\section{Introduction}
Isometric embeddings of metric spaces into infinite-dimensional Banach spaces have a long tradition. 
Classical  results are due to
Fr\'echet, Urysohn,  Kuratowski,  Banach,
 \cite{
  Frechet1910, 
 Urysohn1927,   
Kuratowski1935, 
Banach1955}.
For an introduction to the subject we refer to Heinonen's survey \cite{MR2014506}.
The case of  embeddings into
finite-dimensional Banach spaces is much harder, even when one considers bi-Lipschitz embeddings in place of isometric embeddings.
It is a wide open problem to give an intrinsic characterizations of those metric spaces which admit bi-Lipschitz embeddings into some Euclidean space. See for example
\cite{MR1744622,   
MR1387955,   
Lang-Plaut,   
MR2915474,   
MR3253787}. 

The situation is quite different for quasisymmetric maps (see \cite[Chapter~10-12]{Heinonenbook} for an introduction to the theory of quasisymmetric embeddings).
A metric space quasisymmetrically embeds into some Euclidean spaces if and only if it is doubling (see \cite[Theorem~12.1]{Heinonenbook}).
More specifically, Assouad proved the following result (see \cite{Assouad83}, and also \cite{MR2990137, MR3108866}):
if $(X,d)$ is a doubling metric space and $\alpha\in (0,1)$ then the metric space $(X,d^\alpha)$ admits a bi-Lipschitz embedding into some Euclidean space.
If $d_E$ is the Euclidean distance and $\alpha=\log 2 / \log 3$ then the metric space
$([0,1], d_E^\alpha )$ is bi-Lipschitz equivalent to the von Koch snowflake curve, so
 $(X,d^\alpha)$  is said to be the {\em $\alpha$-snowflake} of  $(X,d)$.

The Assouad Embedding Theorem  is sharp in that there 
  are examples (none of which are trivial) of doubling spaces which do not admit bi-Lipschiz embeddings into any Euclidean space,
  even though each of their $\alpha$-snowflakes do.
See  
\cite{
Semmes,
Semmes2,  
Laakso,  
Cheeger-Kleiner2}. 
We also  stress that it has been known that snowflakes of doubling   spaces
in general do not isometrically  embed  in any Euclidean space.
Indeed,  the space
  $([0,1], d_E^{1/2})$ does not, see
  \cite[Remark~3.16(b)]{MR2014506}.
  
  The main aim of this paper
is to show that if some $\alpha$-snowflake of a metric space
isometrically embedds into a finite dimensional Banach space then the
metric space in question is finite.
 
  Our main result is the following.
  
\begin{theorem}\label{thm:main}
For any $n \in \mathbb N$ and $\alpha \in (0, 1)$ there is an $N \in \mathbb N$ such that if a metric space $(X,d)$ has cardinality at least $N$ then    $(X,d^\alpha)$ does not admit an isometric embedding into any $n$-dimensional normed linear space.
\end{theorem}

The techniques that we use in Section~\ref{sec:thm:main1} for the proof of Theorem~\ref{thm:main} can also be used to study more general notions of snowflakes.
For this purpose, we introduce general   snowflaking functions.
We say that a function $h: \mathbb{R}_{\geq} \to \mathbb{R}_{\geq}$ is a  {\em snowflaking function} if the following hold:
\begin{enumerate}
\item[(S1)] $h(0) = 0$.
\item[(S2)] $h$ is concave.
\item[(S3)] $ \frac{h(t)}{t} \to \infty$, as ${t \to 0}$.
\item[(S4)] $ \frac{h(t)}{t} \to 0$, as ${t \to \infty} $.
\end{enumerate}
Let $h$ be a snowflaking function.
Then function $h$ is  weakly increasing and, if $d$ is a metric on a set $X$ then $h \circ d$ is also a metric on $X$.
Given a snowflaking function $h$ and a metric space $(X,d)$ 
we say that the metric space $(X, h \circ d)$ is the  {\em $h$-snowflake} of $(X,d)$.
If $h(t) = t^\alpha$ for some $\alpha \in (0,1)$ then $(X, h \circ d)$ is the $\alpha$-snowflake of $(X,d)$.
In Section~\ref{sec:thm:main} we prove the following.
\begin{theorem}\label{thm:main1}
Let $h$ be a snowflaking function and $(X,d)$ a metric space.
If the $h$-snowflake of $(X,d)$  admits an isometric embedding into some finite-dimensional Banach space then $X$ is finite.
\end{theorem}
\begin{remark}
Note that for general snowflaking functions there may not be any bound on the number of points one can embed, see Remark \ref{rmk:nobound}.
If one removes either of the requirements (S3) or (S4), then we say that $(X, h \circ d)$ is a degenerate snowflake (at zero or at infinity, respectively). Indeed, in such cases the conclusion of Theorem~\ref{thm:main1} does not hold, in general, see Proposition \ref{prop:necessity}.
\end{remark}
 
We conclude the introduction with few other simple observations about embeddings into Euclidean spaces.
Every $\alpha$-snowflake of $\mathbb{R}^n$, $\alpha\in (0,1)$, isometrically embeds
 into the Hilbert space $\ell^2$ of square summable sequences, see \cite[Remark~3.16(d)]{MR2014506}.
 For any $n\in \mathbb N$, there is a metric space of cardinality $n$ such that for any $\alpha \in (0,1)$ its $\alpha$-snowflake 
can be isometrically embedded into $\mathbb{R}^{n-1}$ (just take the vertices of the standard simplex).
There is a 4-point metric space which has an $\alpha$-snowflake which cannot be isometrically embedded into $\mathbb{R}^4$, 
and so cannot be isometrically embedded into any Euclidean space, (just take the vertices of the (3,1) complete bipartite graph).
Every finite metric space has an $\alpha$-snowflake which admits an isometric embedding into some Euclidean space, see Proposition~\ref{prop:existence} below.















   



\section{Corollaries}


In this section we record a few small results and corollaries.
First of all, note that as a corollary of Theorem \ref{thm:main1},  by Ascoli-Arzel\`a Theorem we immediately obtain:
\begin{corollary}\label{lipschitz}
Suppose that $(X,d)$ is infinite.
For any snowflaking function $h$ and $n \in \mathbb{N}$ there is a $\delta > 0$ such that $(X, h \circ d)$ does not admit a $(1 + \delta)$-bilipschitz embedding into any normed linear space of dimension $n$.
\end{corollary}

Theorem \ref{thm:main} shows that there is a bound (depending on $\alpha$ and $n$) on the cardinality of a metric space whose $\alpha$-snowflake admits an isometric embedding into $\mathbb R^n$.
The next easy result says that any finite metric space can be isometrically embedded in some Euclidean space after some $\alpha$-snowflaking.
\begin{proposition}\label{prop:existence}
If $(X,d)$ is a finite metric space of cardinality $n$, then some $\alpha$-snowflake of $(X,d)$ admits an isometric embedding into the Euclidean space 
 $\mathbb R^n$.
\end{proposition}

Combining Theorem~\ref{thm:main} and Proposition~\ref{prop:existence} we have:\begin{corollary} 
Given a metric space $(X,d)$,
there is an $\alpha$-snowflake of $(X,d)$ that isometrically embedds into some finite-dimensional normed space if and only if $X$ is finite.
\end{corollary}

\begin{proof}[Proof of Proposition \ref{prop:existence}]
We show that $(X,d^\alpha)$ admits an isometric embedding into $\mathbb R^n$ when $\alpha$ is sufficiently close to $1$.
In fact we show that there is an $\epsilon > 0$ such that if $d'$ is any metric on $X$ satisfying $1 - \epsilon < d'(x,y) < 1 +\epsilon$ for all distinct $x,y \in X$ then $(X,d')$ admits an isometric embedding into $\mathbb R^n$.
The proposition follows as $d(x,y)^\alpha \to 1$ as $\alpha \to 0$ for any distinct $x,y \in X$.

Consider the points $p_j:= \tfrac{1}{\sqrt{2}} e_j$ where $e_1, \ldots, e_n$ is the canonical basis of $\mathbb R^n$.
Thus $d(q_i,q_j) = 1$ for all distinct $1 \leq i,j \leq n$.
The proposition is thus a direct consequence of the following claim:

{\bf Claim A.} There is an $\epsilon>0$ such that if $\rho_{ij}\in \mathbb R$, with $1\leq i<j\leq n$ are such that
$|\rho_{ij} - 1 | < \epsilon$, then there exist $q_1, \ldots, q_n\in \mathbb R^n$ such that
$\norm{q_i-q_j} = \rho_{ij}$ for all $1 \leq i < j \leq n$.



To show the claim, consider
the vector subspace $U$ of $\mathbb R^{n\times n}
$
defined by the upper-triangular matrices, i.e., 
the elements of $ U$ are elements of the form $  (a_{ij})_{ij}$ with 
$1\leq i<j\leq n$ and $a_{ij}\in \mathbb R$.
Let
$F:\mathbb R^{n\times n} \to U$ be given by
$$F(q_1, \ldots, q_n) := (\norm{q_i-q_j}^2)_{ij} \quad \text{ for }\,q_1, \ldots, q_n\in \mathbb R^n.$$
Denote by
${\bf p}$ the vector of vectors 
${\bf p}:= (\tfrac{1}{\sqrt{2}} e_1, \ldots, \tfrac{1}{\sqrt{2}} e_n)$.
Notice that
$F({\bf p})=(1)_{ij}$.

Claim A holds if and only if $F(\bf p)$ lies in the interior of the image of $F$.
Thus it suffices to show that the differential $(dF)_{\bf p}$ has maximal rank and apply the inverse function theorem.

Denote by $q_l^k$ the $k$-th component of an $n$-vector $q_l$.
Notice that $e_l^k= \delta_l^k$, where $\delta_l^k$ is the 
Kronecker symbol.
Set $E_{lk} = (\delta_i^l \delta_j^k)_{ij}$ Notice that
$\{E_{lk}, 1\leq i<j\leq n\}$ span the set $U$.
Let us show that any $E_{lk}$ is in the image of $(dF)_{\bf p}$.
Fix $l$ and $k$ so that 
$1\leq l<k\leq n$.
By direct calculation,
\begin{eqnarray*}
\dfrac{\partial F( {\bf q})}{\partial q_l^k}  = 
\left( 2 \langle q_i -q_j , 
\delta_i^l e_k - \delta_j^l e_k\rangle \right)_{ij}
 = 
\left( 2 ( q_i^k -q_j^k )
(\delta_i^l - \delta_j^l ) \right)_{ij}.
\end{eqnarray*}
Evaluating at ${\bf p}$, we get
$$
\left.\dfrac{\partial F( {\bf q})}{\partial q_l^k} \right|_{{\bf q}={\bf p}} = 
\left( 2 \frac{1}{\sqrt{2}}( \delta_i^k -\delta_j^k )
(\delta_i^l - \delta_j^l ) \right)_{ij}.
$$
Since $l\neq k$, we have 
$ \delta_i^k\delta_i^l\equiv 0$ and 
$\delta_j^k\delta_j^l \equiv 0$.
Since $l< k$, and $i<j$, we have 
$ \delta_i^k\delta_j^l \equiv 0$.
Hence 
$$
\left.\dfrac{\partial F( {\bf q})}{\partial q_l^k} \right|_{{\bf q}={\bf p}} = 
\left(  \sqrt{2} ( -\delta_j^k )
\delta_i^l \right)_{ij} = - \sqrt{2}  E_{lk}.
$$
This concludes the proof of Claim A.
\end{proof}

\section{Proofs of the main results}

In the next subsection we first give two simple geometric lemmas which allow us to prove our results for non-euclidean normed vector spaces. The proofs of Theorem~\ref{thm:main} and Theorem~\ref{thm:main1} rely on the Ramsey-theoretic fact that given sufficiently many points in $\mathbb R^n$
there must be a triple that forms as large angle as we want. We recall this result at the end of the subsection in Lemma~\ref{lma:angles}.
In the following two subsections we then prove Theorem~\ref{thm:main} and Theorem~\ref{thm:main1}.
In the final subsection we prove that if $h$ is a degenerate snowflaking function then there exists an infinite metric space whose $h$-snowflake isometrically embeds into $2$-dimensional Euclidean space.

\subsection{Geometric lemmas}

Throughout this section $V$ is an $n$-dimensional normed linear space with norm $\|\cdot\|$. 
For the remainder of this section we fix   an inner product $\langle\cdot,\cdot\rangle$ on $V$ for which John ellipsoid property holds. Namely, we have
\begin{equation}\label{John:eq}
B \subset B_{V} \subset \sqrt{n}B,
\end{equation}
where $B_{V}$ is the $\|\cdot\|$-unit ball and $B$ is the the unit ball in the $l_2$ metric associated to the inner product $\langle\cdot,\cdot\rangle$. For $u,v \in V$ we denote the length given by the inner product by $uv := \sqrt{\langle u-v,u-v\rangle}$.

Given two vectors $u,v \in V$ let $\angle(u,v)$ be their angle with respect to the above inner product $\langle\cdot,\cdot\rangle$:
\[\angle(u,v) := \arccos \dfrac{ \langle u,v\rangle}{ \langle u,u\rangle  \langle v,v\rangle}.\]
Given three points $x,y,z
\in V$ we set
$\angle_y(x,z) := \angle (x-y, z-y)$.

Let $x,y,z \in V$ and let  $z' := x + \langle z-x,y-x \rangle(y-x)$ be the projection of $z$ onto the line through $x$ and $y$.

\begin{lemma}\label{lma:closetoEuclidean}
There exist constants $    \varepsilon , C \in (0, \infty)$ 
 depending only on $n$ such that 

 if $\angle_x(y,z) < \varepsilon$ then
 \[
  \big|\|x-z\| - \|x-z'\|\big| \le C\|x-z'\|\angle_x(y,z).
 \]
\end{lemma}
\begin{proof}
By the John Ellipsoid Theorem, \eqref{John:eq}, there is an angle $\theta$ and a height $h$ only depending on $n$ such that
for any point $p$ on the sphere with respect to $\norm{\cdot}$. 
The cone $C^- (p, \theta, \ell, 0)$ at $p$ with opening $\theta$ and  height $\ell$ in the direction of $0$ is inside the ball, whereas
the  cone $C^+ (p, \theta, \ell, p)$ at $p$ with opening $\theta$ and  height $\ell$ in the direction of $p$ is outside the ball.
 

The inequality that we need to prove is translation and dilation invariant. So we assume that $x=0$ and that $z'$ lies on the unit sphere.
Let $\beta := \angle_x(y,z)$. We shall require $\beta  < \varepsilon < \frac\pi2$ for a small enough $\epsilon$.
Let $t, s, t_1, t_2, t_3$ be as in the picture.
\begin{figure}[!h]
  \centering
  \setlength{\unitlength}{0.1\textwidth}
  \begin{picture}(5,1.9)
  
   \put(0,2){\line(3,-1){6}}  
       \put(0,0){\line(1,0){6}} 
        \put(0,2){\line(0,-2){2}} 
                        \put(0,2){\line(1,0){2}} 
                            \put(2,2){\line(0,-2){2}} 
                            \put(1,2){\line(0,-2){2}} 
                                \put(6,-0.2){$x$}
                                                                \put(4.9,0.1){$\beta$}
           \put(1,0){\line(1,2){1}}
           \put(1,0){\line(-1,2){1}}
    \put(2,-0.22){$t_2$}
        \put(1,-0.22){$z'$}
                \put(0.7,0.05){$\theta$}
                \put(0,-0.22){$t_3$}
                    \put(-0.2,2){$t$}
                                        \put(2.1,2){$t_1$}
    \put(0.8,1.5){$z$}
        \put(1.6,1.5){$s$}
          \end{picture} 
  \caption{The point $w$ is between $t$ and $s$.}
\end{figure}

If $\epsilon$ is small enough, then
$\sin \theta \cos \beta - 2 \cos \theta \sin \beta > {1 \over 2}\sin\theta.$
Setting $r:= tz'$ we have
$$t_2y = t_3y- t_3t_2 = r \dfrac{\sin\theta}{\sin \beta} \cos \beta - 2 r \cos \theta> \dfrac{r\sin\theta}{2\sin\beta}.$$
Also, if $\epsilon$ is small enough, then the point $t$ is in $C^+ (p, \theta, \ell, p)$ and so $s\in C^- (p, \theta, \ell, 0)$.
In particular, between $s$ and $t$ there is a point $w$ such that
$\|x-w\| = \|x-z'\|$.
 
We conclude using, in order, the definition of $w$, the fact that on a line any two norms are a multiple of each other, the triangle inequality, the properties of $s$ and $t$, and the previous bound:
\begin{eqnarray*}
1& \leq &\dfrac{\|x-z\| }{ \|x-z'\|} =
\dfrac{\|x-z\| }{ \|x-w\|} 
= 
\dfrac{xz }{ xw} 
\leq 
\dfrac{xw +wz }{ xw}
=
1+ \dfrac{wz }{ xw} 
\leq  1+ \dfrac{ts }{ sy} \\
&\leq & 1+ \dfrac{tt_1 + tz' }{ t_2y}
\leq  1+ \dfrac{ 2r \cos \theta + r }{  \frac{r}{2}\sin\theta / \sin \beta}
\leq  1+   C_\theta \sin \beta, \\
\end{eqnarray*}
where $C_\theta  =  ( 4  \cos \theta +2) /  \sin\theta    $.
\end{proof}

The following bound is another easy consequence of the John Ellipsoid theorem.
\begin{lemma}\label{lemma234124}
There exists a constant $K>0$ depending only on $n$ such that 
when $  \angle_x(y,z),\angle_y(x,z) \in(0, \frac{\pi}{4})$ we have
 \begin{equation}\label{eq:anglecomparison}
  K^{-1}\frac{\|z'-y\|}{\|z'-x\|} \le \frac{\angle_x(y,z)}{\angle_y(x,z)} \le K \frac{\|z'-y\|}{\|z'-x\|}.
 \end{equation}

\end{lemma}
\begin{proof}
 Set $\alpha =   {\angle_x(y,z)}$, and $ \beta:={\angle_y(x,z)}$. By \eqref{John:eq}
 \begin{eqnarray*}
 \frac{\alpha}{\beta} &\le& \frac{\tan\alpha}{{1 \over 2}\tan\beta} 
  =2 \frac{ zz'/xz' }{  zz'/yz'} 
  =2 \frac{  yz'} { xz' }
 \le 2\frac{\sqrt{n} \|z'-y\|}{\|z'-x\|}.
 \end{eqnarray*}
By symmetry, we get the other inequality.
 \end{proof}
 
Our arguments rely on the following Ramsey-theoretic result. Explicit bounds on the number of points that one can have in $\mathbb R^n$ without forming an angle larger than a given bound can be found in \cite{MR841305}. A proof of Lemma~\ref{lma:angles} can also be found in \cite{MR2737262}.
 
 \begin{lemma}\label{lma:angles}
 For any $n \in \mathbb N$ and $0 < \beta < \pi$ there is an $N \in \mathbb N$ such that if $S \subseteq \mathbb{R}^n$ has cardinality at least $N$ then there are distinct $x,y,z \in S$ such that $\beta \leq \angle(xyz) \leq \pi$.
\end{lemma}
 


 
 

\subsection{Proof of Theorem~\ref{thm:main}}\label{sec:thm:main}

The proof of Theorem~\ref{thm:main} as well as the proof of Theorem~\ref{thm:main1} combines two observations: snowflaking forbids the formation of large angles whereas the fact that we have many points forces such angles to exist.
In the proof of Theorem~\ref{thm:main} the special form of the snowflaking function allows us to directly prove a bound on the cardinality of the snowflaked space that can be embedded in the normed vector space. 


\begin{proof}[Proof of Theorem~\ref{thm:main}]
Fix $n$ and $\alpha$ and let $C,\epsilon, K$ be given by Lemma~\ref{lma:closetoEuclidean} and Lemma~\ref{lemma234124}.
 Now take $\theta < \min\{\epsilon,\frac{\pi}{4},\frac{2-2^\alpha}{3C(2K+2^\alpha)}\}$.

 
We let $N \in \mathbb N$ be the constant in Lemma~\ref{lma:angles} with $\beta = \pi - \theta$. 
 We suppose towards a contradiction that $(X,d)$ has cardinality $N$ and that there is an isometric embedding $\iota$ of $(X,d^\alpha)$ into an $n$-dimensional normed vector space $(V,\|\cdot\|)$. 
 By Lemma~\ref{lma:angles} there exist three isometrically embedded points $x,y,z \in \iota(X)$ such that
 $\pi - \theta < \angle_z(x,y) \leq \pi$. 
 We declare $\delta_z := \angle_z(x,y).$
 We may assume that $\delta_z \ne \pi$. 
We also declare $\delta_x := \angle_x(y,z)$ and $\delta_y := \angle_y(x,z)$.
 Note that $0 < \delta_x,\delta_y < \theta$.
 
 
 Let $z' \in \mathbb R^n$ be the orthogonal projection of $z$ on the line passing through $x$ and $y$, i.e. $z' := x + \langle z-x,y-x \rangle(y-x)$.
Lemma~\ref{lma:closetoEuclidean} yields
 \begin{equation}\label{eq:dy}
  \left|\|x-z\| - \|x-z'\|\right| \le C\|x-z'\|\delta_x  
 \end{equation}
 and
 \begin{equation}\label{eq:dx}
  \left|\|y-z\| - \|y-z'\|\right| \le C\|y-z'\|\delta_y.
  \end{equation}
 
 We now estimate
 \[
  (\|y-z\|^\frac{1}{\alpha} + \|z-x\|^\frac{1}{\alpha})^{2\alpha}
 \]
 from above. 
 Subadditivity of $t \mapsto t^\alpha$ yields:
 \begin{align*}
   (\|y-z\|^\frac{1}{\alpha} + \|z-x\|^\frac{1}{\alpha})^{2\alpha} &= ((\|y-z\|^\frac{1}{\alpha} + \|z-x\|^\frac{1}{\alpha})^2)^{\alpha}\\
   &= (\|y-z\|^\frac2\alpha + \|z-x\|^\frac2\alpha + 2\|y-z\|^\frac1\alpha\|z-x\|^\frac1\alpha)^\alpha\\
   & \le \|y-z\|^2 + \|z-x\|^2 + 2^\alpha\|y-z\|\|z-x\|.
 \end{align*}
 Estimating the obtained terms from above using \eqref{eq:dx} and \eqref{eq:dy} we have
 \begin{align*}
  (\|y-z\|^\frac{1}{\alpha} + \|z-x\|^\frac{1}{\alpha})^{2\alpha} & \le 
 \|y-z'\|^2(1+C\delta_y)^2 + \|z'-x\|^2(1+C\delta_x)^2 \\
 & \quad\, + 2^\alpha \|y-z'\|\|z'-x\|(1+C\delta_x)(1+C\delta_y).
 \end{align*}
 Now we use the fact that $C\delta_x < 1$ and Lemma~\ref{lemma234124} to obtain
 \begin{align*}
  \|y-z'\|^2(1+C\delta_y)^2 & \le \|y-z'\|^2(1+3C\delta_y)
  \le \|y-z'\|^2\left(1+3CK\delta_x\frac{\|z'-x\|}{\|z'-y\|}\right)\\
  & \le \|y-z'\|^2 + 3CK\theta\|y-z'\|\|x-z'\|
 \end{align*}
 and similarly
 \[
  \|x-z'\|^2(1+C\delta_x)^2 \le \|x-z'\|^2 + 3CK\theta\|y-z'\|\|x-z'\|.
 \]
 Again by $C\delta_x < 1$ we get $(1+C\delta_x)(1+C\delta_y) < 1+3C\theta$.
 Collecting the estimates together and using the fact that $6CK\theta + 2^\alpha(1+3C\theta) < 2$ we have:
 \begin{align*}
  (\|y-z\|^\frac{1}{\alpha} + \|z-x\|^\frac{1}{\alpha})^{2\alpha} & \le \|y-z'\|^2 + 3CK\theta\|y-z'\|\|x-z'\|\\
  & \quad\, + \|x-z'\|^2 + 3CK\theta\|y-z'\|\|x-z'\|\\
  & \quad\, + 2^\alpha(1+3C\theta) \|y-z'\|\|z'-x\| \\
  & = \|y-z'\|^2 + \|x-z'\|^2\\
  & \quad\, + (6CK\theta + 2^\alpha(1+3C\theta))\|y-z'\|\|z'-x\|\\
  & < \|x-z'\|^2 + \|z'-y\|^2 + 2\|x-z'\|\|z'-y\|\\
   & = (\|x-z'\|+\|z'-y\|)^2 = \|x-y\|^2.
 \end{align*}

 Therefore
 \begin{align*}
  d(\iota^{-1}(x),\iota^{-1}(y)) & =  \|x-y\|^\frac{1}{\alpha} > \|y-z\|^\frac{1}{\alpha} + \|z-x\|^\frac{1}{\alpha}\\
  & =  d(\iota^{-1}(z),\iota^{-1}(y)) +  d(\iota^{-1}(x),\iota^{-1}(z)).
 \end{align*}
This contradicts the triangle inequality in $(X,d)$.
\end{proof}




\subsection{Proof of Theorem~\ref{thm:main1}}\label{sec:thm:main1}

In the proof of Theorem~\ref{thm:main1} we use the same geometric lemmas (Lemma~\ref{lma:closetoEuclidean} and Lemma~\ref{lemma234124}) as in the proof of Theorem~\ref{thm:main}.
However, in the proof of Theorem~\ref{thm:main1} the choice of a sequence of points giving the contradiction depends not only on the snowflaking function $h$, but also on the first element of the sequence. Therefore no upper bound (depending on $h$ and $n$) on the number of points that can be snowflake embedded can in general be obtained in Theorem \ref{thm:main1}.

\begin{proof}[Proof of Theorem~\ref{thm:main1}]

Suppose to the contrary that $X$ is infinite and that there exists an isometric embedding $\iota \colon (X,h\circ d) \to V$ where $V$ is an $n$-dimensional normed vector space. We divide our proof into two cases.
An infinite bounded subset of $\mathbb{R}^n$ is not discrete, so one of the following holds:
\begin{enumerate}
 \item[(i)] $\iota(X)$ is unbounded;
   \item[(ii)] $\iota(X)$ is not discrete.
\end{enumerate}

If (i) holds we will arrive at a contradiction with the condition (S4) of a snowflaking function. If (ii) holds, a contradiction follows with (S3).

\textbf{Case (i): Suppose $\iota(X)$ is unbounded}\\
Observe that (S4) implies that the existence of a function $T \colon \mathbb R_{\ge} \to \mathbb R_{\ge}$ such 
that for any $t > 0$ and $S \ge T(t)$ we have
\begin{equation}\label{tapio1}
 \frac{t}{h(t)}\frac{h(S)}{S} \le \frac12.
\end{equation}
Combining \eqref{tapio1} with (S1) and (S2) we get 
\begin{equation}\label{tapio2}
 h(S+t) \le h(S) + t\frac{h(S)}{S} = h(S) + \frac{t}{h(t)}\frac{h(S)}{S} h(t)
 \le h(S) + \frac12 h(t).
\end{equation}
Now fix $x_0,x_1 \in X$, $x_0\ne x_1$. Since $(X,d)$ is unbounded, there exists a point $x_2 \in X$ with
 $\angle_{\iota(x_2)}(\iota(x_0), \iota(x_1) )\leq \pi/4$
 and
$d(x_2,x_i) > T(d(x_0,x_1))$ for $i = 0,1$. We continue inductively. Suppose $(x_i)_{i=0}^{N-1} \subset X$ have been chosen.
Now we select $x_N \in X$ satisfying 
\begin{equation}\label{iota_x}
\angle_{\iota(x_N)}(\iota(x_i), \iota(x_j) )\leq \pi/4
\text{  and }
 d(x_i,x_N) > T(d(x_i,x_j)),\quad\forall i,j <N .
 \end{equation}

Let $\epsilon, C , K$ be the constants from Lemma~\ref{lma:closetoEuclidean} and Lemma~\ref{lemma234124}. 
Set 
\begin{equation}\label{eq:delta}
 \delta=\min\{ \epsilon, \pi/4, \frac{1}{2C(1+K)} \}.
\end{equation}
By Lemma~\ref{lma:angles} there exist 
$x,y,z$ in $\{\iota(x_\ell)\}_{\ell\in \mathbb N}$ such that $\angle_{z}(x,y) > \pi - \delta$.
By the condition \eqref{iota_x}, there exist 
$ i, j, k \in \mathbb N$ with $k > \max\{i,j\}$ such that
$x=\iota(x_j)$,
$y=\iota(x_k)$,
$z= \iota(x_i)$.
Let $z'$ be the orthogonal projection of $z$ to the line passing through
$x$ and $y$. On the one hand, we have that
\begin{align*}
 h(d(x_j,x_k)) & = \|x-y\| \\
 & = \|x-z'\| + \|z'-y\|\\
(\text{Using  Lemma~\ref{lma:closetoEuclidean}})\quad & \ge \|x - z\| - C\|x-z'\|\angle_{x}(z,y)\\
 & \hspace{2cm} +  \|z - y\| - C\|y-z'\|\angle_{y}(z,x)\\
 (\text{Using  Lemma~\ref{lemma234124}})\quad & \ge \|x - z\| - C\|x-z'\|\angle_{x}(z,y)\\
 & \hspace{2cm} +  \|z - y\| - CK\|x-z'\|\angle_{x}(z,y)\\
 & = \|x - z\| + \|z - y\| - C(1+K)\|x - z'\|\angle_{x}(z,y)\\
  & \ge \|x - z\| + \|z - y\| - C(1+K)\|x - z\|\angle_{x}(z,y)\\
 & = h(d(x_i,x_k)) + (1-C(1+K) \angle_{x}(z,y)) h(d(x_i,x_j))\\
 (\text{Using  that } \angle_{x}(z,y)< \delta)   \quad & > h(d(x_i,x_k)) + \frac12 h(d(x_i,x_j)).
\end{align*}
On the other hand, 
first notice that since 
$h$ is concave and positive, then $h$ has to be weakly increasing.
Secondly, by the definition of the function $T$ we then have that for any three $ i, j, k \in \mathbb N$ with $k > \max\{i,j\}$  
\begin{equation}\label{eq:a_bit_shorter}
 h(d(x_j,x_k)) \le h(d(x_i,x_k)+d(x_i,x_j)) \le h(d(x_i,x_k)) + \frac12 h(d(x_i,x_j)).
\end{equation}
Therefore we have a contradiction.


\textbf{Case (ii): Suppose $\iota(X)$ is not discrete}\\
This time we observe that (S3) implies the existence of a function $\tilde T \colon \mathbb R_{\ge} \to \mathbb R_{\ge}$ such 
that $\tilde T(r) \le r$ for all $r$ and for any $S > 0$ and $0 < t \le \tilde T(S)$ we have
\eqref{tapio1}, and hence \eqref{tapio2}, using (S1) and (S2). 


Let $y$ be an accumulation point of $X$. First we select $x_0 \in X \setminus \{y\}$.
Next we take a radius $r_0 > 0$ so that for all $y_1,y_2 \in B(y,r_0)$ we have both
$\angle_{\iota(x_0)}(\iota(y_1), \iota(y_2) )\leq \pi/4$
 and
$d(y_1,y_2) < \tilde T(d(x_0,y_i))$ for $i = 0,1$. Now we select a point $x_1 \in B(y,r_1) \setminus \{y\}$.
We continue inductively. Suppose $(x_i)_{i=0}^{N-1} \subset X$ have been chosen.
Now we take a radius $r_{N-1} < r_{N-2}$ such that for all $y_1,y_2 \in B(y,r_0)$ we have
\begin{equation}\label{iota_x_new}
\angle_{\iota(x_i)}(\iota(y_1), \iota(y_2) )\leq \pi/4
\text{  and }
 d(y_1,y_2) < \tilde T(d(x_i,y_j)),\quad\forall i <N, j = 1,2.
 \end{equation}
Then we select a point $x_N \in B(y,r_{N-1}) \setminus \{y\}$.

With the points $\{x_i\}$ we now arrive at a contradiction with the same argument as in the case (i).
Let $\delta$ be as in \eqref{eq:delta}. Again by Lemma~\ref{lma:angles} there exist 
$x,y,z$ in $\{\iota(x_\ell)\}_{\ell\in \mathbb N}$ such that $\angle_{z}(x,y) > \pi - \delta$,
but this time by the condition \eqref{iota_x_new}, there exist 
$ i, j, k \in \mathbb N$ with $k < \min\{i,j\}$ such that
$x=\iota(x_j)$,
$y=\iota(x_k)$,
$z= \iota(x_i)$. Now we continue verbatim the proof in the case (i).
\end{proof}


\subsection{Necessity of (S3) and (S4)}

We end this paper by showing that the conditions (S3) and (S4) of generalized snowflakes are indeed needed for Theorem \ref{thm:main1} to hold.

\begin{proposition}\label{prop:necessity}
 Suppose $h \colon \mathbb R_\ge \to \mathbb R_\ge$ satisfies (S1) and (S2) but fails to satisfy (S3) or (S4). Then there is an
 infinite metric space $(X,d)$ such that $(X,h \circ d)$ admits an isometric embeddeding into $2$-dimensional Euclidean space.
\end{proposition}
\begin{proof}
 We only treat the case where (S4) fails. The case where (S3) fails follows in a similar way.
 
We construct a sequence of points $\{x_i\}_{i = 0}^{\infty}$ in $\mathbb R^2$ such that $(X,d) := (\{x_i\}_{i = 1}^{\infty},h^{-1}\circ d_E)$
 is a metric space. For this purpose we fix a sequence $\{\alpha_i\}_{i=1}^\infty$ of positive angles such that $\sum_{i=1}^\infty \alpha_i < \frac{\pi}{2}$. 
 Depending on the sequence $\{\alpha_i\}_{i = 1}^{\infty}$ and the function $h$ we construct an increasing sequence $\{t_i\}_{i=1}^\infty$ of positive real numbers which determine the Euclidean distance between $x_{i-1}$ and $x_i$. 
 For notational convenience we let $c(t) = \frac{h(t)}t$. Notice that by assumption $c(t) \searrow c > 0$ as $t \to \infty$.
 This allows us to select for every $i \in \mathbb N$ a real number $t_i > 0$ such that for all $s,t \ge t_i$ we have
 \begin{equation}\label{eq:tichoice}
  c(s)(c(s+t)-c) + c(t)(c(s+t)-c) \le 2(c(s)c(t) \cos(\pi-\alpha_i) + c(s+t)^2).
  \end{equation}

 
 Now, using the sequences $\{\alpha_i\}_{i = 1}^{\infty}$ and $\{t_i\}_{i = 1}^{\infty}$ we define the sequence $\{x_i\}_{i \in \mathbb{N}}$ as follows.
 We set 
 $x_0 := (0,0)$, $x_1 := (t_1,0)$, and  inductively for $n \geq 2$ declare
 \[
  x_n := x_{n-1} + \left(t_n\sin(\sum_{j=1}^{n-1}a_j),t_n\cos(\sum_{j=1}^{n-1}a_j)\right).
 \]

 In order to see that $(X,d)$ is a metric space we need to check that the triangle inequality holds. For this purpose let $0 \le i < j < k$ be three integers. Let $d_E$ be the Euclidean metric.
 The only nontrivial inequality that we have to verify is
 \[
  h^{-1}(d_E(x_i,x_k)) \le   h^{-1}(d_E(x_j,x_k)) +   h^{-1}(d_E(x_j,x_k)).
 \]
 Denoting $s :=   h^{-1}(d_E(x_i,x_j))$ and $t :=   h^{-1}(d_E(x_j,x_k))$ the above inequality is equivalent to
 \begin{equation}\label{eq:wewant}
  d_E(x_i,x_k) \le h(s+t).  
 \end{equation}
 By (S1) and (S2) we can estimate
 \begin{equation}\label{eq:cestim}
  c(t) = \frac{h(t)}t \le \frac{h(t+s) - cs}{t} = c(t+s) + (c(t+s) - c)\frac{s}{t}.
 \end{equation}
 Since $s,t \ge t_j$, by applying the law of cosines,
 \eqref{eq:cestim} and \eqref{eq:tichoice} we obtain
 \begin{align*}
  d_E(x_i,x_k)^2 - h(s+t)^2 & = d_E(x_i,x_j)^2 + d_E(x_j,x_k)^2 \\
  & \qquad - 2d_E(x_i,x_j)d_E(x_j,x_k)\cos(\angle_{x_j}(x_i,x_k)) - h(s+t)^2\\
  & \le h(s)^2 + h(t)^2 - 2h(s)h(t)\cos(\pi-\alpha_j) - h(s+t)^2\\
  & = s^2c(s)^2 + t^2c(t)^2 - 2stc(s)c(t) \cos(\pi-\alpha_j) - (s+t)^2c(s+t)^2\\
  & = s^2(c(s)^2-c(s+t)^2) + t^2(c(t)^2 - c(s+t)^2)\\
 & \qquad - 2st(c(s)c(t) \cos(\pi-\alpha_j) + c(s+t)^2)\\
 \text{(Using \eqref{eq:cestim})} \qquad  & \le st\big(c(s)(c(s+t)-c) + c(t)(c(s+t)-c)\\
  & \qquad- 2(c(s)c(t) \cos(\pi-\alpha_j) + c(s+t)^2)\big)\\
  \text{(Using \eqref{eq:tichoice})} \qquad & \le 0
 \end{align*}
 and thus \eqref{eq:wewant} holds.
\end{proof}

\begin{remark}\label{rmk:nobound}
 The proof of Proposition \ref{prop:necessity} can be modified to show that there is a snowflake function $h$ (satisfying all the conditions (S1)--(S4)) such that for every $n \in \mathbb N$ there exists a metric space $(X_n,d_n)$ with cardinality $n$ so that $(X_n,h\circ d_n)$ embeds 
 isometrically into $2$-dimensional Euclidean space.
 
 Indeed, suppose we are given $n \in \mathbb N$ and we have already defined $h$ on $[0,T_{n-1}]$ and that the slope of $h$ at $T_{n - 1}$ is $c_{n-1} > 0$. Then we can define $h$ on an arbitrary long interval $[T_{n-1},T_n]$ as $h(t) = H(T_{n-1}) + c_nt$, where $0 < c_n < c_{n-1}$. By the proof of Proposition \ref{prop:necessity},
 by taking $S$ large enough, there exist $n$ points in $\mathbb R^2$ with the distance between any two of them between $h(T)$ and $h(S)$ so that 
 they are an $h$-snowflake of some metric space. We can define $h$ near $0$ so that it satisfies (S3) and if we take $c_n \searrow 0$ then (S4) is also satisfied. Therefore $h$ will have all the required properties.
\end{remark}


 