I am solving old exams, and I don't have the answer to the following question:
For every $\alpha \lt \kappa$ let $A_\alpha \subset \kappa$ with $|A_\alpha| = \kappa$. Prove that there exists a set $X \subset \kappa$ such that for every $\alpha \lt \kappa$, $|A_\alpha \cap X| = |A_\alpha \setminus X| = \kappa$.
What I tried:
- I've tried to build $X$ using transfinite induction but got stuck in stages where I had to expand $X$ (and no one guarantees that $\cap X_\alpha$ will have the properties I need on limit ordinals).
- I thought about club sets, but didn't find a useful way to use them.
- I am now looking into the idea that if this statement is incorrect, you must have $2^\kappa$ subsets of $\kappa$ that "agree" on $A_\alpha$ such that $|A_\alpha \cap X|$ or $|A_\alpha \setminus X|$ is smaller than $\kappa$. Not sure what to do with that...
Thanks!
Just build $X$ and its complement by transfinite recursion: At stage $\alpha$, we define approximations $X_\alpha$ and $Y_\alpha$ to $X$ and its complement, so that $X_0\subseteq X_1\subseteq\dots$ and $Y_0\subseteq Y_1\subseteq\dots$
At the end (after $\kappa$ stages) $X$ is just the union of the $X_\alpha$, and the union of the $Y_\alpha$ is contained in the complement of $X$.
To do this, first reorder the sets (using that $\kappa\times\kappa\sim\kappa$) as $B_\alpha$, $\alpha<\kappa$, so that each $A_\alpha$ is $B_\gamma$ for $\kappa$ many values of $\gamma$. Consider a typical stage $\alpha$. Recursively, both $\bigcup_{\beta<\alpha}X_\beta$ and $\bigcup_{\beta<\alpha}Y_\beta$ contain precisely $|\alpha|$ many elements. The set $B_\alpha$ is of size $\kappa>|\alpha|$, so there are at least two points $x\ne y$ in $B_\alpha$ that do not appear in $\bigcup_{\beta<\alpha} X_\beta\cup Y_\beta$. Let $X_\alpha=\{x\}\cup\bigcup_{\beta<\alpha}X_\beta$ and $Y_\alpha=\{y\}\cup\bigcup_{\beta<\alpha}Y_\beta$.