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Edit: This is a question from Jech: Set Theory

I have been trying for a few days my luck with the following question: For every stationary $S \subset \omega_1$ and every $\alpha < \omega_1$ there is a closed set of ordinals $A$ of length $\alpha$, such that $A \subset S$.

I can't even understand the clue given below the question..

I would be greatful for any help.

$\mathbf{8.5.}$ For every stationary $S\subset\omega_1$ and every $\alpha<\omega_1$ there is a closed set of ordinals $A$ of length $\alpha$ such that $A\subset s$.

[By induction on $\alpha$: $\forall\gamma\;\exists\text{closed }A\subset S$ of length $\alpha$ such that $\gamma<\min A$. The nontrivial setp: If true for a limit $\alpha$, find a closed $A\subset S$ of length $\alpha$ such that $\sup A\in S$. Let $A_\xi$, $\xi<\omega_1$, be closed subsets of $S$, of length $\alpha$, such that $\lambda_\xi=\sup\bigcup_{\nu<\xi}A_\nu<\min A_\xi$. There is $\xi$ such that $\lambda_\xi\in S$. Let $\xi=\lim_n\xi_n$. Pick initial segments $B_{\xi_n}\subset A_{\xi_n}$ of length $\alpha_n+1$ where $\lim_n\alpha_n=\alpha$. Let $A=\bigcup_{n=0}^\infty B_{\xi_n}$.]

Exercise $8.5$ does not generalize to closed sets of uncountable length. It is not provable in ZFC that given $X\subset\omega_2$, either $X$ or $\omega_2-X$ contains a closed set of length $\omega_1$. On the other hand, this statement is consistent, relative to large cardinals.

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I’ll expand the hint considerably and give a bit more of an overall roadmap while still leaving a lot of the details to you.

The suggestion is to prove (by induction on $\alpha<\omega_1$) the following stronger statement $P(\alpha)$:

For each $\gamma<\omega_1$ there is a closed $A\subseteq S$ of length $\alpha$ such that $\gamma<\min A$.

It’s completely straightforward to show that $P(\alpha+1)$ implies $P(\alpha+2)$ for any $\alpha<\omega_1$. Two cases remain. Suppose that $\alpha<\omega_1$ is a limit ordinal.

Case $\mathbf{1.}$ We must show that if $P(\alpha)$ holds, then so does $P(\alpha+1)$.

The hint deals with this case. $P(\alpha)$ tells us that we can find lots of closed subsets $A$ of $S$ of length $\alpha$. However, to extend such an $A$ to a closed subset of $S$ of length $\alpha+1$, it’s not enough to find an element $\xi$ of $S$ such that $\xi\ge\sup A$. That’s easy, since $S$ is unbounded, but $\alpha$ is a limit ordinal, so in order to make $A\cup\{\xi\}$ closed, we must have $\xi=\sup A$. Thus, we must be able to find a closed $A\subseteq S$ of length $\alpha$ whose supremum is in $S$.

The hint suggests defining an $\omega_1$-sequence $\langle A_\xi:\xi<\omega_1\rangle$ of closed subsets of $S$, each of length $\alpha$, so that the following condition is satisfied:

$\lambda_\xi<\min A_\xi$ for each $\xi<\omega_1$, where $\lambda_\xi=\sup\bigcup_{\nu<\xi}A_\nu$.

This is possible because $\bigcup_{\nu<\xi}A_\nu$ is a countable union of countable sets, so $\lambda_\xi<\omega_1$, and by the hypothesis $P(\alpha)$ there is a closed $A_\xi\subseteq S$ of length $\alpha$ with $\lambda_\xi<\min A_\xi$. Let $\Lambda=\{\lambda_\xi:\xi<\omega_1\}$.

  • Show that $\Lambda$ is a closed, unbounded subset of $\omega_1$, and conclude that there is a $\xi<\omega_1$ such that $\lambda_\xi\in S$.
  • Show further that we may assume that $\xi$ is a limit ordinal.

Now let $\langle\xi_n:n\in\omega\rangle$ be a strictly increasing sequence such that $\xi=\sup_n\xi_n$, and let $\langle\alpha_n:n\in\omega\rangle$ be a sequence in $\alpha$ such that $\sup_n=\alpha$. For each $n\in\omega$ the set $A_{\xi_n}$ is of length $\alpha>\alpha_n+1$, so we can let $B_{\xi_n}$ be the initial segment of $A_{\xi_n}$ of length $\alpha_n+1$. Now let $A=\bigcup_{n\in\omega}B_{\xi_n}$. Clearly $A\subseteq S$.

  • Show that the length of $A$ is $\alpha$.
  • Show that $A$ is closed. (HINT: Each $B_{\xi_n}$ is both open and closed in $A$.)
  • Show that $\sup A=\lambda_\xi\in S$.

Finally, note that for any $\gamma<\omega_1$ we may assume that $\gamma<\min A_0$ and hence that $\gamma<\min A$.

Case $\mathbf{2.}$ We must show that if $P(\beta)$ holds for each $\beta<\alpha$, then $P(\alpha)$ holds.

Use an idea very similar to the construction of $A$ from the sets $B_{\xi_n}$ in Case $\mathbf{1}$.

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  • $\begingroup$ Thank you Brian! I understood most of your proof exapt one thing: In case 1, If $\xi = sup A$ doesn't it contradict the fact that the original $A$ was closed? I mean, If $P(\alpha)$ is true, shouldn't this imply that $A$ contains all it's limit point and therefore contains it's supremum? Thanks again. $\endgroup$
    – user83081
    Commented Feb 16, 2015 at 9:31
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    $\begingroup$ @user83081: You’re welcome. $A$ isn’t required to be closed in $\omega_1$. In fact, if $\alpha$ is a countable limit ordinal, and the length of $A$ is $\alpha$, then it’s impossible for $A$ to be closed in $\omega_1$, since $\sup A\notin A$. The requirement is actually that $A$ be a closed subset of $\sup A$, i.e., that $A$ contain all of its limit points that are less than $\sup A$. It seems that you’re supposed to pick this up from context. If I may ask, what’s the book? $\endgroup$ Commented Feb 16, 2015 at 9:40
  • $\begingroup$ I see now. Thank you!!! The question is from Jech's book in set theory. Page 103 I added the link in my question. $\endgroup$
    – user83081
    Commented Feb 16, 2015 at 10:12
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    $\begingroup$ @user83081: Thanks very much. One of these days I really ought to take a good look at it; somehow I’ve never managed to get around to it, though I have taught from the excellent but much more elementary text that he wrote with Karel Hrbáček. $\endgroup$ Commented Feb 16, 2015 at 10:21
  • $\begingroup$ Good to know he has a more elementary book as well. I will have a look at this one too :) $\endgroup$
    – user83081
    Commented Feb 16, 2015 at 10:23

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