Is a closed set the closure of an open set?

Question

Working in an arbitrary topological space is it necessary that a closed set is the closure of some open set from this topology?

If not then what property does guarantee that any closed set is the closure of an open set?

Answer 1

Just reposting my comment as an answer.

If $X$ is a $T_{1}$ space (a space in which all singletons are closed) that is not discrete, then $X$ has closed sets that are not the closure of an open set. In such a space, let $x$ be a point for which $\{x\}$ is not open. Then $\{x\}$ is a closed set by virtue of $X$ being $T_{1}$, but $\{x\}$ has empty interior and is therefore not the closure of an open set.

Being discrete is a sufficient property for every closed set to (trivially) be the closure of an open set, but I'm unsure of what a necessary and sufficient condition would be.

Answer 2

The observation that any non-open singleton $\{x\}$ in a $T_1$ space gives a counter-example and settles the question is certainly correct.

I would just add that in fact counterexamples are plentiful and not at all sophisticated.

For instance any line in $\mathbb R^2$ or, for that matters, any hyperplane in $\mathbb R^n$ will do.

More generally, let $f(x_1,...,x_n)\in{\mathbb R}[x_1,...,x_n]$ be any non zero polynomal: then the subset $$ Z_f=\{(x_1,...,x_n)\in{\mathbb R}^n\,|\,f(x_1,...,x_n)=0\} $$ if non empty is closed in the standard topology and not the closure of an open set (this is actually an exercise).

Even more generally any subvariety of $\mathbb R^n$ (or $\mathbb C^n$) of dimension $<n$ has the same property.