MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


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.

share|cite|improve this question
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
up vote 10 down vote accepted

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 $F$.

share|cite|improve this answer
Thank you very much! – Stefan Hoffelner Oct 7 '10 at 15:04

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.