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