Constructing a section of an equivalence relation compatible with the intersection

Question

Let $X$ be a (countably) infinite set and define an equivalence relation $\sim$ on the power set $P(X)$ of $X$ by defining two subsets $A$ and $B$ of $X$ to be equivalent if they differ by at most finitely many elements (i.e., $A \sim B$ if the symmetric difference $A \Delta B$ is finite).

Let $[-]\colon P(X) \to P(X)/_{\sim}$ denote the quotient map onto the equivalence classes. Does there exist a section $s \colon P(X)/_{\sim} \to P(X)$ of it with the following three properties?

1. $s([\emptyset]) = \emptyset$.
2. $s([X]) = X$.
3. For every $A$ and $B$ in $P(X)$ we have $s([A]) \cap s([B]) = s([A \cap B])$.

I tried around constructing such a section by using the Lemma of Zorn, but I couldn't figure out how to make the crucial step in the argument work (enlarging the domain of such a section, but which is defined only on a subset of $P(X)/_{\sim}$, by at least one more missing element from $P(X)/_{\sim}$).

Answer 1

No such section exists. The main point in the proof is that there exist uncountably many (in fact continuum many) infinite subsets of $\mathbb N$ such that the intersection of any two of them is finite. (Such sets are called almost disjoint.) In order to preserve $\cap$ and $\emptyset$, a section would have to send the equivalence classes of these almost disjoint sets to genuinely disjoint subsets of $\mathbb N$, and there aren't enough of those.

To produce the claimed almost disjoint sets, it suffices to produce them as subsets of some other countably infinite set rather than of $\mathbb N$. Take the countably infinite set $\mathbb Q$ and associate to each real number $r$ a sequence $A_r$ of distinct rationals converging to $r$. Distinct $r$'s give almost disjoint $A_r$'s.