Remember the Pressing Down Lemma:
Let $k$ be a regular uncountable cardinal, $S\subseteq k$ be a
stationary set and let $f:S\to k$ be such that $f(\gamma)<\gamma$ for
every $\gamma \in S$ (such a function is called a regressive
function). Then there exists an $\alpha<k$ such that $f^{-1}(\{\alpha
\})$ is stationary.
In the article Barely Baire spaces of Fleissner and Kunen the following lemma appears:
Lemma 4 Let $\chi>\omega$ be a regular cardinal. If $K\subseteq {\chi}^{\omega}$ is closed, and $W=\{f^{*}: f\in K \}$ is stationary, then there is $C$ club in $\chi$ such that $C\cap C_{\omega}\chi \subseteq W $
Where, $C_{\omega} \chi$ is the subset of $\chi$ of ordinals of cofinality $\omega$. Also, if $cf (\chi)> \omega$, we can define a map $*:{\chi}^{\omega}\to \chi$, where $*(f)=f^{*}$ is the least $\alpha$ greater than $f(n)$ for all $n\in\omega$.
For the proof, the authors consider $\sigma\in \bigcup_{n\in\omega}\chi^{n}$ and $W_{\sigma}=\{f^{*} : \sigma\subseteq f\in K\}$. Then consider $\Sigma=\{\sigma : W_{\sigma} \hspace{0.1cm}\mbox{is stationary}\hspace{0.1cm}\}$. By hypothesis $\Sigma\not=\emptyset$, because $\emptyset\in \Sigma$.
Then the authors affirm the following,
Using the Pressing Down Lemma one can build a function $\theta: \Sigma \times \chi \to \Sigma$ such that
- $\sigma\subseteq \theta(\sigma, \alpha)$
- $\theta(\sigma, \alpha)\not\in \bigcup_{n\in\omega}\alpha^{n}$
Continuing with the article of Fleissner and Kunen, they consider $C=\{ \gamma < \chi: \theta[ (\Sigma\cap \gamma^{<\omega} ) \times \gamma ]\subseteq \gamma^{<\omega} \}$ and they said that $C$ is a club.
Indeed,
- $C$ is closed.
Let $\gamma\in C^{\prime}$, we will show that $\theta[ (\Sigma\cap \gamma^{<\omega} ) \times \gamma ]\subseteq \gamma^{<\omega}$. Let $(\sigma, \alpha)\in (\Sigma\cap \gamma^{<\omega} ) \times \gamma$, so there is $n_{0}\in\omega$ such that $\sigma\in \gamma^{n_{0}}$, consider $m=\max\{\sigma(n_{0}-1), \alpha \}<\gamma$ then there exists $\beta \in ]m, \gamma+1[ \cap (C \setminus \{\gamma\})$, so $\alpha<\beta<\gamma$ and $\sigma\in \beta^{<\omega}$, then
$\theta(\sigma, \alpha) \in \theta[ (\Sigma\cap \beta^{<\omega} ) \times \beta ]\subseteq \beta^{<\omega} \subseteq \gamma^{<\omega}$. Therefore $C^{\prime} \subseteq C$, that is, $C$ is closed.
- $C$ is unbounded.
For this, define
$$
\begin{array}{lcccl}
f & : & \chi & \longrightarrow & \chi\\
& & \gamma & \longrightarrow & f(\gamma)=\sup\{\theta^{*}(\sigma, \alpha) :\sigma \in \Sigma\cap\gamma^{<\omega}, \alpha<\gamma \},
\end{array}
$$
where $\theta^{*}(\sigma, \alpha)=\sup(ran (\theta(\sigma, \alpha)))$, note that $f$ is well defined, that is, $f(\gamma)=\sup\{\theta^{*}(\sigma, \alpha) :\sigma \in \Sigma\cap\gamma^{<\omega}, \alpha<\gamma \}<\chi$, because $\chi$ is an uncountable regular cardinal.
Previously remember the following fact:
Proposition 1. Let $\kappa$ be an uncountable regular cardinal and $f:\kappa\to\kappa$ be a function. Then $\{\alpha<\kappa : f[\alpha]\subseteq \alpha\}$ is a club in $\kappa$.
Then, by Proposition 1, $\{\gamma<\chi:f[\gamma]\subseteq \gamma\}$ is a club in $\chi$, then $$\tilde{C}= \{\gamma<\chi: \gamma\hspace{0.1cm} \mbox{is a limit ordinal}\hspace{0.1cm} \mbox{and}\hspace{0.1cm} f[\gamma]\subseteq \gamma\}$$ is a club in $\chi$. Note that $\tilde{C}\subseteq C$. Indeed, let $\gamma\in\tilde{C}$ and let $(\sigma, \alpha)\in (\Sigma\cap \gamma^{<\omega})\times \gamma$, as $\gamma$ is a limit ordinal, there is $\alpha<\beta<\gamma$ such that $\sigma\in \beta^{<\omega}$ then $\theta^{*}(\sigma, \alpha)\leq f(\beta)<\gamma$, so $\theta(\sigma, \alpha)\in \gamma^{<\omega}$.
Finally it is commented that $$C\cap C_{\omega}\chi \subseteq W$$
where $C_{\omega}\chi=\{\beta<\chi : cf(\beta)=\omega\}$.
I tried to demonstrate this last part in the following way.
Let $\gamma\in C\cap C_{\omega}\chi$, as $cf(\gamma)=\omega$, there exists a strictly increasing function $g:\omega\to\gamma$ whose range is cofinal in $\gamma$, that's, $\sup\{g(n) :n\in\omega\}=\gamma$.
Also, as $\theta[ (\Sigma\cap \gamma^{<\omega} ) \times \gamma ]\subseteq \gamma^{<\omega}$, choose $\sigma\in \Sigma\cap \gamma^{<\omega}$ and consider $g(0)\in \gamma$ then $$\theta(\sigma, g(0)) \in \gamma^{<\omega}$$
In particular,
- $\sigma\subseteq \theta(\sigma, g(0))$
- $\theta(\sigma, g(0))\not\in g(0)^{<\omega}$
Also, as $W_{\sigma}$ is stationary then $\emptyset\not=[g(1), \chi[\cap W_{\sigma}$ so there exists $f\in K $ such that $g(1)\leq f^{*}$ and $\sigma\subseteq f$
Question. Does anyone have any idea to build this function? My problem is basically how to make that $f$ when it is built belong to $K$ and so far I also don't know how to use that $cf(\gamma)=\omega$.
Thanks
Answer Thanks to @Shervin Sorouri for the solution, I think I can understand it, and here I present it.
We claim that :
\begin{eqnarray*}
C\cap C_{\omega}\chi & = &\{ \gamma < \chi: \theta[ (\Sigma\cap \gamma^{<\omega} ) \times \gamma ]\subseteq \gamma^{<\omega} \} \cap \{\gamma<\chi : cf(\gamma)=\omega\}\\ & \subseteq &
\{f^{*}: f\in K \subseteq
\chi^{\omega} \} = W \\
\end{eqnarray*}
Indeed, let $\gamma\in C\cap C_{\omega}\chi$ then \textcolor{blue}{$cf(\gamma)=\omega$} then there exists a strictly increasing function $g:\omega\to\gamma$ whose range is cofinal in $\gamma$. That's $\sup\{g(n) :n\in\omega\}=\gamma$ and \textcolor{blue}{$\theta[ (\Sigma\cap \gamma^{<\omega} ) \times \gamma ]\subseteq \gamma^{<\omega}$}. The main idea is to iterate $\theta$, and $g$ will help us keep going up to $\gamma$. So inductively build a sequence $\sigma_{n}$ as follows:
- $\sigma_{0}:=\theta(\emptyset, g(0))$ and
- $\sigma_{n+1}:=\theta(\sigma_{n}, g(n))$
Note that $\sigma_{n}\in \Sigma\cap \gamma^{<\omega}$ and for each $n\in\omega$, $\sigma_{n}\subseteq \sigma_{n+1}=\theta(\sigma_{n}, g(n))$. Consider $$f=\bigcup_{n\in\omega}\sigma_{n}$$
We claim that $f\in\chi^{\omega}$ and $f^{*}=\gamma$. In fact,
- $dom(f)=\omega$
Note that for each $n\in\omega$, $dom(\sigma_{n})\in\omega$ then $dom(f)\subseteq\omega$. Also $dom(f)$ is infinite, otherwise,
$dom(f)$ is finite. Then consider $\beta=\max\{f(n) : n\in dom(f) \}$, as $g$ is cofinal, there exists $m\in\omega$ such that $\beta<g(m)$, also consider $\sigma_{m+1}=\theta(\sigma_{m}, g(m))\not \in g(m)^{<\omega}$ so there exists $m^{\prime}\in dom(\sigma_{m+1})\subseteq dom(f)$ such that $g(m)\leq \sigma_{m+1}(m^{\prime})=f(m^{\prime})$ so $\beta<g(m)\leq f(m^{\prime})$, contradiction. Therefore $dom(f)$ is infinite, so $dom(f)$ is unbounded in $\omega$. Then $\omega\subseteq dom(f)$. Indeed, let $m\in\omega$, then there exists $n\in dom(f)$ such that $m< n \in dom(f)$ so $m\in dom(f)$.
- $f^{*}=\sup\{f(m):m\in\omega\}=\gamma$
Let $\beta\in\gamma$, as $ran(g)$ is cofinal in $\gamma$, there is $m\in\omega$ such that $\beta<g(m)$. By construction, $\sigma_{m+1}=\theta(\sigma_{m}, g(m))\subseteq f$ and $\sigma_{m+1}\not\in g(m)^{<\omega}=\bigcup_{n\in\omega}g(m)^{n}$, then there exists $n\in dom(\sigma_{m+1})\subseteq dom(f)$ such that $\beta<\sigma_{m+1}(n)$. Otherwise, $\sigma_{m+1}\in (\beta+1)^{<\omega}\subseteq (g(m))^{<\omega}$, contradiction. Therefore, $\beta<\sigma_{m+1}(n)=f(m)$. On the other hand, note that $\sup\{f(m):m\in\omega\}\subseteq \gamma$, because $\sigma_{m}\in \gamma^{<\omega}$ for each $m\in\omega$.
Finally, note that $f\in K$. Indeed, for each $n\in\omega$ we have that $\sigma_{n}\in \Sigma$, that is, $W_{\sigma_{n}}$ is stationary, in particular $W_{\sigma_{n}}\not=\emptyset$ then there exists $f_{n}\in K$ such that $\sigma_{n}\subseteq f_{n}$. We claim that $f_{n}\xrightarrow{n\to\infty} f$ in $\chi^{\omega}$. In fact, let $f\in N_{s}=\{h\in \chi^{\omega} : s\subseteq h \}$ where $s= (s(0), \cdots, s(n_{s}-1)) \in \chi^{<\omega}$.
As $s\subseteq f$, $n_{s}-1\in dom(f)=\bigcup_{m\in\omega}dom(\sigma_{m})$ then there exists $m_{0}\in\omega$ such that $s\subseteq \sigma_{m_{0}}$. Then, if $m>m_{0}$, $f_{m}\in N_{\sigma}$ therefore $f_{n}\xrightarrow{n\to\infty} f$.
Best Answer
So as you argue, let $\gamma \in C\cap C_{\omega\chi}$ and let $g:\omega \rightarrow \gamma$ be cofinal. The main idea is to iterate $\theta$, and $g$ will help us keep going up to $\gamma$. So inductively build a sequence $\sigma_n$ as follows:
So now as the $\sigma_n$'s are an increasing sequence of finite subsequences of $\gamma$, let $f = \bigcup_{n\in\omega}\sigma_n$. By the closure property of $\gamma$ under $\theta$ and the conditions on $\theta$ and the cofinality of $g$, you can see that $f^* = \gamma$. Also $f \in K$, because for each $\sigma_n$ you can choose $f_n \in K$ with $\sigma_n \subseteq f_n$(you can choose such $f_n$, because $W_\sigma$ is non-empty), and you can see $f_n \rightarrow f$ and because $K$ is closed, you have $f \in K$. So $\gamma \in W$.