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In an attempt to write a proof by contradiction, I end up with a space $X$ with the following properties:

(0) $X$ is nonempty,
(1) $X$ is Hausdorff,
(2) $X$ has no isolated points,
(3) every subspace of $X$ is constructible (finite union of locally closed subsets).

Is this indeed a contradiction?

It would suffice to know that any $X$ with properties (1) and (2) has a dense subset with dense complement: such a set cannot be constructible unless $X=\emptyset$. [Edit: there are counterexamples to this, see the comment by Yves Cornulier]

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A space union of two disjoint dense subsets is called (2-)resolvable. There are Hausdorff perfect counterexamples, see wcomfort.faculty.wesleyan.edu/files/2012/02/wc101N.pdf –  YCor Oct 6 '12 at 16:43
Thanks, Yves! So much for that approach, then. –  Laurent Moret-Bailly Oct 6 '12 at 19:01

1 Answer 1

up vote 6 down vote accepted

No, it is not a contradiction.

A space $X$ is called submaximal if every subset of $X$ is locally closed (and hence constructible). It is easy to see that spaces with only finitely many non-isolated points are submaximal. Finding a submaximal Hausdorff space with no isolated points seems harder but there are plenty of those too.

It is a nice exercise to show that any maximal space is in fact submaximal. A Hausdorff space $(X,\tau)$ is called maximal if it has no isolated points but any finer topology $\tau' \supset \tau$ has isolated points. With a standard use of Zorn´s lemma you can show that any topology without isolated points is contained in a maximal topology. This gives you a way of "constructing" a lot of submaximal Hausdorff spaces with no isolated points. Some of them are even Tychonoff spaces (see E. van Douwen, "Applications of maximal topologies", Topology Appl., 1993), although these are a lot harder to get.

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