Question concerning finite intersecting sets Let $\{X_i\}_{i=1}^{\infty}$, $\{Y_j\}_{j=1}^{\infty}$ be finite sets of cardinality at most $n$. If for any finite $F$, there are $i,j \in \mathbb{N}$ such that $F \cap X_i \cap Y_j = \emptyset$, prove that there are $k,l \in \mathbb{N}$ such that $X_k \cap Y_l = \emptyset$.
 A: Let $\{X_i\}_{i\in\mathbb N},\{Y_i\}_{i\in\mathbb N}$ be families of sets with $|X_i|\leq n$ and $|Y_i|\leq n$ for some positive integer $n$.
If for all $F$ with $|F|<\infty$ there are $i,j\in\mathbb N$ such that $F\bigcap X_i\bigcap Y_j=\emptyset$ then there exists $X\in\{X_i\}_{i\in\mathbb N}$ and $Y\in\{Y_i\}_{i\in\mathbb N}$ such that $X\bigcap Y=\emptyset$.
Proof:
Choose some $X_1\in\{X_i\}_{i\in\mathbb N}$.
Then there exists $k,l\in\mathbb N$ such that,
$$X_1\bigcap X_k\bigcap Y_l=\emptyset$$
Since we can assume that $X_1\bigcap Y_k\neq\emptyset$ and $X_k\bigcap Y_l\neq\emptyset$ it follows that $X_1\bigcap X_k=\emptyset$. 
(If $X_i\bigcap Y_k=\emptyset$ for $i\neq k$ then we would be done, so the upper assumption is valid.)
Define $X_2:=X_k$  with $X_k$ as above and consider $(X_1\bigcup X_2)$.
Then since $(X_1\bigcup X_2)$ is finite there exists again $l,j$ such that,
$$(X_1\bigcup X_2)\bigcap X_l\bigcap Y_j=\emptyset$$
By the same argumentation as above it follows that $X_1\bigcap X_l=\emptyset$ and $X_2\bigcap X_l=\emptyset$.
Proceeding with this we find $n+1$ disjoint sets i.e $\{X_1,...X_{n+1}\}$ with $X_i\bigcap X_j=\emptyset$ where $i\neq j$.
Now taking an arbitrary $Y\in\{Y_i\}_{i\in\mathbb N}$ and since $|Y|\leq n$ it follows that there exists $X_i\in\{X_1,...X_{n+1}\}$ such that $X_i\bigcap Y=\emptyset$.
QED
