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.
Best Answer
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.