Let $(X,\tau)$ be a topological space. We say that $x\neq y\in X$ are *close* if every neighborhood of $x$ intersects every neighborhood of $y$. We associate to $(X,\tau)$ a graph $G(X,\tau)=(V,E)$ with $V = X$ and $$E=\big\{\{x,y\}: x\neq y \in X \text{ and } x,y \text{ are close}\big\}.$$


Given an infinite simple graph $G=(V,E)$, is there a topology $\tau$ on $V$ such that $G(V,\tau) = G$?



Graph associated to a topological space
I think the answer is (sadly) no : take a complete graph $K\_n = (V, E)$ and remove an edge, say between $x$ and $y$. Then assume you have such a topology $\tau$. Take any $U, W\in \tau$ such that $x\in U$, $y \in W$ and $U\cap W=\emptyset$ (two such open sets exist,otherwise $x$ would be close to $y$). Since any point in $U$, different from $x$ is close to $y$, if there were any, say $z$, we would have that $U$ is a neighbourhood of $z$, and $W$ a neighbourhood of $y$, so that $U\cap W \neq \emptyset$. That's a contradiction, so $U=\{x\}$. For sufficiently large $n$, this is obviously impossible ($n=4$ for instance)


Edit :Bof was quicker than me


2nd edit : there was a mistake in my argument : the last part where I say it's impossible : well it's not. There needs to be at least 4 points so that you can remove 2 edges with no common endpoint, apply the same reasoning to another $a$, so that both $\{a\}$ and $\{x\}$ are open sets, and $a,x$ are connected by an edge, which then brings us a contradiction. Sorry about that. (However the answer is still no)


