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

Question

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 $B$.

Answer 1

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.)