Some questions in Set Theory I have some exam questions that were left unanswered for me:


*

*Suppose that for every $\alpha<\kappa$ there is a subset $A_\alpha$ of $\kappa$ of cardinality $\kappa$. Show that there is a subset X of $\kappa$ so that for every $\alpha<\kappa$ : |$A_\alpha\cap$ X|=|$A_\alpha$ \ X|=$\kappa$. 

*{$A_\alpha | \alpha<\aleph_1$} a collection of non-stationary disjoint sets. Suppose that their union is a stationary set. Prove that the set {$minA_\alpha | \alpha<\aleph_1$} is stationary.

*For every couple of sets A,B $\in P(\omega)$ we will denote A $\subseteq^*$ B if $|A\setminus B|<\aleph_0$. Prove that there exist a sequence of length $\aleph_1$ in relation to $\subseteq^*$ .


I'll appreciate any help regarding those questions.
Thanks in advance,
Pavel
 A: Question 3 admits an easy answer, once you check two facts:

$A\subseteq^* B\subseteq^* C$ implies $A\subseteq^* C$.

This is completely straightforward: If $A\setminus B\subseteq m$ and $B\setminus C\subseteq n$, then $A\setminus C\subseteq \max\{m,n\}$.
The second is:

Given a (strictly) $\subseteq^*$-increasing sequence $(A_n\mid n\in\omega)$ (so $A_n\setminus A_{n+1}$ is finite but $A_{n+1}\setminus A_n$ is infinite), there is a $B$ with $A_n\subseteq^* B$ for all $n$ (so, in particular, $B\setminus A_n$ is infinite), and such that $\omega\setminus B$ is infinite as well.

For this: Define two strictly increasing sequences of numbers $m_0<m_1<\dots$ and $$t_{00}<t_{01}<t_{10}<t_{11}<t_{20}<t_{21}<\dots$$ so that  $A_n\setminus A_{n+1}\subseteq m_n$ for all $n$, $[t_{n0},t_{n1})\cap (\omega\setminus A_{n+1})\ne\emptyset$, and $m_n<t_{n0}<t_{n1}<m_{n+1}$ for all $n$. (Check that this is possible.)
Let $$ B=\bigcup_n [m_n,m_{n+1})\cap A_{n+1}. $$ Check that $A_n\setminus B\subseteq m_n$ for any $n$: If $s\in A_n\setminus B$ and $s\ge m_n$ then $s\in A_{n+1}$ (by definition of $m_n$), so $s\ge m_{n+1}$ (or else $s\in B$); inductively, this gives us that $s\ge m_k$ for all $k$, which of course is absurd.  
Moreover, by construction, $\omega\setminus B$ contains at least one element of each interval $[t_{n0},t_{n1})$, so it is infinite.
Using these two facts, a straightforward transfinite construction of length $\omega_1$ gives us the desired $\subseteq^*$ increasing sequence $(A_\alpha\mid\alpha<\omega_1)$:
Start with any $A_0\subseteq\omega$ infinite and coinfinite. Given $A_\alpha$ infinite and coinfinite, let $A_{\alpha+1}$ consist of $A_\alpha$ and an infinite-coinfinite subset of $\omega\setminus A_\alpha$.
At limit stages $\alpha<\omega_1$, given $(A_\beta\mid \beta<\alpha)$, fix $(\beta_n\mid n<\omega)$ increasing and cofinal in $\alpha$ and use the second fact to find $A_\alpha$ with $A_{\beta_n}\subseteq^* A_\alpha$ for all $n$ and $\omega\setminus A_\alpha$ infinite. That $A_\beta\subseteq A_\alpha$ for all $\beta<\alpha$ now follows from the first fact.

For question 2, use Fodor's lemma: Suppose, towards a contradiction that the $A_\alpha$ are pairwise disjoint non-stationary subsets of $\omega_1$ such that $\bigcup_\alpha A_\alpha$ is stationary, and yet $$B=\{{\rm min}(A_\alpha)\mid\alpha<\omega_1\}$$ is non-stationary. Let $A=\bigcup_\alpha A_\alpha\setminus B$. Then $A$ is stationary. Define a regressive $f:A\to\omega_1$ as follows: Given $\beta\in A$, there is a unique $\alpha$ such that $\beta\in A_\alpha$. Then set $f(\beta)=\min(A_\alpha)$.
By Fodor, $f$ is constant on a stationary set. This is, by construction, a subset of a fixed $A_\alpha$ (since the $A_\alpha$ are pairwise disjoint), contradicting that no $A_\alpha$ is stationary.
As a side remark, note that you need something like the assumption that the $A_\alpha$ are disjoint. An easy counterexample otherwise is obtained by setting $A_\alpha=\alpha+1$. Then $\bigcup_\alpha A_\alpha=\omega_1$, yet $B=\{0\}$.
A: Here is a solution for the first problem:
We already know that for every cardinal $\kappa$ we have $\kappa=\kappa\cdot\kappa$. So separate every $A_\alpha$ into $\kappa$ partitions of size $\kappa$ and let's call them $A_{\alpha,\beta}$. Now let $\{B_\alpha : \alpha<\kappa\}$ be an enumeration of all the $A_{\alpha,\beta}$. Observe now that we need to find sets $X,Y$ such that for every $\alpha<\kappa$ we have $B_\alpha\cap X\neq\varnothing\neq B_\alpha\cap Y$ and $X\cap Y=\varnothing$. 
Each of these sets will satisfy the property we want since in every $A_{\alpha,\beta}$ there will exist an element in $X$ and one in $Y$ and therefore $\kappa$ elements of $A_\alpha$ in $X$ and $\kappa$ in $Y$ and furthermore $X\cap Y=\varnothing$ will give $Y\subset\kappa\setminus X$ and $X\subset\kappa\setminus Y$. I will construct two such sets $\mathcal{C}$, $\mathcal{D}$ inductively:
Let $\gamma_0$ be the least element of $B_0$ and make $\mathcal{C}_0=\{\gamma_0\}$. Now let $\delta_0$ be the least element of $B_0\setminus\mathcal{C}_0$ and let $\mathcal{D}_0=\{\delta_0\}$. Assume now that for every $\xi<\alpha$ we have defined $\mathcal{C}_\xi$ and $\mathcal{D}_\xi$ such that each of them have at most $|\xi+1|$ elements.
If $(\bigcup_{\xi<\alpha}\mathcal{C}_\xi)\cap B_\alpha$ is non-empty let $\mathcal{C}_\alpha=\bigcup_{\xi<\alpha}\mathcal{C}_\xi$  Otherwise let $\gamma_\alpha$ be the least element of $B_\alpha\setminus(\bigcup_{\xi<\alpha}\mathcal{D}_\xi)$ (this element has to exist since the union has at most $|\alpha|$ elements while $B_\alpha$ has $\kappa$) and let $\mathcal{C}_\alpha=(\bigcup_{\xi<\alpha}\mathcal{C}_\xi)\cup\{\gamma_\alpha\}$ (observe that this has at most $|\alpha+1|$ elements). Now if $(\bigcup_{\xi<\alpha}\mathcal{D}_\xi)\cap B_\alpha$ is non-empty let $\mathcal{D}_\alpha=\bigcup_{\xi<\alpha}\mathcal{D}_\xi$. Otherwise if $\delta_\alpha$ is the least element of $B_\alpha\setminus\mathcal{C}_\alpha$ (again this exists) let $\mathcal{D}_\alpha=(\bigcup_{\xi<\alpha}\mathcal{D}_\xi)\cup\{\delta_\alpha\}$. Finally let $\mathcal{C}=\bigcup_{\alpha<\kappa}\mathcal{C}_\alpha$ and $\mathcal{D}=\bigcup_{\alpha<\kappa}\mathcal{D}_\alpha$.
Finally, notice that $\mathcal{C}\cap\mathcal{D}=\varnothing$ because of the above construction and for every $\alpha<\kappa$ we have $B_\alpha\cap\mathcal{C}\neq\varnothing$ and $B_\alpha\cap\mathcal{D}\neq\varnothing$.
