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. 



Why is this set stationary?
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 $F$. 


