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

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    $\begingroup$ In a second-countable space, yes, every closed set is a boundary there. In full generality, no idea yet. $\endgroup$ – Daniel Fischer Oct 20 '13 at 23:41
  • $\begingroup$ 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. $\endgroup$ – Niels J. Diepeveen Oct 21 '13 at 14:59
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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:

  1. E. Hewitt, A problem of set-theoretic topology, Duke Math. J. 10 (1943), 309-333.

  2. K. Padmavally, An example of a connected irresolvable Hausdorff space, Duke Math. J. 20 (1953), 513-520.

  3. Douglas R. Anderson, On connected irresolvable Hausdorff spaces, Proc. Amer. Math. Soc. 16 (1965), 463-466.

  4. Guram Bezhanishvili, Ray Mines, and Patrick J. Morandi, Scattered, Hausdorff-reducible, and hereditarily irresolvable spaces, Topology and Appl. 132 (2003) 291-306.

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.

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