# Is a perfect set a boundary?

In a topological space, is a perfect set (i.e. closed with no isolated points) always the boundary of some set?

• In a second-countable space, yes, every closed set is a boundary there. In full generality, no idea yet. – Daniel Fischer Oct 20 '13 at 23:41
• The answer to math.stackexchange.com/questions/151265/… implies that this is true iff every space without isolated points can be partitioned into two dense subsets. – Niels J. Diepeveen Oct 21 '13 at 14:59

A topological space is irresolvable if there is no dense subset whose complement is also dense. If $X$ is irresolvable and dense-in-itself, then $X$ is a perfect subset of $X$ which is not the boundary of any subset of $X$. Irresolvable dense-in-itself spaces have been constructed by Hewitt and others:
For an easy example of an irresolvable T$_1$-space with no isolated points, take an infinite set $X$ and a nonprincipal ultrafilter $\mathcal U$ on $X$, and topologize $X$ by making $\mathcal U$ the collection of all nonempty open sets. This is a door space; clearly, a nonempty door space is irresolvable.