Which sets occur as boundaries of other sets in topological spaces?

Of course the boundary of a set is closed. But is every closed set in a topological space, the boundary of some set in that space?

It is tempting to assert that boundaries have empty interiors, but this is not true, as is shown by the fact that the boundary of $\mathbb{Q}$ in $\mathbb{R}$ is $\mathbb{R}$. In fact it can be seen in general that the boundary of a dense set with empty interior is the whole space. Thus the only boundary in an indiscrete space is the whole set. (This is sort of complementary to the comment of Daniel Fischer below.)

However the intuitive feeling comes right for open sets (and then for closed sets as well): The boundary of an open set cannot contain an open set.

This question concerns all subsets of a topological space.

An alternative question shall be to characterise all topological spaces in which every closed set occurs as a boundary.

Note: I have now asked this on MathOverflow.

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    $\begingroup$ No. For example in discrete spaces, the only boundary is $\varnothing$. $\endgroup$ Jun 23, 2014 at 19:24
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    $\begingroup$ More generally, a boundary contains no isolated points (of the ambient space, not necessarily of itself). $\endgroup$ Jul 9, 2014 at 11:54
  • $\begingroup$ I might be misunderstanding, but doesn't the boundary of $[0,1]$ consist of 2 isolated points? $\endgroup$ Jul 11, 2014 at 5:20
  • $\begingroup$ @DavisYoshida No, the points 0 and 1 are not isolated points of the ambient space R (or [0,1] itself, if you want). They are limit points. What you mean is that they are isolated points of the boundary set {0,1} taken as a space in itself (which happens to be discrete). This is what the parenthesis in Emil's comment is about. $\endgroup$ Jul 11, 2014 at 9:34
  • $\begingroup$ @NUnnikrishnan Ah my mistake. Thanks. $\endgroup$ Jul 12, 2014 at 6:39

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