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In a Noetherian topological space, a constructible set is a finite union of locally closed sets. This is a conclusion on constructible sets:

Every constructible set contains a dense open subset of its closure.

Now neglect the Noetherian condition, and give $\mathbb{R}$ the usual topology. $\{0\}$ is closed, thus "constructible". The only open set contained in $\{0 \}$ is the empty set. It is certain that the empty set is not dense in the closure of $\{0 \}$ which is equal to $\{0 \}$.

So, the Noetherian condition is necessary.

But the result does not seem clear to me. Would you please give me a proof or a reference?

Many thanks.

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up vote 3 down vote accepted

The claim is that if $A$ is constructible, then there is a set $D \subseteq A$ such that $\mathrm{cl}_X D \supseteq A$ and $D$ is relatively open in $\mathrm{cl}_X A$; $D$ need not be open in the whose space $X$. In your example $A = \mathrm{cl}_\mathbb{R} A =$ $\{0\}$, and we may take $D=A$: certainly $D$ is dense in $A$, and $D = \mathrm{cl}_\mathbb{R} A$ is also an open subset of $\mathrm{cl}_\mathbb{R} A$.

The result is Lemma 2.1 of this paper; the proof given there is fairly straightforward.

Edit: I should have mentioned that this result is proved for arbitrary topological spaces; they need not be Noetherian.

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I would have expected just about any reference but a paper by Jinpeng An. He spent a year as a post-doc in Zurich with my advisor Marc Burger a few years ago and he was the office mate of a very good friend of mine. – t.b. Jul 30 '11 at 6:49
Thank you very much. The "open subset" in the statement means "open in the closure of the given constructible set", not "open in the whole space"? I am wondering if there is a counterexample, i.e., a non-constructible subset of a space that doesn't contain an open subset dense in its closure. – ShinyaSakai Jul 31 '11 at 1:58
@ShinyaSakai: Yes, there is. In the reals with the usual topology let $A=\mathbb{Q}\cap [0,1]$; then $\mathrm{cl}A=[0,1]$, so no non-empty subset of $A$ is relatively open in $\mathrm{cl}A$. – Brian M. Scott Jul 31 '11 at 2:04
@Brian: Thank you very much for your reference and counterexample. Now I understand this problem much better. – ShinyaSakai Jul 31 '11 at 17:10

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