Name for topological space where every neighborhood of some point is open? Let $(X,\tau)$ be a topological space such that for every (nonempty) open set $U$, $U \subseteq V$ implies $V$ is open. Is there a name for topological spaces with this property? A quick google search turned nothing up.
 A: I don't know a name for your property, but maybe it will help to describe it in terms of more familiar constructions.
Clearly, a discrete space has your property. Also, if $\mathcal F$ is a proper filter on $X,$ then $\tau=\mathcal F\cup\{\emptyset\}$ is a topology on $X$ which has your property. I claim that these are the only examples.
Claim. If a topological space $(X,\tau)$ has the property "every superset of a nonempty open set is open", then either $\tau=\mathcal P(X)$ (i.e., the discrete topology), or else $\tau=\mathcal F\cup\{\emptyset\}$ where $\mathcal F$ is some proper filter on $X.$
Proof. Suppose $\tau\ne\mathcal P(X);$ I have to show that $\tau\setminus\{\emptyset\},$ the collection of all nonempty open sets, is a proper filter on $X.$ The only nontrivial step is showing that the intersection of two nonempty open sets is nonempty. Let $U,V$ be any two nonempty open sets. Choose a set $S\in\mathcal P(X)\setminus\tau.$ Then $S\cup U$ and $S\cup V$ are open sets, and so $S\cup(U\cap V)=(S\cup U)\cap(S\cup V)$ is open. Since $S$ is not open, we can conclude that $S\cup(U\cap V)\ne S,$ and so $U\cap V\ne\emptyset.$
