Uncountable family of pairwise disjoint non-stationary subsets

Question

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 $\omega_1$.

Answer 1

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.

Answer 2

For a short solution, see Exercise 25 in chapter 20 of Problems and Theorems in Classical Set Theory by Komjath and Totik.

As explained in Mirko's answer, we can assume the family of sets $\{A_\alpha:\alpha\in\omega_1\}$ is enumerated by strictly increasing values of $\min(A_\alpha)$ (this is where we use the fact that the $A_\alpha$ are disjoint). We have $\min(A_\alpha)\ge\alpha$ (easily proved by induction, for any strictly increasing function). We claim that the union $$A=\bigcup\{A_{\alpha+1}:\alpha<\omega_1\}$$ is nonstationary. To show this, consider the function $f:A\to\omega_1$ defined by $f(x)=\alpha$ for $x\in A_{\alpha+1}$. This is a regressive function since $\min(A_{\alpha+1})\ge\alpha+1>\alpha$, and the inverse image of every element in its range is nonstationary (it's one of the $A_{\alpha+1}$). Hence by Fodor's lemma, the domain of the function, namely $A$, is nonstationary.