Closeness graph of a topological space Let $(X,\tau)$ be a topological space. We say that $x, y \in X$ are close if for every neighborhood $U$ of $x$ and $V$ of $y$ we have $U\cap V \neq \emptyset$. Let $E$ be the set of $\{x,y\}$ where $x,y\in X$ with $x\neq y$ and $x,y$ are close. We say $G(X,\tau) = (X,E)$ is the closeness graph of $(X,\tau)$.
Given any simple undirected graph $G$, is there a topological space $(X,\tau)$ such that $G\cong G(X,\tau)$?
 A: There is no topology on 4 elements whose closeness graph is the $4$-cycle.
(Proof: Let's call the elements $\{A,B,C,D\}$. Define a directed graph with these elements as vertices and a directed edge $u\to v$ whenever every open set containing $u$, also contains $v$, this has to be transitive and contain all self loops, lets call this directed graph $H(X,\tau)$. Two vertices $u,v$ in the closeness graph are connected iff there is a $w$ such that $u\to w$ and $v\to w$ in $H$. Now, we see that we can't have edges between $A$ and $C$ or between $B$ and $D$, however any orientation of the $4$-cycle $ABCD$ either includes a diagonal in the transitive closure, or it has edges $A\to B$, $C\to B$, or it has edges $B\to C$, $D\to C$. In each case the closeness graph must contain one of the diagonals.)
Update: Every undirected graph $G$ can be realized as a connected component of the closeness graph of some topological space. To prove this, let $Y=V(G)\sqcup E(G)$, and let $\tau$ be the topology generated by the open sets $\{e\}$ for all edges $e$, and $\{v,e\}$ whenever $v$ is an endpoint of $e$. Then $G$ is a connected component of $G(Y,\tau)$.
