Closed set which does not have open subsets

Question

I wonder if there is an example of a set $A$ that is closed and that does not contain any nonempty open set.

Answer 1

Here is an example of such a set with infinitely many elements:

The set of integers in $\mathbb R$ is closed, and any nonempty subset of integers is closed as well.

Answer 2

In any Hausdorff space, any finite set is closed.

So in any connected Hausdorff space a singleton does contain any non-empty open set.

(I mentioned connectedness as then the singleton cannot be open since it is already closed.)

Answer 3

The empty set is closed and does not contain any nonempty open set. Probably you meant to ask for a nonempty closed set which does not contain any nonempty open set. The following theorem answers this question in a general topological space X.

Theorem. If in the space $X$ there is a closed set which is not open, then there is a nonempty closed set which does not contain any nonempty open set.

Proof. If there is a closed set which is not open, then its complement, call it $U,$ is an open set which is not closed. Of course $U\ne\emptyset,$ since $\emptyset$ is closed. Assuming the axiom of choice, we can extend $\{U\}$ to a maximal collection $\mathcal U$ of pairwise disjoint nonempty open sets. Then the set $A=X\setminus\bigcup\mathcal U$ is a closed set which does not contain any nonempty open set. I claim that $A\ne\emptyset.$

If we had $A=\emptyset,$ it would follow that $\bigcup\mathcal U=X$ and that $\bigcup(\mathcal U\setminus\{U\})=X\setminus U,$ so that $X\setminus U$ would be an open set, contradicting the fact that $U$ is not closed.