4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ This would make the topology a filter, but I'm not sure that this has a special name. $\endgroup$ – user259242 Nov 7 '16 at 9:00
  • $\begingroup$ @user259242 Filters have the property that any two elements have nonempty interseciton. $\endgroup$ – Crostul Nov 7 '16 at 9:05
  • $\begingroup$ @bof Right you are. Good pick up. $\endgroup$ – user259242 Nov 7 '16 at 11:47
  • $\begingroup$ @Crostul If we follow Bourbaki in defining a filter to be an upwardly closed collection of subsets closed under finite intersections and not including the empty set, then you're correct. Others (including myself and bof) call such gagdets proper filters and allow filters to contain the empty set. $\endgroup$ – user259242 Nov 7 '16 at 11:49
2
$\begingroup$

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 $U\cap V\ne\emptyset.$

$\endgroup$
  • $\begingroup$ This was not trivial. Very elegant, though, +1. $\endgroup$ – Crostul Nov 7 '16 at 12:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.