Show there is a set which has a finite intersection with any set from a countable family

Let $X$ and $Y$ be countable sets such that $(\bigcup X)\cup(\bigcup Y)$ is countable, and such that for each $x\in X$ and $y\in Y$, $x\cap y$ is finite. Prove that there exists some $z$ such that $x\cap z$ is finite for each $x\in X$ and $y\backslash z$ is finite for each $y\in Y$.

This question is a slightly modified version of question question 19.1(b) from Discovering Modern Set Theory volume 2.

The original question had $X$ and $Y$ elements of $[[\omega]^{\aleph_0}]]^{<2^{\aleph_0}}$ and asked us to prove the result by assuming the continuum hypothesis. (The background set theory is ZFC)

I understood that CH gives us that X and Y are countable. The question comes from a chapter Martin's Axiom, but once we have X and Y are countable, we no longer need Martin's Axiom to show the existence of a generic filter.

The question states that this would be unprovable if we didn't assume CH, which means what ever approach we use to solve this must somehow use the fact that X and Y are countable.

The standard approach is to define a partial order $\mathbb{P} = \{ \langle s, F \rangle: s\in[\omega]^{<\aleph_0}, F\in[A]^{<\aleph_0}\} $ Where $A$ is either $X$ or $X\cup Y$ With the ordering $\langle s,F \rangle \leq \langle s_1, F_1\rangle$ iff $s\supseteq s_1$, $F\supseteq F_1$ and $(s\backslash s_1)\cap\bigcup F_1=\varnothing$

You then define a family of dense sets for each $x\in X$ $$D_x=\{\langle s,F\rangle\in\mathbb{P}: x\in F\} $$ If we define a filter $\mathbb{G}$ which is generic over $D_x$ and we can then define $$z=\bigcup \{s:\exists F(\langle s,F\rangle\in\mathbb{G}\}$$ Which will have the property that $z\cap x$ is finite for each $x\in X$

The part I am stuck at is we need to define a second family of dense sets somehow based on the elements of $Y$ in-order to assure that $z$ defined in the same way for the new generic filter also has the property that $y\backslash z$ is finite for each $y\in Y$.

I am not sure how to define the second family though, although I know we will somehow have to use the fact that $X$ and $Y$ are countable, as well as the fact that $x\cap y$ is finite, as neither fact has been used anywhere yet.

Solution 1:

1. If $\cup X=\emptyset$ then let $z=\cup Y.$

2. If $\cup Y=\emptyset\ne\cup X$ then let $z=\emptyset.$

3. If $\cup X\ne\emptyset\ne \cup Y$ then let $X=\{x_n: n\in\Bbb N\}$ and $Y=\{y_n: n\in\Bbb N\}$. It does not matter whether $x_n=x_m$ or $y_n=y_m$ for some $n\ne m.$

For $n\in\Bbb N$ let $z_n=y_n\setminus (\cup_{j<n}x_j).$ Let $z=\cup_{n\in\Bbb N}z_n.$

Now for any $k\in \Bbb N:$

$x_k$ is disjoint from $z_n$ for all $n> k$. So $x_k\cap z= \cup_{j\le k}(x_k\cap z_j)\subseteq \cup_{j\le k}(x_k\cap y_j)$ which is finite.

$y_k\setminus z\subseteq y_k\setminus z_k\subseteq \cup_{j<k}(y_k\cap x_j)$ which is finite.