# Why is this set stationary?

Hi

I really need a proof for the following statement by Baumgartner:

There exists a stationary subset of $[\omega_2]^{\omega}$ of size $\aleph_2$.

This is Exercise 38.15. in Jechs Book (2003) and you can find a hint there which goes like this: For each $\alpha < \omega_2$, let $f_{\alpha} : \alpha \to \omega_1$ be one to one. If $\alpha < \omega_2$ and $\xi < \omega_1$ set $X_{\alpha, \xi} =$ { $\beta < \alpha : f_{\alpha} (\beta) < \xi$ }. Then $S:=$ { $X_{\alpha, \xi} : \alpha < \omega_2, \xi < \omega_1$} is our desired stationary subset.

But so far my attempts to proof this didn't work, because the sequence of the $f_{\alpha}$s doesn't have any nice regularity properties. Thank you.

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You only have to show, that if $F:[\omega_2]^{<\omega}\longrightarrow[\omega_2]^{\aleph_0}$ then some element of $S$ is closed under $F$. – Péter Komjáth Oct 7 '10 at 8:16
Also, you can use my book, Komjath-Totik: Problems and theorems in classical set theory, Springer, 2006. This is problem 29.19. – Péter Komjáth Oct 7 '10 at 8:19
Thank you very much – Stefan Hoffelner Oct 7 '10 at 8:38
(Did you just seriously complain, to the author, that no one has made illegal copies of his book available on the internet?) Have you tried looking for the book in a library? – Willie Wong Oct 7 '10 at 10:46
@Willie: given that it is not unheard of for authors to publish downloadable copies of their books online, I don't think his question was out of line... – dvitek Oct 7 '10 at 11:30

Given $F:[\omega_2]^{<\omega}\to[\omega_2]^{\aleph_0}$ as above, we first claim the existence of an ordinal $\omega_1\leq\alpha<\omega$ that is closed under $F$, i.e., $s\in [\alpha]^{<\omega}$ implies $F(s)\subseteq\alpha$. For this, let $\alpha$ be the limit of the sequence $\omega_1=\alpha_0<\alpha_1<\cdots$ where $\alpha_{n+1}$ is sufficiently large that $F(s)\subseteq \alpha_{n+1}$ for $s\in [\alpha_{n}]^{<\omega}$.
Given $\alpha$ as above, construct similarly the ordinal $\omega\leq\xi<\omega_1$ so that $f^{-1}_\alpha[\xi]$, that is, $\{\beta<\alpha:f_\alpha(\beta)<\xi\}$, is closed under $F$. This can be done similarly: let $\xi$ be the limit of the sequence $\omega=\xi_0<\xi_1<\cdots$ where $\xi_{n+1}$ is chosen so that if $s$ is a finite subset of $\{\beta<\alpha:f_\alpha(\beta)<\xi_n\}$, then $F(s)$ (which is a subset of $\alpha$) is a subset of $\{\beta<\alpha:f_\alpha(\beta)<\xi_{n+1})\}$. Now $X_{\alpha,\xi}$ is closed under $F$.