Every open set has a proper open subset. What spaces satisfy this property?

Suppose $X$ is a topological space satisfying the following property: every (nonempty) open set of $X$ has a (nonempty) proper open subset.

Does this property have a name?

What are some spaces with this property?

What are some properties that imply this property (eg. Hausdorff, separable, second countable, etc.)?

The intuition is that, if you have an open set $U \subseteq X$, you can "zoom in" at any point of $U$, forever.

Example. If $X$ has the discrete topology, then it doesn't have this property, because singletons $\{x\} \subseteq X$ are open sets that don't have (nonempty) proper open subsets.

Example. If $X$ is ${\mathbb R}^n$ with the Euclidean topology, then it does have this property, because, for any open ball $B$ of radius $\epsilon$ around $x \in X$, you can take the open ball of radius $\epsilon/2$, which is a (nonempty) proper open subset of $B$.

• Any space which has open sets who contain only finitely many elements can't have this property, as an extension of the argument for the discrete topology. – Duncan Ramage Nov 18 '17 at 21:53
• Also, any infinite space with a finite topology can't have this property. – étale-cohomology Nov 19 '17 at 6:44
• As stated in the title, the Answer is "None" because the empty set is always open and has no proper subset. Clearly you meant "every non-empty open set". – DanielWainfleet Nov 19 '17 at 12:54
• @DanielWainfleet The title is vague. The main text is more precise. – étale-cohomology Nov 19 '17 at 12:59
• A $T_1$ space with no isolated points has no non-empty subset-minimal open sets. – DanielWainfleet Nov 19 '17 at 13:00

It's not implied by a lot of your properties as a countable discrete space lacks it, but such a space is second countable Hausdorff, separable etc.

It implies some other properties: $X$ has infinitely many open sets, all of which are infinite. (if $X$ would have finitely many open sets, their (open!) intersection would be minimal; If $U$ is open non-empty, define $U_0 = U$ and $\emptyset \neq U_{n+1} \subsetneq U_n$ whenever $X$ has this property, and then $\cup_n (U_{n}\setminus U_{n+1})$ is an infinite subset of $U$, so $U$ is infinite.

Spaces that are dense in themselves (i.e. have no isolated points) and are also $T_0$ will have this property: if $U$ is open then $U$ is not a singleton, so we have $x,y \in X$ with $x \neq y$, By $T_0$-ness we find an open set $V$ containing $x$ but not $y$ (or reversely), and then $U \cap V$ is strictly smaller and open too.

Equivalently, if $X$ fails the condition, there is some minimal open subset $U$. If $U$ has one point this point is isolated. If it has more, the $T_0$ condition fails for points from this set. So your condition is implied by "$X$ is dense in itself and $T_0$".

As @bof remarked in the comments: if $X$ has the "no minimal open set property", and $Y$ is any space whatsoever, then $X \times Y$ also obeys it ($O$ non-empty open in $X \times Y$ contains a basic non-empty open set $U \times V\subseteq O$ and $U$ is not minimal so $\emptyset \neq U' \subsetneq U$ open exists and then $U' \times V$ is smaller and non-empty open inside $O$ so $O$ is not minimal.)

• Could be clearer. Of course "no point has a minimal open set containing it" is only a sufficient condition for "every nonempty open set has a nonempty open proper subset", but the casual reader might think you're saying it's a necessary and sufficient condition. – bof Nov 19 '17 at 0:00
• By the way, $T_0$ is just as good as $T_1$ here. – bof Nov 19 '17 at 0:02
• @bof sure, it's equivalent with $T_0$ added. I expanded. – Henno Brandsma Nov 19 '17 at 7:02
• The OP's condition is not equivalent to "$X$ is dense in itself and $T_0$" because it doesn't imply $T_0$. If $X$ has the OP's property, then $X\times Y$ also has the property, but $Y$ could have the "indiscrete" topology. – bof Nov 19 '17 at 9:16