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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).

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]

2 added 77 characters in body

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).

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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# Constructible sets in Hausdorff spaces

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).

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 $X=\emptyset$.