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,
Claim 4.1 Using the Pressing Down Lemma one can build a function $\theta: \Sigma \times \chi \to \Sigma$ such that
- $\sigma\subseteq \theta(\sigma, \alpha)$
- $\theta(\sigma, \alpha)\not\in \bigcup_{n\in\omega}\alpha^{n}$
Question 1. Does anyone have any idea to build this function?, at first I was trying as follows, let $\sigma\in\Sigma$, then $W_{\sigma}$ is stattionary. I don't have an initial idea to be able to define a regressive function $g_{\sigma}:W_{\sigma}\to \chi$.
@Shervin Sorouri, managed to demonstrate this part, you can see the answer in the first comments.
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 2. 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
Best Answer
For each $\sigma$ and $\alpha$ let $P = W_\sigma - \alpha$ which is still stationary. Now for each $f^* \in P$, let $g_\sigma(f^*)$ be $f(n)$ where $n$ is least such that $f(n) \ge \alpha$. Now using pressing down lemma and the pigeonhole principle, you can find some fixed $n$ and some fixed $\gamma \ge \alpha$ such that $\{ f^* \in P: f(n) = \gamma\}$ is stationary. Now if $n\le |\sigma|$, you are done. Else to fill the gap between $|\sigma|$ and $n$ you can use repeated applications of the pressing down lemma, to get the desired $\theta$.
EDIT: [This edit will try to complete the below answer you provided.][Disclaimer: I am using your notation.]
As in your answer, let $S = \{f^* \in P: f(m) = \gamma\}$ and suppose $m \gt |\sigma|$. By the fact that $P \subseteq W_\sigma$, we have $f||\sigma| = \sigma$, for any $f^* \in S$. First we inductively choose a finite sequence of stationary sets $\langle S_0, \dots, S_{m-|\sigma|-1}\rangle$ and a finite sequence of ordinals $\langle \beta_0, \dots, \beta_{m-|\sigma|-1}\rangle$, such that $S_0 \subseteq S$, $S_{i+1} \subseteq S_i$, for $i \lt m-|\sigma|-1$. Also we make sure that for each $f^* \in S_i$, $f(i+|\sigma|) = \beta_i$.
This can be done easily, using the pressing down lemma. For the base case $i = 0$, consider $g(f^*) = f(|\sigma|)$ and by the pressing down lemma you have some stationary $S_0 \subseteq S$ and some ordinal $\beta_0$ such that $g"S_0 = \{\beta_0\}$. At the $i$th step just look at $g(f^*) = f(i+|\sigma|)$, and construct $S_i$ and $\beta_i$ as above.
So we wish to build a $\theta \in \chi^{m+1}$ that satisfies the conditions in the question. First let $\theta||\sigma| = \sigma$ and $\theta(m) = \gamma$. Now for $|\sigma| \le i \lt m$, let $\theta(i) = \beta_{i-|\sigma|}$. Now you can see that $W_\theta$ is stationary as it contains $S_{m-|\sigma|-1}$. And also because of $\gamma$ you have $\theta \not \in \cup_{n\in\omega} \alpha^n$.