# Uncountable family of pairwise disjoint non-stationary subsets

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 $$\omega_1$$.

• 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 $\mathcal A$). – Mirko Sep 12 '19 at 1:49
• 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. – taupi Sep 12 '19 at 2:14
• 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 $\omega_1)\}$ – Mirko Sep 12 '19 at 2:20
• 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. – Mirko Sep 12 '19 at 2:29
• Thank you. I will take a look at it and try to do this question again. – 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.

• +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 $\cup B=\omega_1.$ – DanielWainfleet Sep 15 '19 at 3:14