb{N}$, $\mathbb{M}\subseteq\mathbb{N}$ and $\mathcal{C}(n,\mathbb{M})$ be a CoP admitting aligned embedding.
If holds $\vert\mathcal{C}(n,\mathbb{M})\vert\geq 4$
then the CoP $\mathcal{C}(n,\mathbb{M})$ admits an infinite chain of its descendants.
\end{proposition}
\begin{proof}
Due to $\vert\mathcal{C}(n,\mathbb{M})\vert\geq 4$ there is an axis point $[u]\in\mathcal{C}(n,\mathbb{M})$ with 
\[
u>\min\left(\Vert[w]\Vert\mid [w]\in\mathcal{C}(n,\mathbb{M})\right)
\] 
and a point $[v]\in\mathcal{C}(n,\mathbb{M})$ of another axis with $u+v=m>n$. It follows that there exists an axis $\mathbb{L}_{[u],[v]}~ \hat{\in}~ \mathcal{C}(m,\mathbb{M})$. Ergo holds $[u]\in\mathcal{C}(m,\mathbb{M})$. Appealing to Proposition \ref{P-child} the CoP $\mathcal{C}(m,\mathbb{M})$ is a child of the CoP $\mathcal{C}(n,\mathbb{M})$.
Since $m>n$ and $\mathcal{C}(n,\mathbb{M})$ admits aligned embedding, it  
holds 
\[
\mathcal{C}(n,\mathbb{M})\subset\mathcal{C}(m,\mathbb{M}).
\]
Now we choose a point $[w]$ of $\mathcal{C}(m,\mathbb{M})$ and the latter changes its role to be a parent.
With the same procedure as above we produce an axis $\mathbb{L}_{[u],[w]}~ \hat{\in}~ \mathcal{C}(r,\mathbb{M})$ with $[u],[w]\in\mathcal{C}(r,\mathbb{M})$ and
\[
\mathcal{C}(n,\mathbb{M})\subset\mathcal{C}(m,\mathbb{M})
\subset\mathcal{C}(r,\mathbb{M}).
\]
This procedure can be repeated infinitely many often.  We obtain an infinite chain of descendants of the CoP $\mathcal{C}(n,\mathbb{M})$ as its prime father.
\end{proof}

\begin{proposition}\label{breaker}
Let $\mathbb{M}\subseteq \mathbb{N}$ and $\mathcal{C}(n,\mathbb{M})$ be a parent of a complete family. Then $\mathcal{C}(n,\mathbb{M})$ partitions the offspring into two incomplete families of equal sizes.
\end{proposition}

\begin{proof}
In virtue of Proposition \ref{P-child} two points of distinct axes of the CoP $\mathcal{C}(n,\mathbb{M})$ generates a child of it. Let 
\[
\mathbb{L}_{[u],[v]}, \mathbb{L}_{[x],[y]}\mid u<x
\]
two arbitrary axes of $\mathcal{C}(n,\mathbb{M})$. Because $[u],[v]$ and $[x],[y]$ are axis points holds
\begin{align*}
&n=u+v=x+y\mbox{ and therefore}\\
&v=x-u+y\mbox{ and because of }x>u\\
&v>y
\end{align*}
Hence we get
\begin{align*}
&u<x<y<v\mbox{ and therefore}\\
&s_1:=u+x<s_2:=u+y<n=x+y\mbox{ and}\\
&t_1:=v+y>t_2:=v+x>n=v+u
\end{align*}
and a chain of children
\begin{align*}
\mathcal{C}(s_1,\mathbb{M}),\mathcal{C}(s_2,\mathbb{M})
&,\mathcal{C}(n,\mathbb{M}),
\mathcal{C}(t_2,\mathbb{M}),\mathcal{C}(t_1,\mathbb{M})
\mbox{ with}\\
s_1<s_2&<n<t_2<t_1.
\end{align*}
Therefore holds that for all two axes $4$ children are generated, two on the left side of $\mathcal{C}(n,\mathbb{M})$ and two on the right side in a chain of children. Because $\mathcal{C}(n,\mathbb{M})$ for all two axes is located in the middle of the chain, the parent CoP $\mathcal{C}(n,\mathbb{M})$ partitions its offspring in two halves, the incomplete families of equal sizes.   
\end{proof}

\begin{proposition}\label{P-child-embedding}
If the parent CoP admits embedding then their children admit aligned embedding.
\end{proposition}
\begin{proof}
We look at the last proof and choose $[u]$ as the first point of the parent CoP $\mathcal{C}(n,\mathbb{M})$
\[
u:=\min \left(w\in\Vert\mathcal{C}(n,\mathbb{M})\Vert\right).
\]
Then holds
\begin{align*}
&[u]\in\mathcal{C}(s_1,\mathbb{M})\mbox{ and }
[u]\in\mathcal{C}(s_2,\mathbb{M})\mbox{ and}\\
&\max\left(w\in\Vert\mathcal{C}(s_1,\mathbb{M})\Vert\right)=x<y
=\max\left(w\in\Vert\mathcal{C}(s_2,\mathbb{M})\Vert\right)\\
&\mbox{and hence}\\
&\mathcal{C}(s_1,\mathbb{M})\subset\mathcal{C}(s_2,\mathbb{M})
\mbox{ under }s_1<s_2.
\end{align*}
Because $\mathcal{C}(n,\mathbb{M})$ admits embedding holds
\begin{align*}
\mathcal{C}(s_1,\mathbb{M})\subset\mathcal{C}(s_2,\mathbb{M})
&\subset \mathcal{C}(n,\mathbb{M})\subset\mathcal{C}(t_2,\mathbb{M})
\subset\mathcal{C}(t_1,\mathbb{M})\mbox{ under}\\
s_1<s_2&<n<t_2<t_1.
\end{align*}
\end{proof}

\begin{corollary}\label{C-children-chain}
If the CoP $\mathcal{C}(n,\mathbb{M})$ admits embedding and has $2k$ children, then it follows
by virtue of Propositions \ref{breaker} and \ref{P-child-embedding} for its complete family
\[
\mathcal{C}(s_1,\mathbb{M})\subset\ldots\subset\mathcal{C}(s_k,\mathbb{M})
\subset\mathcal{C}(n,\mathbb{M})\subset
\mathcal{C}(t_k,\mathbb{M})\subset\ldots\subset\mathcal{C}(t_1,\mathbb{M})
\]
and we have the following symmetry
\[
s_2-s_1=t_1-t_2,\ldots,s_k-s_{k-1}=t_{k-1}-t_k,n-s_k=t_k-n.
\]
\end{corollary}
\begin{proof}
The embedding chain is a direct consequence of the Propositions \ref{breaker} and \ref{P-child-embedding}.\\
Now we prove the symmetry of the differences of the children generators.
We look again at the proof of Proposition \ref{breaker} with
\begin{align*}
&u<x<y<v\mbox{ and therefore}\\
&s_1:=u+x<s_2:=u+y<n=x+y\mbox{ and}\\
&t_1:=v+y>t_2:=v+x>n=v+u
\end{align*}
for two arbitrary axes $\mathbb{L}_{[u],[v]},\mathbb{L}_{[x],[y]}$ of $\mathcal{C}(n,\mathbb{M})$. Then is
\[
s_1<s_2<n<t_2<t_1
\]
and we get
\begin{align*}
s_2-s_1=u+y-u-x&=\mathbf{y-x}\mbox{ and }n-s_2=x+y-u-y=\mathbf{x-u}\\
t_1-t_2=v+y-v-x&=\mathbf{y-x}\mbox{ and }t_2-n=v+x-v-u=\mathbf{x-u}.
\end{align*}
Because the two axes are arbitrary this symmetry around the generator $n$ holds for all axes. From this follows the claim.
\end{proof}

\begin{theorem}\label{T-numberofchildren}
Let $\mathbb{M}\subseteq \mathbb{N}$ and $\mathcal{C}(n,\mathbb{M})$ be a CoP with $|\mathcal{C}(n,\mathbb{M})|=k$. Then the number of children in the family with parent $\mathcal{C}(n,\mathbb{M})$ satisfies the upper bound
\begin{align}
    \leq 2\left \lfloor\frac{k}{2}\right \rfloor \bigg(\left \lfloor \frac{k}{2}\right \rfloor-1\bigg)\nonumber
\end{align}
and the lower bound
\begin{align*}
\geq 2\left(n_a-2\right)=4\left(\left\lfloor\frac{k}{2}\right\rfloor-1\right)
\mbox{ with }n_a=2\left\lfloor\frac{k}{2}\right\rfloor.
\end{align*}
\end{theorem}

\begin{proof}
At first we prove the upper bound.
The CoP  $\mathcal{C}(n,\mathbb{M})$ with $|\mathcal{C}(n,\mathbb{M})|=k$ contains $\lfloor \frac{k}{2}\rfloor$ different axes. Each axis contains two points of the parent $\mathcal{C}(n,\mathbb{M})$ and  determines children with at most $\lfloor \frac{k}{2}\rfloor-1$ number of axes. The upper bound follows from this counting argument.\\
Now we prove the lower bound. In virtue of Corollary \ref{ordered_weights} the weights of the points of $\mathcal{C}(n,\mathbb{M})$ are strictly totally ordered. Now we remove from this sequence the weight of the center if it exists. It remains $n_a=2\left\lfloor\frac{k}{2}\right\rfloor$ weights. We enumerate them as
\[
x_1<x_2<\ldots<x_{n_a-1}<x_{n_a}
\]
and form the following sequences
\begin{align*}
s_1:=x_1+x_2<s_2:=x_1+x_3<&\ldots<s_{n_a-2}:=x_1+x_{n_a-1}<x_1+x_{n_a}=n\\
&\mbox{and}\\
t_1:=x_{n_a}+x_{n_a-1}>t_2:=x_{n_a}+x_{n_a-2}>&\ldots
>t_{n_a-2}:=x_{na}+x_2>x_{n_a}+x_1=n.
\end{align*}
Hence we obtain
\[
s_1<\ldots<s_{n_a-2}<n<t_{n_a-2}<\ldots<t_1
\]
and have at least $2(n_a-2)$ different generators for children 
of $\mathcal{C}(n,\mathbb{M})$.
\end{proof}

\begin{remark}
We observe that if a CoP contains not more than $3$ points then the CoP has no children. We call these CoPs \textbf{childless}. And if a CoP has two axes then the CoP has $4$ children. Therefore there are no CoPs with only one child or only two or three children.
\end{remark}

\begin{proposition}\label{P-notallMchildless}
Let $\mathbb{M}\subseteq\mathbb{N}$. There is no parent CoP $\mathcal{C}(n,\mathbb{M})$ admitting embedding with $\vert\mathcal{C}(n,\mathbb{M})\vert\geq 4$ such that all its children are childless.
\end{proposition}
\begin{proof}
Because of Proposition \ref{P-child-embedding} the children admit aligned embedding. And since the parent CoP has at least $4$ children there are because of Proposition \ref{breaker} at least $2$ children with generators $>n$. Because the children admit aligned embedding then holds for a child $\mathcal{C}(s,\mathbb{M})$ with $s>n$
\[
\mathcal{C}(s,\mathbb{M})\supset\mathcal{C}(n,\mathbb{M})
\mbox{ and therefore }
\vert\mathcal{C}(s,\mathbb{M})\vert >\vert\mathcal{C}(n,\mathbb{M})\vert
\geq 4.
\]
Hence there are at least two children with more than $4$ own children and hence not childless.
\end{proof}
\bigskip


\begin{remark}
Next we launch an important result that will certainly have significant offshoots throughout our studies. Very roughly, it tells us that we can always partition any complete family into incomplete families with equal dilation between the members.
\end{remark}

\begin{lemma}[Regularity lemma]\label{regular}
The offspring of a CoP $\mathcal{C}(n,\mathbb{M})$ can be partitioned into incomplete families with equal scale dilation between sequence of successive embedding.
\end{lemma}

\begin{proof}
If there exist no embedding among the children of the parent $\mathcal{C}(n,\mathbb{M})$, then obviously we have a partition into a one member incomplete family and the dilation in each family is trivial.  Let us assume $\mathcal{C}(s_1,\mathbb{M})\subset \mathcal{C}(s_2,\mathbb{M})\subset \cdots \subset \mathcal{C}(s_k,\mathbb{M})$ for $k\geq 2$ be a sequence of children of the  parent $\mathcal{C}(n,\mathbb{M})$ with equal scale dilation between successive embedding.
If the sequence is all of the children of the parent $\mathcal{C}(n,\mathbb{M})$ then the parent must be inserted
in virtue of Corollary \ref{C-children-chain} in the middle of the offset chain.
Now let us remove from the chain the parent $\mathcal{C}(n,\mathbb{M})$ with the two closest children. Then we obtain a partition of collection of children in the embedding into two sub-chains of embedding with equal scale dilation between successive children, those to the left of the children closest to the parent $\mathcal{C}(n,\mathbb{M})$ and to the right of the children closest to the parent $\mathcal{C}(n,\mathbb{M})$. For the sequence removed from the a priori sequence of children given below
\begin{align}
\mathcal{C}(s_i,\mathbb{M})\subset \mathcal{C}(n,\mathbb{M})\subset  \mathcal{C}(s_{i+1},\mathbb{M})\nonumber
\end{align}
we remove the parent $\mathcal{C}(n,\mathbb{M})$ and we obtain a third partition of offspring with equal scale dilation
\begin{align}
\mathcal{C}(s_i,\mathbb{M})\subset \mathcal{C}(s_{i+1},\mathbb{M}).\nonumber
\end{align}
For the case where not all children are contained in the a priori embedding,  then we have already obtained a partition of collection of children into an incomplete family with equal scale dilation between successive members. The remaining collection of children can also be partitioned into incomplete families by choosing  an embedding with equal scale dilation between successive children.
\end{proof}


\begin{theorem}\label{connectoffspring}
The number of pairs of connected children in any complete family is lower bounded by
\[
\geq \frac{n_a(n_a-2)(n_a-3)}{2}=
2\left\lfloor\frac{k}{2}\right\rfloor\left(\left\lfloor\frac{k}{2}\right\rfloor-1\right)(n_a-3)
\geq 2\left\lfloor\frac{k}{2}\right\rfloor\left(\left\lfloor\frac{k}{2}\right\rfloor-1\right)
\]
if the parent CoP has $n_a$ axis points and $n_a=2\left\lfloor\frac{k}{2}\right\rfloor>3$.
\end{theorem}
\begin{proof}
In virtue of Proposition \ref{P-common_points} two CoPs are connected if and only if they have a common point. And the children are generated by pairs of points on different axes.
Each such point $[x]$ of the parent CoP occurs therefore in $n_a-2$ children at least. Hence there are $\frac{(n_a-2)(n_a-3)}{2}$ pairs of children containing the point $[x]$. There are $n_a$ axis points. Therefore this number of pairs must be multiplied by $n_a$. This results the formula of the lower bound.\\
\end{proof}

In comparison with Theorem \ref{T-numberofchildren} we observe that the number of pairs of connected children of a complete family is always greater or equal to the number of its children. From the proof of Theorem \ref{connectoffspring} we see that each child is connected with another child of the same family.

\begin{example}
We take as parent CoP\\
$\mathcal{C}(22,\mathbb{P})=\lbrace [3],[5],[11],[17],[19]\rbrace\rightarrow k=5,n_a=4$.
In virtue of Theorem 9.1. it has maximal
\[
2\cdot 2\cdot 1=4
\]
children and in virtue of Theorem \ref{connectoffspring} at least
\[
\frac{4\cdot 2 \cdot 1}{2}=4
\]
pairs of connected children. As children we get
\begin{align*}
\mathcal{C}(8,\mathbb{P})&=\lbrace\textbf{[3]},\textbf{[5]}\rbrace\\
\mathcal{C}(20,\mathbb{P})&=\lbrace\textbf{[3]},[7],[13],\textbf{[17]}\rbrace\\
----&----------\\
\mathcal{C}(24,\mathbb{P})&=\lbrace\textbf{[5]},[7],[11],[13],[17],\textbf{[19]}\rbrace\\
\mathcal{C}(36,\mathbb{P})&=\lbrace[5],[7],[13],\textbf{[17]},\textbf{[19]},[23],[29],[31]\rbrace.
\end{align*}
We see that $[3]$ occurs in the children $2$ times. With it there is $1$ pair of children containing the point $[3]$. $[5]$ occurs $3$ times and is hence contained in $3$ pairs. $[17]$ occurs $3$ times too and $[19]$ occurs $2$ times and is contained in $1$ pair. All together we have $8>6$ pairs of connected children with respect to the points of the parent CoP. But we see that more than these points are common points in the offset. Hence there are $6$ further pairs of connected children.
The principal axes are marked as boldface.\\
The CoP $\mathcal{C}(24,\mathbb{P})$ contains $6$ axis points and has therefore at most $12$ children with $36$ pairs of connected children at least.
\end{example}

\begin{theorem}\label{T-nonembedding}
Let $\mathbb{M}\subseteq\mathbb{N}$ and $\mathcal{C}(n,\mathbb{M})$,  $\mathcal{C}(m,\mathbb{M})$ be two CoPs and $\hat{\mathcal{O}}_n$,  $\hat{\mathcal{O}}_m$ their complete families. If
$\vert\hat{\mathcal{O}}_n\vert<\vert\hat{\mathcal{O}}_m\vert$ and 
there exists a child $\mathcal{C}(s,\mathbb{M})\in\hat{\mathcal{O}}_n$ with  $\mathcal{C}(s,\mathbb{M})\not\in\hat{\mathcal{O}}_m$ then holds 
\[
\mathcal{C}(n,\mathbb{M})\not\subset\mathcal{C}(m,\mathbb{M})
\mbox{ and }
\mathcal{C}(n,\mathbb{M})\not\supset\mathcal{C}(m,\mathbb{M}),
\]
which means that these $\mathcal{C}(n,\mathbb{M})$ and $\mathcal{C}(m,\mathbb{M})$ not admit embedding.
\end{theorem}
\begin{proof}
Due to $\mathcal{C}(s,\mathbb{M})$ is a child of $\mathcal{C}(n,\mathbb{M})$ there are two points $[x],[u]\in\mathcal{C}(n,\mathbb{M})$ with $x+u=s$. Then is $\mathbb{L}_{[x],[u]}$ the principal axis of $\mathcal{C}(s,\mathbb{M})$. And due to $\mathcal{C}(s,\mathbb{M})$ is not a child of $\mathcal{C}(m,\mathbb{M})$ there are no points in $\mathcal{C}(m,\mathbb{M})$ with a weight sum equals $s$. Therefore the points $[x],[u]$ belong not to $\mathcal{C}(m,\mathbb{M})$. Hence holds $\mathcal{C}(n,\mathbb{M})\not\subset\mathcal{C}(m,\mathbb{M})$.\\
Because of $\vert\hat{\mathcal{O}}_n\vert<\vert\hat{\mathcal{O}}_m\vert$ there is a child of $\mathcal{C}(m,\mathbb{M})$ which is not a child of $\mathcal{C}(n,\mathbb{M})$. Therefore holds $\mathcal{C}(n,\mathbb{M})\not\supset\mathcal{C}(m,\mathbb{M})$.  
\end{proof}


\begin{theorem}\label{T-embedding_child}
Let $\mathbb{M}\subseteq\mathbb{N}$ and $\mathcal{C}(n,\mathbb{M})$ and $\mathcal{C}(m,\mathbb{M})$ be two CoPs admitting aligned embedding.
Without loss of generality we assume
\[
\mathcal{C}(n,\mathbb{M})\subset\mathcal{C}(m,\mathbb{M}).
\]
Then $\mathcal{C}(n,\mathbb{M})$ is a child of $\mathcal{C}(m,\mathbb{M})$.
If there is a chord $\mathcal{L}_{[x],[y]}$ of $\mathcal{C}(n,\mathbb{M})$ with $x+y=m$
then $\mathcal{C}(m,\mathbb{M})$ is also a child of $\mathcal{C}(n,\mathbb{M})$.
Additionally the complete family $\hat{\mathcal{O}}_n$ of $\mathcal{C}(n,\mathbb{M})$ is a subset of the complete family $\hat{\mathcal{O}}_m$ of $\mathcal{C}(m,\mathbb{M})$.   
\end{theorem}

\begin{proof}
Due to $\mathcal{C}(n,\mathbb{M})\subset\mathcal{C}(m,\mathbb{M})$ by virtue of Definition \ref{aligned_embedding} hold $n<m$ and 
\[
\min\left(x\mid [x]\in\mathcal{C}(n,\mathbb{M})\right)=
\min\left(u\mid [u]\in\mathcal{C}(m,\mathbb{M})\right).
\]
All chords $\mathcal{L}_{[x],[y]}$ of $\mathcal{C}(n,\mathbb{M})$ are also chords of $\mathcal{C}(m,\mathbb{M})$ excluding the chords between points $[x],[y]\in\mathcal{C}(n,\mathbb{M})$ with $x+y=m$. By exploiting the underlying embedding, we notice that all chords of $\mathcal{C}(m,\mathbb{M})$ which are axes of $\mathcal{C}(n,\mathbb{M})$  generate all the same child, the CoP $\mathcal{C}(n,\mathbb{M})$. Hence the CoP $\mathcal{C}(n,\mathbb{M})$ is a child of the CoP $\mathcal{C}(m,\mathbb{M})$, and if there is no chord $\mathcal{L}_{[x],[y]}$ of $\mathcal{C}(n,\mathbb{M})$ with $x+y=m$ then all children of $\mathcal{C}(n,\mathbb{M})$ are children of $\mathcal{C}(m,\mathbb{M})$ too. Hence the complete family $\hat{\mathcal{O}}_n$ is a subset of the complete family $\hat{\mathcal{O}}_m$ in this case.\\
If such a chord of $\mathcal{C}(n,\mathbb{M})$ exists then this chord is an axis of $\mathcal{C}(m,\mathbb{M})$, so that $\mathcal{C}(m,\mathbb{M})$ is a child of $\mathcal{C}(n,\mathbb{M})$. Because the parents belong to its complete family holds that the complete family $\hat{\mathcal{O}}_n$ is a subset of $\hat{\mathcal{O}}_m$ in this case too.
\end{proof}


\section{\textbf{Isomorphic Circles of Partition}}
In this section we introduce and study the notion of isomorphism between CoPs. 

\begin{definition}\label{D-isomor}
Let $\mathbb{M}\subseteq \mathbb{N}$ and let $\mathcal{C}(n,\mathbb{M})$ and $\mathcal{C}(m,\mathbb{M})$ be parents with the complete families $\hat{\mathcal{O}}_n$ and $\hat{\mathcal{O}}_m$, respectively. Then we say the parents $\mathcal{C}(n,\mathbb{M})$ and $\mathcal{C}(m,\mathbb{M})$ are isomorphic if 
\begin{align}
    \hat{\mathcal{O}}_m \cap \hat{\mathcal{O}}_n\neq \emptyset. \nonumber
\end{align}
We call the number $|\hat{\mathcal{O}}_m \cap \hat{\mathcal{O}}_n|$ the degree of isomorphism. We denote this isomorphism by $\mathcal{C}(n,\mathbb{M})\cong \mathcal{C}(m,\mathbb{M})$. We say the degree of isomorphism is high if at least one of the following equality holds
\begin{align}
    \frac{|\hat{\mathcal{O}}_m \cap \hat{\mathcal{O}}_n|}{|\hat{\mathcal{O}}_n|}=1\nonumber
\end{align}
or 
\begin{align}
     \frac{|\hat{\mathcal{O}}_m \cap \hat{\mathcal{O}}_n|}{|\hat{\mathcal{O}}_m|}=1.\nonumber
\end{align}
Otherwise, we say the degree of isomorphism is low.
\end{definition}
\bigskip

\begin{proposition}\label{connect isomorphism}
Let $\mathcal{C}(n,\mathbb{M})$ and $\mathcal{C}(m,\mathbb{M})$ be two CoPs. If $\mathcal{C}(n,\mathbb{M})$ is connected to $\mathcal{C}(m,\mathbb{M})$ by at least three distinct paths then $\mathcal{C}(n,\mathbb{M})\cong \mathcal{C}(m,\mathbb{M})$.
\end{proposition}

\begin{proof}
First let us assume the CoPs $\mathcal{C}(n,\mathbb{M})$ is connected to the CoP $\mathcal{C}(m,\mathbb{M})$ by at least three distinct paths. Then it follows that there exist some distinct points $[x],[y],[z]\in \mathcal{C}(m,\mathbb{M})$ such that $\mathbb{L}_{[x],[u]},\mathbb{L}_{[y],[v]},\mathbb{L}_{[z],[w]}~ \hat{\in}~  \mathcal{C}(m,\mathbb{M})$. It follows that there exist at least the following chords $\mathcal{L}_{[x],[y]},\mathcal{L}_{[x],[z]},\mathcal{L}_{[y],[z]}~ \hat{\in}~  \mathcal{C}(m,\mathbb{M})$. It follows from the pigeonhole principle that at least one of the following lines $\mathcal{L}_{[x],[y]}, \mathcal{L}_{[x],[z]},\mathcal{L}_{[y],[z]}$ must be a chord of the CoP $\mathcal{C}(n,\mathbb{M})$. It follows that the families $\mathcal{O}_n\cap \mathcal{O}_m\neq \emptyset$.
\end{proof}

\begin{theorem}\label{isomor1}
Let $\mathcal{C}(n,\mathbb{M})$ and $\mathcal{C}(m,\mathbb{M})$ be two parent CoPs admitting aligned embedding. Then holds
\[
\mathcal{C}(n,\mathbb{M})\cong \mathcal{C}(m,\mathbb{M})
\]
with a high degree.
\end{theorem}

\begin{proof}
Without loss of generality let us assume that $\mathcal{C}(n,\mathbb{M})\subset \mathcal{C}(m,\mathbb{M})$. Then by virtue of Theorem \ref{T-embedding_child} all children of $\mathcal{C}(n,\mathbb{M})$ are children of $\mathcal{C}(m,\mathbb{M})$ too. Hence holds $\hat{\mathcal{O}}_n\subset\hat{\mathcal{O}}_m$ and therefore
\[
\frac{\vert\hat{\mathcal{O}}_n\cap\hat{\mathcal{O}}_m\vert}{\vert\hat{\mathcal{O}}_n\vert}=1.
\]
\end{proof}



\section{\textbf{Compatible and Incompatible Circles of Partition}}
In this section we introduce the notion of compatibility and incompatibility of circles of partition. We launch the following formal language.

\begin{definition}
Let $\mathbb{M}\subseteq \mathbb{N}$ and $\mathcal{C}(n,\mathbb{M})$ and $\mathcal{C}(m,\mathbb{M})$ be any two CoPs. Then we say the CoPs $\mathcal{C}(n,\mathbb{M})$ and $\mathcal{C}(m,\mathbb{M})$ are  \textbf{compatible} if there exists some CoP $\mathcal{C}(r,\mathbb{M})$ satisfying
\begin{align}
    \mathcal{C}(n,\mathbb{M})\cup \mathcal{C}(m,\mathbb{M})\subseteq \mathcal{C}(r,\mathbb{M})\nonumber
\end{align}such that for each $[x]\in \mathcal{C}(n,\mathbb{M})\cup \mathcal{C}(m,\mathbb{M})$ with $2x\neq n$ there exist some $[y]\in \mathcal{C}(n,\mathbb{M})\cup \mathcal{C}(m,\mathbb{M})$ so that 
\begin{align}
    \mathbb{L}_{[x],[y]}~ \hat{\in}~  \mathcal{C}(r,\mathbb{M}).\nonumber
\end{align}
We denote the compatibility by $\mathcal{C}(n,\mathbb{M})\diamond \mathcal{C}(m,\mathbb{M})$. We call the CoP $\mathcal{C}(r,\mathbb{M})$ the \textbf{cover} of this compatibility.
\end{definition}
\bigskip

\begin{proposition}\label{P-compatible_embedding}
Let $\mathcal{C}(n,\mathbb{M})$ and $\mathcal{C}(m,\mathbb{M})$ be any two CoPs admitting aligned embedding. Then $\mathcal{C}(n,\mathbb{M})\diamond \mathcal{C}(m,\mathbb{M})$.
\end{proposition}

\begin{proof}
 W.l.o.g. we assume
\[
\mathcal{C}(n,\mathbb{M})\subset\mathcal{C}(m,\mathbb{M}).
\]
Then holds
\begin{align*}
\mathcal{C}(n,\mathbb{M})\cup\mathcal{C}(m,\mathbb{M})
&=\mathcal{C}(m,\mathbb{M})\mbox{ and therefore}\\
\mathcal{C}(n,\mathbb{M})&\diamond\mathcal{C}(m,\mathbb{M}).
\end{align*}
\end{proof}
\bigskip

\begin{theorem}\label{rigornocompatible}
Let $\mathbb{M}\subseteq \mathbb{N}$. Then there exists no CoPs of the forms $\mathcal{C}(n,\mathbb{M})$ and $\mathcal{C}(m,\mathbb{M})$ with all axes points concentrated at their center and additionally that $\mathcal{C}(n,\mathbb{M})\cap \mathcal{C}(m,\mathbb{M})=\emptyset$ for  $|\mathcal{C}(n,\mathbb{M})|>2$ and $|\mathcal{C}(m,\mathbb{M})|>2$ with 
\begin{align*}
\nu(n,\mathbb{M})\neq \nu(m,\mathbb{M})
\end{align*}
such that
\begin{align}
    \mathcal{C}(n,\mathbb{M})\diamond \mathcal{C}(m,\mathbb{M})\nonumber
\end{align}with a cover whose axes points are away from the center.
\end{theorem}

\begin{proof}
Let us suppose there exists at least a pair of CoPs of the form $\mathcal{C}(n,\mathbb{M})$ and $\mathcal{C}(m,\mathbb{M})$ with $m\neq n$ such that $\mathcal{C}(n,\mathbb{M})\cap \mathcal{C}(m,\mathbb{M})=\emptyset$ for $|\mathcal{C}(n,\mathbb{M})|,|\mathcal{C}(m,\mathbb{M})|>2$ and additionally that 
\begin{align*}
\nu(n,\mathbb{M})\neq\nu(m,\mathbb{M}) 
\end{align*}
so that $\mathcal{C}(n,\mathbb{M})\diamond \mathcal{C}(m,\mathbb{M})$. It follows that there exists some CoP $\mathcal{C}(s,\mathbb{M})$ such that 
\begin{align}
    \mathcal{C}(n,\mathbb{M})\cup \mathcal{C}(m,\mathbb{M})\subseteq \mathcal{C}(s,\mathbb{M})\nonumber
\end{align}so that for each $[x]\in \mathcal{C}(n,\mathbb{M})\cup \mathcal{C}(m,\mathbb{M})$ there exists some $[y]\in \mathcal{C}(n,\mathbb{M})\cup \mathcal{C}(m,\mathbb{M})$ such that 
\begin{align}
    \mathbb{L}_{[x],[y]}~ \hat{\in}~  \mathcal{C}(s,\mathbb{M}).\nonumber
\end{align}Under the conditions 
\begin{align*}
\nu(n,\mathbb{M})\neq \nu(m,\mathbb{M}) 
\end{align*}
and 
\begin{align}
    \mathcal{C}(n,\mathbb{M})\cap \mathcal{C}(m,\mathbb{M})=\emptyset \nonumber
\end{align}it follows from the \textbf{pigeon-hole} principle and the \textbf{uniqueness} of the axes of CoPs there exists some $\mathbb{L}_{[x],[y]}~ \hat{\in}~  \mathcal{C}(s,\mathbb{M})$ such that $[x],[y]\in \mathcal{C}(n,\mathbb{M})$ or $[x],[y]\in \mathcal{C}(m,\mathbb{M})$. Without loss of generality let us assume that $[x],[y]\in \mathcal{C}(n,\mathbb{M})$. By virtue of the embedding 
\begin{align}
    \mathcal{C}(n,\mathbb{M})\subset \mathcal{C}(s,\mathbb{M})\nonumber
\end{align}the line $\mathcal{L}_{[x],[y]}~ \hat{\in}~  \mathcal{C}(n,\mathbb{M})$ is such that $\mathcal{L}_{[x],[y]}\neq \mathbb{L}_{[x],[y]}~ \hat{\in}~  \mathcal{C}(n,\mathbb{M})$. It follows that the line $\mathcal{L}_{[x],[y]}$ must be a chord in $\mathcal{C}(n,\mathbb{M})$ and $\mathcal{C}(s,\mathbb{M})$ must be a \textbf{child} of the parent $\mathcal{C}(n,\mathbb{M})$. Now let us locate all the remaining \textbf{chords} $\mathcal{L}_{[u],[v]}\neq \mathcal{L}_{[x],[y]}$ in the parent $\mathcal{C}(n,\mathbb{M})$. We claim that each chord $\mathcal{L}_{[u],[v]}$ must be an axis of the child $\mathcal{C}(s,\mathbb{M})$. Let us assume to the contrary that some chord $\mathcal{L}_{[u],[v]}~ \hat{\in}~  \mathcal{C}(n,\mathbb{M})$ is also a chord in the child $\mathcal{C}(s,\mathbb{M})$. Then there exist some axes
\begin{align}
    \mathbb{L}_{[u],[a]},\mathbb{L}_{[v],[b]}~ \hat{\in}~  \mathcal{C}(n,\mathbb{M}).\nonumber
\end{align}
By virtue of the underlying embedding, it follows that the lines
\begin{align}
    \mathcal{L}_{[u],[a]}, \mathcal{L}_{[v],[b]}\nonumber
\end{align}
cannot be axes of the CoP $\mathcal{C}(s,\mathbb{M})$ so that $\mathcal{L}_{[u],[a]}$ and $\mathcal{L} _{[v],[b]}$ are chords in $\mathcal{C}(s,\mathbb{M})$ with 
\begin{align}
\Gamma([u],[b])=\Gamma([v],[a]).\label{chord1}
\end{align}
It follows that at least one of $\mathcal{L}_{[u],[b]}$ and $\mathcal{L}_{[v],[a]}$ must be chords in $\mathcal{C}(s,\mathbb{M})$. Otherwise, it would mean both lines $\mathcal{L}_{[u],[b]}=\mathbb{L}_{[u],[b]}~ \hat{\in}~  \mathcal{C}(s,\mathbb{M})$ and $\mathcal{L}_{[v],[a]}=\mathbb{L}_{[v],[a]}~ \hat{\in}~  \mathcal{C}(s,\mathbb{M})$, which in relation to \eqref{chord1} is absurd for axes points of CoPs. Without loss of generality let us assume $\mathcal{L}_{[u],[b]}$ is a chord then so is $\mathcal{L}_{[v],[a]}$ under the condition $\mathbb{L}_{[u],[a]},\mathbb{L}_{[v],[b]}~ \hat{\in}~  \mathcal{C}(n,\mathbb{M})$. Otherwise it would imply the chord $\mathcal{L}_{[a],[v]}$ must be an axis of $\mathcal{C}(s,\mathbb{M})$. Since all the axes points of $\mathcal{C}(n,\mathbb{M})$ are concentrated around the center, it certainly follows that 
\begin{align}
\frac{n}{2}=\frac{a+u}{2}\approx a \quad \mathrm{and} \quad \frac{n}{2}=\frac{a+u}{2}\approx u\label{conc1}
\end{align}and 
\begin{align}
\frac{n}{2}=\frac{b+v}{2}\approx b \quad \mathrm{and} \quad \frac{n}{2}=\frac{b+v}{2}\approx v\label{conc2}
\end{align}so that we have $a\approx b \approx u\approx v$ and we deduce that the co-axis point $[a],[v]$ of the cover CoP $\mathcal{C}(s,\mathbb{M})$ is close to the center by the relation 
\begin{align}
\frac{s}{2}=\frac{a+v}{2}\approx a \approx v\nonumber
\end{align} 
which contradicts the requirement of the proximity of the axes points of the cover $\mathcal{C}(s,\mathbb{M})$. It follows that $\mathcal{L}_{[u],[v]}$ and $\mathcal{L}_{[a],[b]}$ are also chords in $\mathcal{C}(s,\mathbb{M})$ with 
\begin{align}
    \Gamma([u],[v])=\Gamma([a],[b])\label{chord2}
\end{align}
since the lines $\mathbb{L}_{[u],[a]}, \mathbb{L}_{[v],[b]}~ \hat{\in}~  \mathcal{C}(n,\mathbb{M})$ tied with the embedding $\mathcal{C}(n,\mathbb{M})\subset \mathcal{C}(s,\mathbb{M})$.  It follows from \eqref{chord1} and \eqref{chord2}
\begin{align}
    \mathbb{L}_{[u],[a]},\mathbb{L}_{[v],[b]}~ \hat{\in}~  \mathcal{C}(s,\mathbb{M})\nonumber
\end{align}
so that $n=u+a=v+b=s$ and $\mathcal{C}(n,\mathbb{M})=\mathcal{C}(s,\mathbb{M})$, thereby contradicting the embedding 
\begin{align}
    \mathcal{C}(n,\mathbb{M})\subset \mathcal{C}(s,\mathbb{M}).\nonumber
\end{align}
Thus each \textbf{chord} in $\mathcal{C}(n,\mathbb{M})$ must be a axis of the \textbf{child} $\mathcal{C}(s,\mathbb{M})$. The upshot is that the parent has only one child $\mathcal{C}(s,\mathbb{M})$, which is impossible since $|\mathcal{C}(n,\mathbb{M})|>2$.
\end{proof}

\begin{conjecture}\label{compatible conjecture}
Let $\mathcal{C}(n,\mathbb{M})$ and $\mathcal{C}(m,\mathbb{M})$ be parents CoPs with the offspring $\mathcal{O}_n$ and $\mathcal{O}_m$, respectively. Then $\mathcal{C}(n,\mathbb{M})\diamond \mathcal{C}(m,\mathbb{M})$ if and only if there exists some $\mathcal{C}(s,\mathbb{M})\in \mathcal{O}_m$ and $\mathcal{C}(t,\mathbb{M})\in \mathcal{O}_n$ such that 
\begin{align}
    \mathcal{C}(s,\mathbb{M})\diamond \mathcal{C}(t,\mathbb{M}).\nonumber
\end{align}
\end{conjecture}

\noindent
For a CoP there are two possibilities:
\begin{itemize}
\item
The CoP admits embedding. Then holds Proposition \ref{P-compatible_embedding} for the parents and for their children and Conjecture \ref{compatible conjecture} is valid.
\item
The CoP do not admit embedding. An example for such CoPs is $\mathcal{C}(n,\mathbb{P})$. The following example demonstrates that Conjecture \ref{compatible conjecture} not holds for such CoPs.
\end{itemize}

\begin{example}
We consider the weights of the CoPs 
\begin{align*}
&\Vert\mathcal{C}(16,\mathbb{P})\Vert=\lbrace 3,5,11,13\rbrace\mbox{ and}\\
&\Vert\mathcal{C}(18,\mathbb{P})\Vert=\lbrace 5,7,11,13\rbrace\mbox{ and}\\
&\Vert\mathcal{C}(24,\mathbb{P})\Vert=\lbrace 5,7,11,13,17,19\rbrace
\mbox{ as child of }\mathcal{C}(16,\mathbb{P})\mbox{ and}\\
&\Vert\mathcal{C}(12,\mathbb{P})\Vert=\lbrace 5,7\rbrace
\mbox{ as child of }\mathcal{C}(18,\mathbb{P}).
\end{align*}
Then we obtain
\begin{align*}
\mathcal{C}(24,\mathbb{P})\cup\mathcal{C}(12,\mathbb{P})
&=\mathcal{C}(24,\mathbb{P})\mbox{ and therefore}\\
\mathcal{C}(24,\mathbb{P})&\diamond\mathcal{C}(12,\mathbb{P}).
\end{align*}
On the other hand is
\begin{align*}
&\Vert\mathcal{C}(16,\mathbb{P})\cup\mathcal{C}(18,\mathbb{P})\Vert=
\lbrace 3,5,7,11,13\rbrace.
\end{align*}
Because $3,5,7$ are the only $3$ primes which have a distance of $2$ between each other there exists no CoP $\mathcal{C}(n,\mathbb{P})$ for which holds that the last $3$ weights have a distance of $2$ between each other. Hence $\mathcal{C}(16,\mathbb{P})$ and $\mathcal{C}(18,\mathbb{P})$ are not compatible although they have children which are compatible.
\end{example}

We mind that we obtain a CoP from the union of $\mathcal{C}(16,\mathbb{P})$ and $\mathcal{C}(18,\mathbb{P})$ if we remove $[3]$ or $[7]$ from the union. In the first case we get $\mathcal{C}(18,\mathbb{P})$ and in the second case $\mathcal{C}(16,\mathbb{P})$. In both cases we have a so called \textit{weak compatibility} $\mathcal{C}(16,\mathbb{P})\circ\mathcal{C}(18,\mathbb{P})$.   

\begin{definition}
Let $\mathbb{M}\subseteq \mathbb{N}$ and $\mathcal{C}(n,\mathbb{M})$ and $\mathcal{C}(m,\mathbb{M})$ be any two CoPs. Then we say the CoPs $\mathcal{C}(n,\mathbb{M})$ and $\mathcal{C}(m,\mathbb{M})$ are  \textbf{weakly compatible} if there exist some CoP $\mathcal{C}(r,\mathbb{M})$ and a point $[z]\in\mathcal{C}(n,\mathbb{M})\cup \mathcal{C}(m,\mathbb{M})$ satisfying
\begin{align*}
    \mathcal{C}(n,\mathbb{M})\cup \mathcal{C}(m,\mathbb{M})\setminus\lbrace[z]\rbrace\subseteq \mathcal{C}(r,\mathbb{M})
\end{align*}
such that for each $[x]\in \mathcal{C}(n,\mathbb{M})\cup \mathcal{C}(m,\mathbb{M})$ with $2x\neq n$ there exist some $[y]\in \mathcal{C}(n,\mathbb{M})\cup \mathcal{C}(m,\mathbb{M})$ so that 
\begin{align}
    \mathbb{L}_{[x],[y]}~ \hat{\in}~  \mathcal{C}(r,\mathbb{M}).\nonumber
\end{align}
We denote the weak compatibility by $\mathcal{C}(n,\mathbb{M})\circ \mathcal{C}(m,\mathbb{M})$. We call the CoP $\mathcal{C}(r,\mathbb{M})$ also the \textbf{cover} of this compatibility.
\end{definition}

Another example for weakly compatible CoPs are 
\begin{align*}
\Vert\mathcal{C}(28,\mathbb{P})\Vert&=\lbrace 5,11,17,23\rbrace\mbox{ and}\\
\Vert\mathcal{C}(30,\mathbb{P})\Vert&=\lbrace 7,11,13,17,19,23\rbrace
\mbox{ and}\\
\mathcal{C}(28,\mathbb{P})\cup\mathcal{C}(30,\mathbb{P})&\setminus\lbrace[5]\rbrace=\mathcal{C}(30,\mathbb{P})\\
\mbox{and }&\mbox{therefore}\\
\mathcal{C}(28,\mathbb{P})&\circ\mathcal{C}(30,\mathbb{P}).
\end{align*}
\bigskip

Conjecture \ref{compatible conjecture} could have several ramifications if it turns out to be true. Yet we believe it is very hard to establish as we found it far-fetched with the current tools developed thus far. Any progress on this conjecture would require an expansion on the notion of compatibility and their interplay with other concepts.

\section{\textbf{Extended Circles of Partition}}
In this section we introduce the notion of extended CoPs and demonstrate some important properties. Finally we give a partial proof of the binary Goldbach conjecture.

\begin{definition}\label{D-extended_CoP}
Let $\mathbb{M}\subset\mathbb{N}$ and $\mathcal{C}(n)$ be a CoP with $\mathbb{N}$ as base set. Then we denote
\[
\mathcal{C}^*(n,\mathbb{M}):=\lbrace [x]\in\mathcal{C}(n)\mid\lbrace x,n-x\rbrace\cap\mathbb{M}\neq\emptyset, x>2\rbrace
\]
as \textbf{extended Circle of Partition}. We abbreviate it as \textbf{xCoP}.\\
We denote by $\mathcal{O}_n^*(\mathbb{M})$ the extended family of the xCoP $\mathcal{C}^*(n,\mathbb{M})$ as collection of all children $\mathcal{C}^*(s,\mathbb{M})$ with principal axes $\mathbb{L}_{[x],[y]}~ \hat{\in}~ \mathcal{C}^*(s,\mathbb{M}),s=x+y\neq n$ whereby $[x]$ or $[y]$ is not the center of the parent xCoP. We denote by $\hat{\mathcal{O}}_n^*(\mathbb{M})$ the union of the parent xCoP with its extended family as complete extended family. With $\mathbb{F}_n(\mathbb{M})$ we denote the set of the generators of a complete extended family.\\
We call an axis whose both axis points are members of $\mathbb{M}$ full--$\mathbb{M}$ axis and in the other case half--$\mathbb{M}$ axis. If $\mathbb{M}=\mathbb{P}$ then we say full--prime axis resp. half--prime axis.
\end{definition}
While all axes of a CoP $\mathcal{C}(n,\mathbb{M})$ are full--$\mathbb{M}$ axes the axes of an xCoP $\mathcal{C}^*(n,\mathbb{M})$ are full-- or half--$\mathbb{M}$ axes. 
\bigskip

\begin{proposition}\label{P-ext_subset}
It holds  $ \mathcal{C}(n,\mathbb{M})\subseteq\mathcal{C}^*(n,\mathbb{M})$ for all $n\in\mathbb{N}$ with $\mathcal{C}(n,\mathbb{M})\neq\emptyset$.
\end{proposition}
\begin{proof}
If holds $x\in\mathbb{M}$ and $n-x\in\mathbb{M}$ then is $[x]\in\mathcal{C}(n,\mathbb{M})$ as well as $[x]\in\mathcal{C}^*(n,\mathbb{M})$. If is only $x\in\mathbb{M}$ and $[n-x]\not\in\mathbb{M}$ or vice versa then is $[x]\in\mathcal{C}^*(n,\mathbb{M})$ but $[x]\not\in\mathcal{C}(n,\mathbb{M})$.
\end{proof}
\bigskip

\begin{corollary}\label{C-ext_Psubset}
Let $\mathbb{P}$ be the set of all prime numbers.
Due to Proposition \ref{P-ext_subset} holds for all $n\in2\mathbb{N}$ with $\mathcal{C}(n,\mathbb{P})\neq\emptyset$
\[
\mathcal{C}(n,\mathbb{P})\subset\mathcal{C}^*(n,\mathbb{P}).
\]
\end{corollary}
\bigskip

\begin{proposition}\label{P-ext_progression}
Let $\mathbb{M}_{a,d}\subset\mathbb{N}$ the set like in \eqref{Mad}. Then is for all $n\in\mathbb{M}_{2a,d}$
\[
\mathcal{C}(n,\mathbb{M}_{a,d})\equiv\mathcal{C}^*(n,\mathbb{M}_{a,d}).
\]
\end{proposition}
\begin{proof}
Because of $n\in\mathbb{M}_{2a,d}$ for each member $x$ of $\mathbb{M}_{a,d}$ holds that also $n-x$ is a member of $\mathbb{M}_{a,d}$. Therefore there exists no $x\in\mathbb{M}_{a,d}$ with $n-x\notin\mathbb{M}_{a,d}$ and vice versa. But these would be the extensions of $\mathcal{C}(n,\mathbb{M}_{a,d})$. 
\end{proof}
\bigskip

Next we look at xCoPs with the set $\mathbb{P}$ of all prime numbers as base set.
\bigskip

\begin{proposition}\label{P-ext_primes}
Let be $n\in 2\mathbb{N}, n\geq 8$.
Then the xCoP $\mathcal{C}^*(n,\mathbb{P})$ contains all odd primes not greater than $n-3$ and it holds
\[
\vert\mathcal{C}^*(n,\mathbb{P})\vert\geq\pi(n-3)-1\geq 2.
\]
\end{proposition}
\begin{proof}
The xCoP $\mathcal{C}^*(n,\mathbb{P})$ contains only odd numbers because if $x$ would be even $>2$ then must also $n-x$ be even and therefore both are not prime. The first member of each such xCoP is $[3]$. Its axis partner is $[n-3]$ and is a member of the xCoP because $3$ is prime. Therefore contains $\mathcal{C}^*(n,\mathbb{P})$ prime numbers between $3$ and not greater than $n-3$. Not contained in $\mathcal{C}^*(n,\mathbb{P})$ are all axis points $[u]\in\mathcal{C}(n)$ and their axis partner $[n-u]$ which both are not prime. Because $ \Vert\mathcal{C}(n)\Vert $  contains all natural numbers $1\leq x\leq n-1$ then $ \Vert\mathcal{C}(n)\Vert $ contains also all primes $3\leq p \leq n-3$. And these will be not excluded by the condition $ \lbrace x,n-x\rbrace\cap\mathbb{P}\neq\emptyset $ .
Therefore $\mathcal{C}^*(n,\mathbb{P})$ contains all primes between $3$ and less or equal $n-3$ and hence it holds 
\[
\vert\mathcal{C}^*(n,\mathbb{P})\vert\geq\pi(n-3)-1\geq 2
\]
because $\pi(8-3)-1=2$.
\end{proof}

\begin{corollary}\label{C-ext_nonempty}
From Proposition \ref{P-ext_primes} follows that for all $n\in 2\mathbb{N}, n\geq 8$ holds
\[
\mathcal{C}^*(n,\mathbb{P})\neq\emptyset
\]
since the all contain the points $[3]$ and $[5]$ at least.
\end{corollary}

\begin{corollary}\label{C-ext_cap}
From Proposition \ref{P-ext_primes} follows that for all $n\in 2\mathbb{N}, n\geq 8$ with $\mathcal{C}(n,\mathbb{P})\neq\emptyset$ holds
\[
\mathcal{C}(n,\mathbb{P})\cap\mathcal{C}^*(n,\mathbb{P})\neq\emptyset
\]
because also $\mathcal{C}(n,\mathbb{P})$ contains primes not greater than $n-3$, but not all necessarily.
\end{corollary}
\bigskip

\noindent
Additional we need the following lemmata.

\begin{lemma}\label{L-ext_symmetrCoP}
Let $\mathbb{M}\subseteq\mathbb{N}$ and $\mathcal{C}^*(n,\mathbb{M})$ an xCoP. Then the weights of the xCoP are symmetrically distributed around $\frac{n}{2}$. 
\end{lemma}
\begin{proof}
This is true because
\[
x_{i+1}-x_i = n-n+x_{i+1}-x_i=(n-x_i)-(n-x_{i+1})
\mbox{ for } i=1,2,\ldots,\left\lfloor\frac{k}{2}\right\rfloor
\]
if $x_1,x_2,\ldots,x_k$ are the weights of the points of xCoP.
\end{proof}

\begin{corollary}\label{C-ext_number}
From Lemma \eqref{L-ext_symmetrCoP} follows immediately that if $\frac{n}{2}$ is a member of $\mathbb{M}$ then has $\mathcal{C}^*(n,\mathbb{M})$ an odd number of points. Else the number is even.
\end{corollary}
\bigskip

\begin{lemma}\label{L-ext_symmetrFam}
Let $\mathbb{M}\subseteq\mathbb{N}$ and $\mathcal{C}^*(n,\mathbb{M})$ an xCoP. Then the generators of its children are symmetrically distributed around the generator of the parent xCoP. 
\end{lemma}
\begin{proof}
Analogously to the proof of Proposition \ref{breaker} we get for two arbitrary axes 
\[
\mathbb{L}_{[u],[v]},\mathbb{L}_{[x],[y]}~ \hat{\in}~ \mathcal{C}^*(n,\mathbb{M}), u<x
\]
of the xCoP the following children generators
\begin{align*}
s_1=u+x<s_2=u+y<~ &n<t_2=v+x<t_1=v+y\\ 
\mbox{and the }&\mbox{following distances}\\
s_2-s_1=u+y-u-x &\mbox{ and }t_1-t_2=v+y-v-x\\
\mbox{and }&\mbox{hence}\\
s_2-s_1=y &-x = t_1-t_2.
\end{align*}
Because the chosen axes are arbitrary and always the parent generator is located in the middle of such inequalities the distances between the children generators are symmetrically distributed around $n$.
\end{proof}
\noindent
The statements of both lemmata are also true for unextended CoPs.

\begin{corollary}\label{C-ext_evenChildren}
From the proof of Lemma \ref{L-ext_symmetrFam} concludes that the number of children of any xCoP is even and hence every complete extended family has an odd number of members.
\end{corollary}
\bigskip

\begin{theorem}\label{T-ext_embFams}
Let $m,n\in 2\mathbb{N}$ and $\mathcal{C}^*(m,\mathbb{P})$ and $\mathcal{C}^*(n,\mathbb{P})$ be two xCoPs with $m<n$ and $m\geq 16$.
If holds $\mathcal{C}(k,\mathbb{P})\neq\emptyset$ for $k\in 2\mathbb{N}\mid 8\leq k <n$ 
then $\mathbb{F}_n(\mathbb{P})$ contains all even integers $8\leq x\leq 2n-8$ and it
holds
\[
\mathbb{F}_m(\mathbb{P})\subset\mathbb{F}_n(\mathbb{P}).
\]
Additionally holds $\vert\mathbb{F}_n(\mathbb{P})\vert=n-7$. This means that $\mathcal{C}^*(n,\mathbb{P})$ has $n-8$ children.
\end{theorem}
\begin{proof}
Because the first three odd numbers $\geq 3$ all are primes the first three points which not lies on common axes or on a degenerated axis of all xCoPs
are $[3],[5],[7]$ for $m\geq 16$. 
Ergo the first three generators of children of all such xCoPs are
\[
f_1=3+5=8,
f_2=3+7=10\mbox{ and }
f_3=5+7=12.
\]
Because of Lemma \ref{L-ext_symmetrFam} we have only to prove that all even integers $8\leq x<n$ are members of $\mathbb{F}_n(\mathbb{P})$.
Let $t$ be the cardinality of $\mathbb{F}_n(\mathbb{P})$ and
\[
s:=\frac{t+1}{2}.
\]
By virtue of Corollary \ref{C-ext_evenChildren} $t$ is an odd number and
$\mathbb{F}_n(\mathbb{P})$ has the parent generator $n$ in the middle and even children generators left and equal ones on the right side. Therefore $n$ is the $s^{th}$ element of $\mathbb{F}_n(\mathbb{P})$
\[
f_s=n.
\]
Since the second weight of all xCoPs $5$ is prime the point $[n-5]$ is a member of $\mathcal{C}^*(n,\mathbb{P})$ and $\mathcal{L}_{[3][n-5]}$ is not an axis and hence a principal axis for the child $\mathcal{C}^*(n-2,\mathbb{P})$. Hence the greatest member of $\mathbb{F}_n(\mathbb{P})$ less than $n$ is   
\[
f_{s-1}=n-2.
\]
Because of $\mathcal{C}(k,\mathbb{P})\neq\emptyset$ for $k\in 2\mathbb{N}\mid 8\leq k <n$ all such even numbers $k$ have at least one representation as sum of two primes. And by virtue of Proposition \ref{P-ext_primes} the xCoP $\mathcal{C}^*(n,\mathbb{P})$ contains all primes between $3$ and not greater than $n-3$. Therefore every even number between $8$ and $n-2$ has a representation as sum of two weights of $\mathcal{C}^*(n,\mathbb{P})$ whose points don't form an axis of $\mathcal{C}^*(n,\mathbb{P})$. Therefore these even numbers are generators of children of the parent xCoP $\mathcal{C}^*(n,\mathbb{P})$, hence the elements $f_1=8,f_2=10,\ldots,f_{s-1}=n-2\in\mathbb{F}_n(\mathbb{P})$.
This are 
\[
s-1=\frac{n-2-8}{2}+1=\frac{n}{2}-4
\]
members. With it we can calculate 
\[
t=\vert\mathbb{F}_n(\mathbb{P})\vert=2s-1=n-6-1=n-7.
\]
By virtue of Lemma \ref{L-ext_symmetrFam} the members from $f_{s+1}$ until $f_t$ are the $s-1$ even numbers $>n$.
The greatest children generator is
\[
f_t =n-3+n-5=2n-8
\]
Therefore all even numbers between $8$ and $2n-8$ are the members of $\mathbb{F}_n(\mathbb{P})$ and it holds $\mathbb{F}_m(\mathbb{P})\subset\mathbb{F}_n(\mathbb{P})$ because of $\vert\mathbb{F}_m(\mathbb{P})\vert<\vert\mathbb{F}_n(\mathbb{P})\vert$. 

Between $8$ and $2n-8$ there are exectly $n-7$ numbers. Hence $\mathcal{C}^*(n,\mathbb{P})$ has $n-8$ children.
Out of the even numbers between $8$ and $2n-8$ there cannot be a child generator.
\end{proof}


It is easy to check that the xCoPs $\mathcal{C}^*(2,\mathbb{P})$ until $\mathcal{C}^*(10,\mathbb{P})$ have no children.


\begin{corollary}\label{C-ext_isomorphic}
Because of $\mathbb{F}_m(\mathbb{P})\subset\mathbb{F}_n(\mathbb{P})$ by $m<n$ all xCoPs $\mathcal{C}^*(n,\mathbb{P})$ with $n\geq 12$
and $\mathcal{C}(k,\mathbb{P})\neq\emptyset\mid 8\leq k<n$  
holds
\[
\hat{\mathcal{O}}^*_m\subset\hat{\mathcal{O}}^*_n\mbox{ if }m<n,
\]
which means that all such two xCoPs are isomorphic with a high degree (see Definition \ref{D-isomor}). 
\end{corollary}
\bigskip

\begin{corollary}\label{C-ext_connected}
Due to $[3]$ is a point of all xCoPs $\mathcal{C}^*(n,\mathbb{P})$ for $n\geq 6$ all the xCoPs are connected (see Proposition \ref{P-common_points}).
\end{corollary}
\bigskip

\begin{theorem}\label{T-ext_axis}
Let $n\in 2\mathbb{N}$, $\mathcal{C}^*(n,\mathbb{P})$ be a xCoP with $n\geq 8$ and $\mathcal{C}(n,\mathbb{P})\neq\emptyset$. Then exists an axis $\mathbb{L}_{[x],[n-x]}~ \hat{\in}~ \mathcal{C}^*(n,\mathbb{P})$ such that $\mathbb{L}_{[x],[n-x]}$ is also an axis of $\mathcal{C}(n,\mathbb{P})$.
\end{theorem}
\begin{proof}
By virtue of Corollary \ref{C-ext_cap} there exists a point $[x]$ which is member of $\mathcal{C}^*(n,\mathbb{P})$ as well as of $\mathcal{C}(n,\mathbb{P})$. Because $\mathcal{C}(n,\mathbb{P})$ contains only points with prime weights $x$ must be prime and also $[n-x]$ is member of $\mathcal{C}(n,\mathbb{P})$ and has a prime weight. But also $[n-x]$ must be a member of $\mathcal{C}^*(n,\mathbb{P})$ because $x$ is prime and $x+(n-x)=n$. Hence we have an axis $\mathbb{L}_{[x],[n-x]}$ which belongs to $\mathcal{C}(n,\mathbb{P})$ as well as to $\mathcal{C}^*(n,\mathbb{P})$.
\end{proof}
\bigskip

\begin{definition}\label{D-ext_nrofaxes}
Let $\mathcal{C}^*(n,\mathbb{M})$ be a nonempty xCoP. We denote by
\[
\nu^*(n,\mathbb{M}):=\#\lbrace\mathbb{L}{[x],[y]}~ \hat{\in}~ \mathcal{C}^*(n,\mathbb{M})\rbrace
\]
the number of real axes in the xCoP $\mathcal{C}^*(n,\mathbb{M})$. 
\end{definition}
\bigskip

\begin{corollary}\label{C-ext_nrofaxes}
Let $\mathbb{M}\subset\mathbb{N}$ and $\mathcal{C}^*(n,\mathbb{M})$ be a nonempty xCoP. Then holds
\begin{align*}
\nu^*(n,\mathbb{M})&=\#\lbrace\mathbb{L}{[x],[y]}~ \hat{\in}~ \mathcal{C}^*(n,\mathbb{M})\rbrace\\
&=\#\lbrace\mathbb{L}{[x],[y]}~ \hat{\in}~ \mathcal{C}(n)\mid\lbrace x,y\rbrace\cap\mathbb{M}\neq\emptyset\rbrace\\
&=\nu(n,\mathbb{M})+\#\lbrace\mathbb{L}_{[x],[y]}~ \hat{\in}~ \mathcal{C}(n)\mid x\in\mathbb{M},y\not\in\mathbb{M}\rbrace\\
&=\nu(n,\mathbb{M})+\bar{\nu}(n,\mathbb{M})
\end{align*}
with $\bar{\nu}(n,\mathbb{M}):=\#\lbrace\mathbb{L}_{[x],[y]}~ \hat{\in}~ \mathcal{C}(n)\mid x\in\mathbb{M},y\not\in\mathbb{M}\rbrace$
as the number of half--$\mathbb{M}$ axes of the xCoP $\mathcal{C}^*(n,\mathbb{M})$ and $\nu(n,\mathbb{M})$ like defined in Definition \ref{axis} the number of full--$\mathbb{M}$ axes.
\end{corollary}
\bigskip

\begin{theorem}\label{T-ext_density}
For the density of points by virtue of Definition \ref{pointdensity} with $\mathbb{M}=\mathbb{N}$ and $\mathbb{H}=\mathbb{P}$ holds
\[
\mathcal{D}\left(\mathbb{P}_{\mathcal{C}(\infty)}\right)\sim\frac{1}{\log n}.
\]
\end{theorem}
\begin{proof}
By virtue of Definition \ref{pointdensity} and Corollary \ref{C-ext_nrofaxes} holds with $\mathbb{M}=\mathbb{N}$ and $\mathbb{H}=\mathbb{P}$
\begin{align*}
\mathcal{D}\left(\mathbb{P}_{\mathcal{C}(\infty)}\right)
&\sim\frac{\#\lbrace\mathbb{L}_{[x],[n-x]}~ \hat{\in}~ \mathcal{C}(n)\mid\lbrace x,n-x\rbrace\cap\mathbb{P}\neq\emptyset\rbrace}{\nu(n,\mathbb{N})}\\
&\sim\frac{\#\lbrace\mathbb{L}_{[x],[n-x]}~ \hat{\in}~ \mathcal{C}^*(n)\rbrace}{\left\lfloor\frac{n-1}{2}\right\rfloor}\\
&\sim\frac{\nu^*(n,\mathbb{P})}{\frac{n-1}{2}}
\mbox{ and in virtue of Proposition \ref{P-ext_primes}}\\
&\sim\frac{\pi(n-3)-1}{n-1}
\sim\frac{1}{\log n}.
\end{align*}
\end{proof}

\begin{theorem}[Binary Goldbach Conjecture]\label{T-BGC}
For all $n\in 2\mathbb{N}, n\geq 6$ 
there exists at least one representation as sum of two primes.
\end{theorem}
\begin{proof}
First we note that the following statement is an analogon to the claim:
\begin{center}
\textit{
There is no xCoP $\mathcal{C}^*(n,\mathbb{P})$
for $n\in 2\mathbb{N}, n\geq 6$
with only half--prime axes
},
\end{center}
because in this case there is no empty CoP for $n\in 2\mathbb{N}, n\geq 6$ and hence for each such $n$ there exists at least one representation as sum of two primes (weights of axis points).

By virtue of Proposition \ref{P-ext_primes} the weights of an xCoP contains all primes between $3$ and not greater than $n-3$. Let us assume contrarily that  $\mathcal{C}^*(n_o,\mathbb{P})$ is an xCoP with only half--prime axes and $n_o$ is the least generator of such xCoPs. This means that for $6\leq k \leq n_o-2$ holds $\mathcal{C}(k,\mathbb{P})\neq\emptyset$.
Because in virtue of our assumtion the weights of the last three points $[n_o-3]$, $[n_o-5]$ and $[n_o-7]$ cannot be prime we find in this case in $\Vert\mathcal{C}^*(n_o,\mathbb{P})\Vert$ only the primes between $3$ and not greater than $n_o-9$.\\
$\ldots$\\
On the other hand there is always an $[x]\in\mathcal{C}^*(n_o,\mathbb{P})$ with $x\in\mathbb{P}$ and an 
$r\in 2\mathbb{N}\mid 2\leq r \leq x-3$  
such that $[x-r]\in\mathcal{C}(n_o-r,\mathbb{P})$. Then also holds 
\[
\mathbb{L}_{[x-r],[n_o-r-(x-r)]}~ \hat{\in}~ \mathcal{C}(n_o-r,\mathbb{P}).
\]
And because of Theorem \ref{central} holds for the point $[n_o-x]\in\mathcal{C}^*(n_o,\mathbb{P})$, the axis partner of $[x]$,
that $n_o-x$ is prime. But this contradicts the premise that all axes are half-prime axes since also $x$ is prime. 
\end{proof}
This proof must be completed in order to remove ''$\ldots$''.

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