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 $P(X)/\_{\sim}$).



Constructing a section of an equivalence relation compatible with the intersection
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.


