An application of the Pressing Down Lemma (Part 1)

Question

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,

1. $\sigma\subseteq \theta(\sigma, g(0))$
2. $\theta(\sigma, g(0))\not\in g(0)^{<\omega}$

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

Answer 1

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 $\theta$.

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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 $\theta \not \in \cup_{n\in\omega} \alpha^n$.

Answer 2

This demonstration was done by @Shervin Sorouri, in this part I am dividing your demonstration with some previous lemmas.

Lemma 1 Let $\kappa$ be a regular uncountable cardinal and let $\alpha \in \kappa$. If $S$ is stationary in $\kappa$, then $S\setminus \alpha$ is stationary in $\kappa$.

Lemma 2 Suppose that $\kappa$ is a regular uncountable cardinal and that $\gamma\in \kappa$. Let $\langle S_{\alpha} : \alpha\in\gamma \rangle$ be a $\gamma$-sequence of subsets of $\kappa$. Suppose that the set $\bigcup_{\alpha\in\gamma}S_{\alpha}$ is stationary in $\kappa$. Then $S_{\alpha}$ is stationary, for some $\alpha\in\gamma$.

Proof of Claim 4.1 Indeed, let $\sigma \in \bigcup_{n\in\omega}\chi^{n}$ and $\alpha<\chi$, consider $P=W_{\sigma}\setminus \alpha$, by Lemma 1, $P$ is stationary in $\chi$. Define $$ \begin{array}{lcccl} g_{\sigma} & : & P & \longrightarrow & \chi\\ & & f^{*} & \longrightarrow & g_{\sigma}(f^{*})=f(n), \end{array} $$ where $n=\min\{n\in\omega : f(n)\geq\alpha \}$. Note that $g_{\sigma}(f^{*})<f^{*}$, for all $f^{*}\in P$, so by Pressing Down Lemma, there is $\gamma<\chi$ such that $g_{\sigma}^{-1}(\{\gamma \})=\{f^{*}\in P : g_{\sigma}(f^{*})=f(n)=\gamma\}$ is stationary, note that $\gamma\geq\alpha$. Finally, define $$ \begin{array}{lcccl} h & : & g_{\sigma}^{-1}(\{\gamma \}) & \longrightarrow & \omega\\ & & f^{*} & \longrightarrow & h(f^{*})=n, \end{array} $$ where $n\in\omega$ is such that $g_{\sigma}(f^{*})=f(n)$. Note that $g_{\sigma}^{-1}(\{\gamma \})=\bigcup_{n\in\omega}h^{-1}(\{n\})$, then by Lemma 2, there is $m\in\omega$ such that $h^{-1}(\{m\})=\{f^{*}\in g_{\sigma}^{-1}(\{\gamma \}):h(f^{*})=m\}=\{f^{*}\in P : f(m)=\gamma\}$ is stationary.

• If $m\leq |\sigma|$, then $\theta(\sigma, \alpha)=\sigma$, in this case $\theta(\sigma, \alpha)\not\in \bigcup_{n\in\omega}\alpha^{n}$, because $\sigma(m)=f(m)=\gamma\geq\alpha$.

• If $m>|\sigma|$,

Claim 4.1.1 There are a finite sequence of stationary sets $\langle S_{0} , \cdots, S_{m-|\sigma|-1}\rangle$ and a finite sequence of ordinals $\langle \beta_{0} , \cdots, \beta_{m-|\sigma|-1}\rangle$ such that $S_{0}\subseteq S$ and, for $i<m-|\sigma|-1$, then $S_{i+1}\subseteq S_{i}$ and if $f^{*}\in S_{i}$ then $f(i+|\sigma|)=\beta_{i}$.

Proof of Claim 4.1.1 In fact, for $i=0$, consider $$ \begin{array}{lcccl} g_{0} & : & S & \longrightarrow & \chi\\ & & f^{*} & \longrightarrow & g_{0}(f^{*})=f(|\sigma|)<f^{*}, \end{array} $$ By the Pressing down lemma, there exists $\beta_{0}<\chi$ such that $g^{-1}_{0}(\{\beta_{0}\})=S_{0}$.

For $0<i<m-|\sigma|-1$, consider $$ \begin{array}{lcccl} g_{i} & : & S_{i-1} & \longrightarrow & \chi\\ & & f^{*} & \longrightarrow & g_{i}(f^{*})=f(|\sigma|+i)<f^{*}, \end{array} $$ By the Pressing down lemma, there exists $\beta_{i}<\chi$ such that $g^{-1}_{i}(\{\beta_{i}\})=S_{i}\subseteq S_{i-1}$.

Note that, if $f^{*}\in S_{i}$ then $f(i+|\sigma|)=\beta_{i}$.

Now we will build $\theta\in \chi^{m+1}$, let $\theta|_{|\sigma|}=\sigma$ and $\theta(m)=\gamma$. Then, if $|\sigma|\leq i<m$, define $\theta(i)=\beta_{i-|\sigma|}$.

Finally, note that $S_{m-|\sigma|-1}\subseteq W_{\theta}$. In fact, let $f^{*}\in S_{m-|\sigma|-1}$, in particular, $\sigma\subseteq f\in K$ and $f(m)=\gamma$. By Claim 4.1.1, $f(i+|\sigma|)=\beta_{i}$ for $i<m-|\sigma|-1$, so $f\in W_{\theta}$.