Let Lim$(\omega_1)$ denote the set of limit ordinals in $\omega_1$. Suppose $A_\alpha$ is a sequence where $\alpha \in$ Lim$(\omega_1)$ that satisfies the following.

i) For every $\alpha \in$ Lim$(\omega_1),A_\alpha$ is an unbounded subset of $\alpha$ of order type $\omega$.

ii) For every uncountable $A \subseteq \omega_1$, there exists $\alpha \in$ Lim$(\omega_1)$ such that $A_\alpha \subseteq A$.

Show that for every uncountable $A \subseteq \omega_1$, the set $\{\alpha \in Lim(\omega_1): A_\alpha \subseteq A\}$ is stationary in $\omega_1$.

I dont know how to start with this problem. Help very much appreciated.

  • $\begingroup$ $A_\omega$? Do you mean $A_\alpha$? $\endgroup$
    – Asaf Karagila
    Sep 12, 2019 at 7:33
  • $\begingroup$ Yes thanks for highlighting the mistake. I have edited it. $\endgroup$
    – taupi
    Sep 12, 2019 at 7:36
  • $\begingroup$ I edited your question, only adding tag "infinitary-combinatorics". Your question is closely related to so-called club guessing sequences which are part of the area called infinitary-combinatorics. Construction of club guessing sequences often involves extra axioms in addition to the usual ZFC, and your problem does not need extra axioms, yet the statements involved are indeed very similar. (Feel free of course to revert to the version before my edit, if you wish.) $\endgroup$
    – Mirko
    Sep 17, 2019 at 15:26
  • $\begingroup$ I wonder you you have any questions and comments on my answer, do you follow it, does it answer your question? $\endgroup$
    – Mirko
    Oct 29, 2019 at 2:41

1 Answer 1


First I copy the statement of the problem, fixing some notation as well.

Let $L=\mathrm{Lim}(\omega_1)$ denote the set of all limit ordinals in $\omega_1$. Suppose $\{A_\alpha:\alpha\in L\}$ is a (transfinite) sequence of subsets of $\omega_1$ that satisfies the following.

(i) $A_\alpha$ is an unbounded subset of $\alpha$ of order type $\omega$, for each $\alpha\in L$.

(ii) For every uncountable $A\subseteq\omega_1$, there exists $\alpha\in L$ such that $A_\alpha \subseteq A$.

Fix any uncountable $U \subseteq \omega_1$, and let $S=\{\alpha \in L: A_\alpha \subseteq U\}$. We need to show that $S$ is stationary in $\omega_1$. Fix any club (closed unbounded set) $C$. We need to show $\emptyset\neq S\cap C$.

The main idea is to apply (ii) repeatedly to suitable tails of $U$ (going with bigger and bigger ordinals) so that we build an uncountable $B\subseteq U$ such that $B$ interleaves with $C$, so that in the end if $A_{\alpha(B)}\subseteq B$ with $\alpha(B)\in L$ we would have $\alpha(B)\in S$ (since $B\subseteq U$), and $\alpha(B)=\sup(A_{\alpha(B)})\in C$, since $A_{\alpha(B)}$ interleaves with $C$.
(One may use this as a hint, before reading the details below. I ended up writing a version 2 of my answer, both versions enclosed, below, essentially the same, version 2 may be shorter and slightly more general.)

Version 1.
Pick any $\alpha_0\in S$ and any $\gamma_0\in C$ with $\gamma_0>\alpha_0$. For any $\beta<\omega_1$ let $U_\beta=U\cap(\beta,\omega_1)$, clearly each $U_\beta$ is uncountable. Using (ii), there is $\alpha_1\in L$ such that $A_{\alpha_1}\subseteq U_{\gamma_0}$. Clearly $\alpha_1\in S$ (since $U_{\gamma_0}\subseteq U$), and $\gamma_0<\alpha_1$ (since $\gamma_0+1=\min(U_{\gamma_0})\le\min(A_{\alpha_1})<\alpha_1$, where $\min(P)$ in general denotes the smallest element of any given non-empty set $P$ of countable ordinals).

Recursively define $\alpha_\beta$ and $\gamma_\beta$ for each $\beta<\omega_1$ as follows.
(1) if $\alpha_\beta$ is defined, pick any $\gamma_\beta\in C$ with $\gamma_\beta>\alpha_\beta$,
(2) (continuing (1)), then, using (ii), there is $\alpha_{\beta+1}\in L$ such that $A_{\alpha_{\beta+1}}\subseteq U_{\gamma_\beta}$,
(3) if $\delta$ is limit and $\alpha_\beta$ were defined for all $\beta<\delta$ then using (ii), there is there is $\alpha_\delta\in L$ such that $A_{\alpha_\delta}\subseteq U_{\sup\{\beta<\delta:\gamma_\beta\}}$. (Following (1), also pick $\gamma_\delta\in C$ with $\gamma_\delta>\alpha_\delta$.)

Note that by construction we have $\alpha_\beta\in S$ for each $\beta$, and $\alpha_{\beta+1}>\gamma_\beta$. Moreover, if $\mu_\beta=\min(A_{\alpha_\beta})$ then $\mu_{\beta+1}>\gamma_\beta$, for each $\beta<\omega_1$. If $\delta$ is limit then $\mu_\delta>\gamma_\beta$ for all $\beta<\delta$.

Let $B=\{\mu_\beta:\beta<\omega_1\}$. Clearly $B$ is uncountable and $B\subseteq U$ (since $\mu_\beta=\min(A_{\alpha_\beta})\in A_{\alpha_\beta}\subseteq U$). By (ii), there is $\alpha(B)\in L$ such that $A_{\alpha(B)}\subseteq B$. Then $\alpha(B)\in S$ (using that $B\subseteq U$), and we need to show that also $\alpha(B)\in C$. Each element of $A_{\alpha(B)}$ is of the form $\mu_\beta$, for some $\beta$. Hence there is an increasing sequence $\{\beta_n:n<\omega\}$ such that $A_{\alpha(B)}=\{\mu_{\beta_n}:n<\omega\}$. Then $\alpha(B)=\sup(A_{\alpha(B)})=\sup_{n<\omega}\mu_{\beta_n}=\sup_{n<\omega}\gamma_{\beta_n}$, where the last equality holds since $\mu_{\beta_n}<\alpha_{\beta_n}<\gamma_{\beta_n}<\mu_{({\beta_n+1})}\le\mu_{\beta_{n+1}}$. Since $C$ is closed we have $\alpha(B)=\sup_{n<\omega}\gamma_{\beta_n}\in C$, which completes the proof.

It doesn't look like that the condition that "the order-type of each $A_\alpha$ is $\omega\ $" is essential (as long as $\sup(A_\alpha)=\alpha$). Indeed, instead of $A_{\alpha(B)}=\{\mu_{\beta_n}:n<\omega\}$ we could just take increasing $\{\mu_{\beta_n}:n<\omega\}\subseteq A_{\alpha(B)}$ with $\sup(\{\mu_{\beta_n}:n<\omega\})=\sup(A_{\alpha(B)})=\alpha(B)$.

Version 2.
This version is proved using a formally weaker condition in place of (i), namely:
(i') $A_\alpha$ is a non-empty unbounded subset of $\alpha$ , i.e. $\sup(A_\alpha)=\alpha$, for each $\alpha \in L$. (The difference with condition (i) is that we allow $A_\alpha$ to have the order-type of any countable limit ordinal, not necessarily $\omega$. Condition (ii) remains unchanged.)

Lemma. The set $S=\{\alpha \in L: A_\alpha \subseteq U\}$ is unbounded (for any given uncountable $U\subseteq\omega_1$). Moreover, given any $\beta<\omega_1$ there is $\alpha\in S$ such that $\beta<\min(A_\alpha)<\alpha$.
Proof. Observe that $U_\beta:=U\cap(\beta,\omega_1)$ is uncountable, and by (ii) there is $\alpha \in L$ with $A_\alpha\subseteq U_\beta\subseteq U$ (and hence $\alpha \in S$). Then $\beta<\beta+1\le\min(A_\alpha)<\alpha$.

Using the lemma, we may construct by recursion transfinite sequences $\{\alpha_\beta:\beta<\omega_1\}$ and $\{\gamma_\beta:\beta<\omega_1\}$, and let $\mu_\beta=\min(A_{\alpha_\beta})$ such that:
(1') $\mu_\beta=\min(A_{\alpha_\beta})<\alpha_\beta=\sup(A_{\alpha_\beta})\le\gamma_\beta<\mu_{\beta+1}$ and $\alpha_\beta\in S$ (i.e $A_{\alpha_\beta}\subseteq U$), and $\gamma_\beta\in C$, for each $\beta$ (where $C$ is a club we have fixed, and need to prove $S\cap C\neq\emptyset$),
(2') if $\delta\in L$ then $\sup_{\beta<\delta}\gamma_\beta<\mu_\delta=\min(A_{\alpha_\delta})<\alpha_\delta=\sup(A_{\alpha_\delta})\le\gamma_\delta<\mu_{\delta+1}$ and $\alpha_\delta\in S$ and $\gamma_\delta\in C$.

Let $B=\{\mu_\beta:\beta<\omega_1\}$. Clearly $B$ is uncountable and $B\subseteq U$ (since $\mu_\beta=\min(A_{\alpha_\beta})\in A_{\alpha_\beta}\subseteq U$). By (ii), there is $\alpha(B)\in L$ such that $A_{\alpha(B)}\subseteq B$. Since $B\subseteq U$ we have $A_{\alpha(B)}\subseteq U$, hence $\alpha(B)\in S$. It remains to show that $\alpha(B)\in C$. Since $A_{\alpha(B)}\subseteq B$ we have that each element of $A_{\alpha(B)}$ is of the form $\mu_\beta$, for some $\beta<\omega_1$. Since the order type of $A_{\alpha(B)}$ is a countable limit ordinal, there is an increasing sequence $\{\beta_n:n<\omega\}$ such that $\{\mu_{\beta_n}:n<\omega\}\subseteq A_{\alpha(B)}$ and $\sup\{\mu_{\beta_n}:n<\omega\}=\alpha(B)$. Then $\alpha(B)=\sup_{n<\omega}\mu_{\beta_n}\le$ $\sup_{n<\omega}\alpha_{\beta_n}\le$ $\sup_{n<\omega}\gamma_{\beta_n}\le$ $\sup_{n<\omega}\mu_{(\beta_n+1)}\le$ $\sup_{n<\omega}\mu_{\beta_{n+1}}=\alpha(B)$. It follows that $\alpha(B)=\sup_{n<\omega}\gamma_{\beta_n}\in C$ (using that $\gamma_{\beta_n}\in C$ for each $\beta$, and $C$ is closed). Hence $\alpha(B)\in S\cap C\neq\emptyset$, which completes the proof.


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