Suppose $\mathcal A$ is an uncountable family of pairwise disjoint non-stationary subsets of $\omega_1$. Show that there exists $\mathcal B \subseteq \mathcal A$ such that $\mathcal B$ is uncountable and $\bigcup \mathcal B$ is non-stationary subset of $\omega_1$.

This is what I have gotten.

$\bullet$ Pairwise disjoint: $\forall \beta<\alpha<\omega_1, X_\beta \cap X_\alpha= \varnothing$.

$\bullet Non-stationary: \forall \alpha < \omega_1,$ $\exists C \in Club(\omega_1), C \cap X_\alpha=0$

$\bullet Club(\omega_1)=\{C\subseteq \omega_1:(\exists D\subseteq C)(D $ is closed and unbounded in $\omega_1)\}$

Would it work if i take out countably many $X_a$ from $\mathcal A$ and call that set $\mathcal B$. $\mathcal B$ would be uncountable but I dont think that $\bigcup \mathcal B$ would be a subset of $\omega_1$.

  • $\begingroup$ that $\alpha>\omega_1$ seems misplaced. Also, that $\forall C$ looks like it should be $\exists C$. Try to think of this and perhaps show progress and edit question. Also, notation $Club(\mathcal A)$ confuses me, isn't the family of all club sets defined independent of $\mathcal A$? Like supersets of sets that are closed and UNbounded in $\omega_1$ (not $\mathcal A$). $\endgroup$ – Mirko Sep 12 '19 at 1:49
  • $\begingroup$ Thanks for your comments. Yes the $\forall C$ should be $\exists C$ and my previous definition of the Club filter was wrong. But I don't understand why $\alpha > \omega_1$ is misplaced since $\mathcal A$ is an uncountable family. $\endgroup$ – taupi Sep 12 '19 at 2:14
  • $\begingroup$ perhaps you mean (without loss of generality) that $\mathcal A=\{X_\alpha: \alpha<\omega_1\}$. Note here indexing is with $\alpha<\omega_1$, note $\omega_1$ is uncountable. Note also in the club definition bounded should be UNbounded. (Club is an abbreviation of closed UNbounded). $\bullet$ Pairwise disjoint: $\forall \beta<\alpha<\omega_1,X_{\beta}\cap X_{\alpha}=\emptyset$ $\bullet$ Non-stationary: $\forall \alpha<\omega_1, \exists C \in Club(\omega_1), C \cap X_\alpha=0$ $\bullet Club(\omega_1)=\{C\subseteq\omega_1:(\exists D\subseteq C)(D $ is closed and UNbounded in $\omega_1)\}$ $\endgroup$ – Mirko Sep 12 '19 at 2:20
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    $\begingroup$ I believe the answer might have something to do with the diagonal intersection of a family of club sets (which one gets naturally here taking a club $C_\alpha$ disjoint from $X_\alpha$), see math.stackexchange.com/q/380626 Note that the condition in your question that the non-stationary sets are disjoint is essential. There is an easy counterexample, if they were not required to be disjoint, take $X_\alpha=\alpha=[0,\alpha)=\{\beta:\beta<\alpha\}$. I need to think of the details, but at the moment I do not see what else could be relevant, but the diagonal intersection. $\endgroup$ – Mirko Sep 12 '19 at 2:29
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    $\begingroup$ Thank you. I will take a look at it and try to do this question again. $\endgroup$ – taupi Sep 12 '19 at 2:45

There are two closely related questions (and answers) on MSE. Diagonal intersection of club sets, URL (version: 2013-07-30): Diagonal intersection of club sets , as well as Diagonal union of non-stationary sets, URL (version: 2016-05-14): Diagonal union of non-stationary sets .

The above two questions are closely related, as they involve "dual" notions. In the second of the above links there is a proof (using the first link) that the diagonal union of non-stationary sets is non-stationary. I will use this result to answer the present question.

We are given that $\mathcal A$ is an uncountable family of pairwise disjoint non-stationary (and non-empty) subsets of $\omega_1$. Since the sets in $\mathcal A$ are disjoint we have that $|\mathcal A|=\omega_1$, hence we may list $\mathcal A=\{A_\alpha:\alpha<\omega_1\}$ where $A_\alpha\not=A_\beta$ if $\alpha<\beta<\omega_1$.

Let $\delta_0=\min\{\alpha:\min(A_\alpha)>0\}$ (where $\min(A_\alpha)$ denotes the smallest element of $A_\alpha$, using that $\omega_1$ is well-ordered). Recursively, if $\gamma<\omega_1$ and $\delta_\beta$ have been defined for all $\beta<\gamma$, define $\delta_\gamma=\min\{\alpha\ge\sup_{\beta<\gamma}(\delta_\beta+1):\min(A_\alpha)>\gamma\}$. Note that for any given $\gamma<\omega_1$ there are uncountably many $\alpha$ such that $\min(A_\alpha)>\gamma$: Indeed, otherwise there will be some $\gamma$ with $\min(A_\alpha)\le\gamma$ for uncountably many $\alpha$, and hence there will be some $\nu\le\gamma$ with $\min(A_\alpha)=\nu$ for uncountably many $\alpha$ contradicting that $\mathcal A$ is a disjoint family.

Therefore we define $\delta_\gamma$ as above, for all $\gamma<\omega_1$. Let $\mathcal B=\{B_\gamma:\gamma<\omega_1\}$, where $B_\gamma=A_{\delta_\gamma}$ for each $\gamma$. Clearly $\mathcal B$ is uncountable, $\mathcal B\subseteq\mathcal A$. Note that, by the above construction, $\min(B_\gamma)=\min(A_{\delta_\gamma})>\gamma$, that is $B_\gamma\subseteq(\gamma,\omega_1)$ for all $\gamma<\omega_1$. We have that:
$\cup\mathcal B=\cup_{\gamma<\omega_1}B_\gamma=$ $\cup_{\gamma<\omega_1}\bigl((\gamma,\omega_1)\cap B_\gamma\bigr)=$ $\nabla_{\gamma<\omega_1}B_\gamma$. The last expression denotes the diagonal union, and, as indicated earlier, it is non-stationary.

Remark. We could have also assumed that $\mathcal A=\{A_\alpha:\alpha<\omega_1\}$ where $\min(A_\alpha)<\min(A_\beta)$ if $\alpha<\beta$. Indeed, if $M=\{\min(A):A\in \mathcal A\}$ then $M$ is an uncountable subset of $\omega_1$ and hence order-isomorphic to $\omega_1$ (that is, there is an order-preserving bijection between $M$ and $\omega_1$, which could be used to list $\mathcal A$ as in this remark).

Edit. If we were to work with club sets and diagonal intersection directly (instead of the dual notions), we may define $B_\gamma$ as earlier and then finish off the proof as follows. For each $\gamma$ fix a club (closed unbounded) $C_\gamma$ disjoint from $B_\gamma$. Then $[0,\gamma]\cup C_\gamma$ is a club disjoint from $B_\gamma$, so (replacing $C_\gamma$ with $[0,\gamma]\cup C_\gamma$ if necessary) we may assume that $C_\gamma$ is a club disjoint form $B_\gamma$, and that $[0,\gamma]\subseteq C_\gamma$. Then the set $C:=\cap_{\gamma<\omega_1}C_\gamma$ is clearly disjoint from $\cup\mathcal B$ (using De Morgan's laws). On the other hand $C=\cap_{\gamma<\omega_1}\bigl([0,\gamma]\cup C_\gamma\bigr)$ $=\Delta_{\gamma<\omega_1}C_\gamma$ , where the latter expression is the diagonal intersection, and is hence a club. Thus the club $C$ is disjoint from $\cup\mathcal B$, hence $\cup\mathcal B$ is non-stationary.

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    $\begingroup$ +1... A counter-example if $A$ is not assumed to be pair-wise disjoint, is $A=\omega_1.$ If $B$ is an uncountable subset of $A=\omega_1$ then $\cup B=\omega_1.$ $\endgroup$ – DanielWainfleet Sep 15 '19 at 3:14

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