TITLE: Some questions in Set Theory
QUESTION [5 upvotes]: 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

REPLY [5 votes]: 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 $B_\alpha\cap\mathcal{D}\neq\varnothing$.