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Jul 13, 2014 at 17:49 vote accept N Unnikrishnan
Jul 12, 2014 at 21:14 history edited Will Sawin CC BY-SA 3.0
Just fixing your description of what I pointed out : )
Jul 12, 2014 at 5:57 comment added nomen I wish I had known that stuff about $V = L$ when I was writing my undergrad thesis. I spun my wheels for a long, long time trying to find a bottom to the recursion $C = \partial(\partial C \cup (C^\circ \cap A))$ for a particular kind of $A$. It turns out Baire category isn't enough to impose a bottom.
Jul 11, 2014 at 21:49 comment added Christian Remling Equivalently, we can say that every closed set in $X$ is a boundary precisely if $X$ is a boundary (this is, of course, a trivial reformulation of this very neat answer, but I kind of like it).
Jul 11, 2014 at 6:20 comment added N Unnikrishnan @WillSawin I am in a dilemma as only one answer can be accepted. The title question is answered technically by Will Sawin, though, as he himself notes, the crucial idea comes from this elegant post of Joseph. Of course, this is because I have asked two questions in one post.
Jul 11, 2014 at 6:15 comment added N Unnikrishnan Special thanks for all the extra information on resolvable and maximally resolvable spaces!
Jul 11, 2014 at 4:35 history edited Joseph Van Name CC BY-SA 3.0
added 1289 characters in body
Jul 11, 2014 at 3:09 history answered Joseph Van Name CC BY-SA 3.0