\section{Introduction}
Pcf theory was developed by Shelah, see \cite{shelah},  to answer some cardinal arithmetic questions, though it latter found wider applications in
set theory and other parts of mathematics. A key concept in pcf theory is that of $\pcf(A)$, for a set $A$ of regular cardinals, which is defined as
\[
\pcf(A)=\{\cf(\prod A, <_D): D \text{~an ultrafilter on~} A   \}.
\]
The set $\pcf(A)$ behaves very nice if the set $A$ is an interval of regular cardinal and is regressive, meaning that $\min(A) > |A|.$ In the absence of such an assumption, the set
$\pcf(A)$ lacks some of its nice properties, see for example \cite{jech} and \cite{kenta}.

Let $\Reg$ denote the class of regular cardinals. For a set $A \subseteq \Reg$ let us define the increasing sequence
 $\langle  \pcf^\alpha(A): \alpha \in \Ord     \rangle$ by induction on $\alpha$ as follows:
\begin{center}
 $\pcf^{\alpha}(A)=$ $\left\{
\begin{array}{l}
         A \hspace{3.5cm} \text{ if } \alpha=0,\\
         \pcf(\pcf^\beta(A))  \hspace{1.5cm} \text{ if } \alpha=\beta+1,\\
         \bigcup_{\beta < \alpha} \pcf^\beta(A) \hspace{1.4cm} \text{ if } \lim(\alpha).
     \end{array} \right.$
\end{center}
In \cite{kenta}, Tsukuura showed that if there are infinitely many supercompact cardinals, then in a forcing extension, for a set $A$ of regular cardinals, the sequence $\langle  \pcf^n(A): n<\omega     \rangle$
can be strictly increasing and he raised the following problem.

\begin{question}
\label{a1} Is it a theorem of ZFC that for every $A \subseteq \Reg$ there is an $\alpha$ such that
$\pcf^{\alpha+1}(A)=\pcf^\alpha(A)$.
\end{question}
In this paper we give a consistent negative answer to this question, by proving the following theorem.
\begin{theorem}
\label{a2} Assume $\langle \kappa_n: n<\omega \rangle $ is an increasing sequence of indestructible  supercompact cardinals  and  the set $A=\{\alpha < \kappa_0: \alpha$ is supercompact    $ \}$ belongs to any normal measure on $\kappa_0$. Then we can find a model of set theory in which the sequence
$\langle  \pcf^\alpha(A): \alpha \in \Ord     \rangle$
is strictly increasing.
\end{theorem}



\section{Some preliminaries}
\label{s1}
In this section we present a few lemmas that we will need for the proof of
Theorem \ref{a2}.

\begin{lemma} (Laver \cite{laver})
\label{a3} Suppose $\kappa$ is a supercompact cardinal and $\lambda < \kappa$.
Then there exists a $\lambda^+$-directed closed $\kappa$-c.c. forcing notion
$\mathbb{L}(\lambda, \kappa)$ of size $\kappa$
such that in $V^{\mathbb{L}(\lambda, \kappa)}$, the supercompactness of $\kappa$
remains indestructible under $\kappa$-directed closed forcing notions.
\end{lemma}
The next lemma shows that for a measurable cardinal $\kappa$, the set $\pcf(\kappa \cap \Reg)$ is already large and
also discusses the structure of $\pcf(A)$ in nice forcing extensions.
\begin{lemma} (see \cite{kenta})
\label{a4}
\begin{enumerate}
\item  If $\kappa$ is a measurable cardinal, then $\pcf(\kappa \cap \Reg)= (2^\kappa)^+ \cap \Reg$,

\item Suppose $A \subseteq \Reg$, $\lambda=\min(A)$ and $\bbQ$ is a $\lambda$-c.c.c forcing notion. Then
$\pcf^V(A) \subseteq \pcf^{V[\bold G_{\bbQ}]}(A).$
\end{enumerate}
\end{lemma}
The next lemma is due to Shelah.
\begin{lemma} (see \cite{shelah})
\label{a5} Suppose $\lambda$ is a singular cardinal. Then $\lambda^+ \in \pcf(\lambda \cap \Reg)$.
\end{lemma}




\section{Proof of Theorem \ref{a2}}
In this section we prove our main result by building a model of ZFC in which the sequence
$\langle  \pcf^\alpha(A): \alpha \in \Ord     \rangle$ is strictly increasing.

Thus suppose that $\text{GCH}$ holds and  $\langle \kappa_n: n<\omega   \rangle $
is an increasing sequence of supercompact cardinals and let $A$ be defined as Theorem \ref{a2}.   For notational simplicity set
 $\kappa=\kappa_0$.
 \\
 {\bf Step 1.} By a suitable preparation (see for example \cite{apter}), there exists a full support iteration of the form
 \[
\bar{\bbP}^1= \langle \langle  \bbP^1_n: n \leq \omega       \rangle, \langle  \name{\bbQ}^1_n: n<\omega     \rangle\rangle,
\]
where
\begin{enumerate}
\item[(1-1)] $\bbP^1_1 \simeq \bbQ^1_0$ is an iteration  such that in the generic extension by it,
the set
$A'=\{\alpha < \kappa: \alpha$ is $< \kappa$-supercompact and indestructible under $\alpha$-directed closed forcing notions from
$V_\kappa \}$
belongs to any normal measure on $\kappa$,


\item[(1-2)] after forcing with  $\bbP^1_1,$ $\kappa$ becomes supercompact and indestructible and $2^\kappa=\kappa^+$,


\item[(1-3)] for $n<\omega, $  $\Vdash_{\bbP^1_n}$``$\name{\bbQ}^1_{n+1}=\mathbb{L}(\kappa_{n}, \kappa_{n+1})$'', where $\mathbb{L}(\kappa_{n}, \kappa_{n+1})$ is as in Lemma \ref{a3}.
\end{enumerate}
Let $\bold V^1=\bold V[\bold G_{\bbP^1_\omega}]$. Thus in $\bold V^1$,
each $\kappa_n, n<\omega$, is supercompact and Laver indestructible and  $2^{\kappa_n}=\kappa_n^+$.
\\
{\bf Step 2.} Work in $\bold V^1$ and force with the full support product
$\bbP^2=\prod_{n<\omega} \Add(\kappa_n, \kappa_{n+1})$. In the generic extension $\bold V^2=\bold V^1[\bold G_{\bbP^1}]$, the cardinal $\kappa$
still remains supercompact and  $2^{\kappa_n}=\kappa_{n+1}$ for each  $n<\omega$. Furthermore for a measure one set $A''$
 of $\beta < \kappa$ we can assume that:
\begin{enumerate}
\item[(2-1)] $\beta$ is supercompact in $\bold V,$

\item[(2-2)]  we can write $\bbP^1_1$ as
\[
\bbP^1_0 \simeq \bbP^{1}_0(< \beta) \ast \name{\bbP}^1_0(\beta) \ast \name{\bbP}^1_{0}(> \beta),
\]
where
\begin{enumerate}
\item $\bbP^1_0(<\beta)$ is $\beta$-c.c. of size $\beta$,

\item $\bbP^1_0(< \beta)$ forces $\beta$ is $< \kappa$-supercompact and indestructible under $\beta$-directed closed forcing notions from
$V_\kappa$,
\item $\Vdash_{\bbP^1_0(<\beta)}$``$\name{\bbP}^1_0(\beta)=\prod_{n<\omega}\Add(\zeta_{\beta, n}, \zeta_{\beta, n+1})$'',
where $\langle \zeta_{\beta, n}: n<\omega     \rangle$ is an increasing sequence of element of $A$ with $\zeta_{\beta, 0}=\beta$,

\item $\Vdash_{\bbP^1_0(<\beta)}$``the measurability of $\zeta_{\beta, n}$ is indestructible under $\zeta_{\beta, n}$-directed closed forcing notions from $V_\kappa$'',


\item $\Vdash_{\bbP^{1}_0(< \beta) \ast \name{\bbP}^1_0(\beta)}$``$\name{\bbP}^1_{0}(> \beta)$ is
$\mu$-directed closed, where $\mu \in A$ is above $\sup_{n<\omega}\zeta_{\beta, n}$.
\end{enumerate}
\end{enumerate}
Note that this is true for the sequence $\langle  \kappa_n: n<\omega   \rangle$,
and so by supercompactness of $\kappa$, those properties reflect done to a large set.
\\
{\bf Step 3.} Working in the model $\bold V^2$, by \cite{foreman},
there exists a forcing notion $\bbR$ such that the following properties are hold:
\begin{enumerate}
\item $(\bbR, \leq, \leq^*)$ is a Prikry type forcing notion,

\item $\bbR$ is $\kappa^+$-c.c.,

\item $\kappa$ remains inaccessible in $\bold V^2[\bold G_{\bbR}],$

\item forcing with $\bbR$ adds a club $C$ of $\kappa$ consisting of elements of $A''$,

\item suppose $\alpha < \beta$ are two successive points in $C$. Then $\beth_5(\alpha)=2^{\beth_4(\alpha)}=\beta,$

\item (factorization property)\footnote{Indeed the claim as stated is not quite exact, as we need to consider $\bbR/p$ for a suitable chosen $p$, but we ignore it as it does not affect the truth of our result.} suppose $\alpha \in C$ and $\beta =\min(C \setminus (\alpha+1))$. Then
\[
\bbR \simeq \bbR^{< \alpha} \times \prod_{i<4} \Add(\beth_i(\alpha), \beth_{i+1}(\alpha)) \times \Add(\beth_4(\alpha), \beta) \times \bbR^{>\beta},
\]
where
\begin{enumerate}
\item $\bbR^{< \alpha}$ is $\alpha^+$-c.c.,

\item $(\bbR^{> \beta}, \leq^*)$ is $\beta^+$-closed
\end{enumerate}

\end{enumerate}
Let $\bold V^3=\bold V^2[\bold G_{\bbR}]$.
\\
{\bf Step 3.}
We will need the following  lemma.
\begin{lemma} (in $\bold V^3$)
\label{a7} Suppose $\beta$ is a singular limit point of $C$. Then
\[
\pcf( \beta \cap \Reg) = \beth_1(\beta)^+ \cap \Reg.
\]
\end{lemma}
The proof is essentially the same as in \cite{jech} ( see also \cite{kenta}).
\\
{\bf Step 4.} In this step we complete the proof of Theorem \ref{a2}.
Let $\langle  \theta_\xi: \xi < \kappa    \rangle$
be an increasing enumeration of $C$ and set $\theta_\kappa=\kappa.$ Let $\sigma(*) \leq \kappa$
be the least ordinal bigger than $0$ such that $\theta_{\sigma(*)}$ is an inaccessible cardinal. We show that the model
 $\bold M= \bold V^3_{\theta_{\sigma(*)}}$
is as required.
To see this, let  $A=\theta_0 \cap \Reg.$ It suffices to show that
\[
\langle  \pcf^\alpha(A): \alpha < \sigma(*)   \rangle
\]
is strictly increasing.


Suppose $\alpha < \sigma(*)$  and write it as $\alpha=\delta+n$, where $\delta < \sigma(*)$ is a limit ordinal and $n<\omega$.
We show by induction that:
\begin{itemize}
\item if $n >0$, then
\[
\pcf^\alpha(A)= \beth_{n}(\theta_{\delta})^+ \cap \Reg.
\]
\item
if $\alpha$ is a limit ordinal (including $\alpha=0$),
then
\[
\pcf^\alpha(A)= \theta_\alpha \cap \Reg.
\]
\end{itemize}
The claim is true for $\alpha=0$  and if $\alpha$ is a limit ordinal, then by the induction hypothesis,
\[
\pcf^\alpha(A)=  \bigcup_{\xi < \alpha, \lim(\xi)}\bigcup_{n<\omega} \pcf^{\xi+n}(A) = \bigcup_{\xi < \alpha, \lim(\xi)}\bigcup_{n<\omega}
\beth_{n}(\theta_{\xi})^+ \cap \Reg=\theta_\alpha \cap \Reg.
\]
Now suppose the claim for $\alpha$. If $\alpha$ is a limit ordinal, then by our choice of $\sigma(*)$, the cardinal $\theta_\alpha$ is singular and hence  by Lemma \ref{a7},
\[
\pcf^{\alpha+1}(A) =\pcf(\theta_\alpha \cap \Reg) = \beth_{1}(\theta_{\alpha})^+ \cap \Reg.
\]
Otherwise $\alpha=\delta+n$ for some $n>0$
and using Lemma \ref{a4}, and the factorization property (6) we can easily conclude, as in \cite{kenta}, that
\[
\pcf^{\alpha+1}(A) =\pcf(\beth_{n}(\theta_{\delta})^+ \cap \Reg) = \beth_{n+1}(\theta_{\delta})^+ \cap \Reg.
\]
Indeed, let $n=5m+k$, where $k<5$.  Then $\beth_n(\theta_\delta)=\beth_k(\theta_{\delta+m})$. Let $\beta=\theta_{\delta+m}$, so 
\begin{itemize}
\item $\beta \in C,$
 \item for $i \leq 4$, $\beth_{i}(\theta_{\delta+m})=\zeta_{\beta, i}$, \footnote{See (2-2)(c).}
 \item $\beth_{5}(\theta_{\delta+m})=\theta_{\delta+m+1}$,
 
 \item we can write $\bbP^1_\omega \ast \bbP^2 \ast \bbR$ as
 \[
 \bbP^1_\omega \ast \bbP^2 \ast \bbR \simeq \bbP^1_0(< \beta) \ast \bigg(\bbR^{< \theta_{\delta+m}} \times \big(\prod_{i<4} \Add(\zeta_{\beta, i}, \zeta_{\beta, i+1}) \times
\Add(\zeta_{\beta+4}, \theta_{\delta+m+1}) \big) \bigg) \ast \name{\mathbb{K}},
 \]
 where $\name{\mathbb{K}}$ is forced to add no new subsets to $\beth_{n}(\theta_{\delta})^+.$
\iffalse
 \item   we have
$$\bbR \simeq \bbR^{< \theta_{\delta+m}} \times \bigg(\prod_{i<4} \Add(\zeta_{\beta, i}, \zeta_{\beta, i+1}) \times
\Add(\zeta_{\beta+4}, \theta_{\delta+m+1}) \bigg) \times \bbR^{> \theta_{\delta+m+1}},$$
where $\bbR^{< \theta_{\delta+m}}$ is $\theta_{\delta+m}$-c.c. and forcing with $\bbR^{> \theta_{\delta+m+1}}$
does not add any new bounded subsets to $\theta_{\delta+m+1}$.
\fi
\end{itemize}
Let us first  take care of the forcing notion
$$\bbP^1_0(< \beta) \ast \bigg(\bbR^{< \theta_{\delta+m}} \times \big(\prod_{i<4} \Add(\zeta_{\beta, i}, \zeta_{\beta, i+1}) \times
\Add(\zeta_{\beta+4}, \theta_{\delta+m+1}) \big) \bigg).$$
Now in $\bold V^1[\bold G_{\bbP^1_0(< \beta) \ast  (\prod_{k\leq i<4} \Add(\zeta_{\beta, i}, \zeta_{\beta, i+1}) \times
\Add(\zeta_{\beta+4}, \theta_{\delta+m+1}))}]$,
the cardinal $\zeta_{\beta+k}$ is measurable and  hence by Lemma \ref{a4}, in the above model,
\[
\pcf(\beth_{n}(\theta_{\delta})^+ \cap \Reg)=\pcf(\zeta_{\beta, k}^+ \cap \Reg)=\zeta_{\beta, k+1}^+  \cap \Reg = \beth_{n+1}(\theta_{\delta})^+ \cap \Reg
\]
if $k<4$ and
\[
\pcf(\beth_{n}(\theta_{\delta})^+ \cap \Reg)=\pcf(\zeta_{\beta, 4}^+ \cap \Reg) = \theta_{\delta+m+1}^+ \cap \Reg= \beth_{n+1}(\theta_{\delta})^+ \cap \Reg
\]
if $k=4.$ As the forcing notion
 $\bbP^1_0(< \beta) \ast \big(\bbR^{< \theta_{\delta+m}} \times (\prod_{i<k} \Add(\zeta_{\beta, i}, \zeta_{\beta, i+1}))$
 is $\zeta_{\beta, k}$-c.c., by Lemma \ref{a4}, the above equalities continue to hold in 
 $$\bold V[\bold G_{\bbP^1_0(< \beta) \ast (\bbR^{< \theta_{\delta+m}} \times (\prod_{i<4} \Add(\zeta_{\beta, i}, \zeta_{\beta, i+1}) \times
\Add(\zeta_{\beta+4}, \theta_{\delta+m+1}) ))}].$$
Finally as $\mathbb{K}$ has enough closure properties, the above equalities still continue to hold in
$\bold V^3.$
We are done.
\begin{remark}
The cardinal $\theta_0$ can be taken to be $(\theta_0+2)$-strong, so given any regular cardinal $\eta < \theta_0$,
if we force with Magidor forcing to change the cofinality of
$\theta_0$ into $\eta$,
then still the above results hold in the extension and in the resulting model,
$\theta_0$ is a singular cardinal of cofinality $\eta.$
\end{remark}










