nonempty interiors can't be defined by their infinite behavior Show that there is no topology with the property that the interior of any set is nonempty if and only if the set is infinite. 
 A: The empty topological space is the only counterexample. Assume for a contradiction that there is a nonempty topological space $X$ in which any set has nonempty interior if and only if it is infinite. Choose a countably infinite set $Y\subseteq X$ and an uncountable family $(Z_t:t\in\mathbb R)$ of almost-disjoint infinite subsets of $Y$. Each set $Z_t$, being infinite, has nonempty interior. Those interiors must be pairwise disjoint, since their pairwise intersections, being open and finite, must be empty. Thus we have an uncountable family of pairwise disjoint nonempty subsets of $Y$, contradicting the fact that $Y$ is countable.
P.S. The argument above uses a weak form of the Axiom of Choice, namely, that every infinite set has a countably infinite subset. This use of the Axiom of Choice can not be avoided. Consider an infinite space $X$ with the cofinite topology, i.e., the open sets are the cofinite sets and the empty set. Clearly, a subset of $X$ has nonempty interior if and only if it cofinite. Now, it is consistent with ZF (Zermelo-Fraenkel set theory without the Axiom of Choice) that there exists a so-called amorphous set, i.e., an infinite set which can not be partitioned into two infinite sets. If $X$ is such a set, then "cofinite" = "infinite" for subsets of $X$.
