Finite set as intersection of two (or more) infinite sets. I stumbled upon an interesting question and I can't seem to find the answer, although it seems easy.
So, if we have an infinite set $X$ and a finite subset $A \subseteq X$ can we find two (or more) infinite subsets $A_1, A_2 \subset X$ such that $A=A_1 \cap A_2$? In either case, can you provide a proof sketch or a counterexample? Thank you in advance! 
 A: Yes. This is because any infinite set can be partitioned into two infinite subsets. So given such an $X$ and $A$, let $Y_1\sqcup Y_2$ be a partition of $X\setminus A$ into disjoint infinite sets, and let $A_1=Y_1\cup A, A_2=Y_2\cup A$.
This proof, however, uses the axiom of choice. Without the axiom of choice, we can have infinite sets which cannot be paritioned into two disjoint infinite subsets; these are called amorphous sets.
To prove (using choice) that you can always partition an infinite set into two disjoint infinite subsets, the most direct approach is via Zorn's lemma. Let $X$ be a fixed infinite set, and let $\mathbb{P}$ be the poset whose elements are pairs $(p, q)$ with 


*

*$p, q$ disjoint subsets of $X$, and

*either $p$ and $q$ are both infinite, or they are both finite and have the same cardinality;
we order $\mathbb{P}$ by $$(p, q)\ge (p', q')\iff p\supseteq p'\mbox{ and } q\supseteq q'.$$ It's not hard to show that every chain in $\mathbb{P}$ has an upper bound (HINT: take unions), so by Zorn $\mathbb{P}$ has a maximal element $(p, q)$; but since $X$ is infinite, we can't have $p, q$ finite (if we did, since $X$ is infinite $X\setminus(p\cup q)$ is also infinite; let $x, y\in X\setminus(p\cup q)$, and consider $(p\cup\{x\}, q\cup\{y\})$), so $p$ and $q$ are infinite subsets of $X$. It's then easy to show that $X=p\sqcup q$, so we're done.
