Lemma 5.18.5. Let $X$ be Jacobson. The following types of subsets $T \subset X$ are Jacobson:
Open subspaces.
Closed subspaces.
Locally closed subspaces.
Unions of locally closed subspaces.
Constructible sets.
Any subset $T \subset X$ which locally on $X$ is a union of locally closed subsets.
In each of these cases closed points of $T$ are closed in $X$.
Proof. Let $X_0$ be the set of closed points of $X$. For any subset $T \subset X$ we let $(*)$ denote the property:
Every nonempty locally closed subset of $T$ has a point closed in $X$.
Note that always $X_0 \cap T \subset T_0$. Hence property $(*)$ implies that $T$ is Jacobson. In addition it clearly implies that every closed point of $T$ is closed in $X$.
Suppose that $T=\bigcup _ i T_ i$ with $T_ i$ locally closed in $X$. Take $A\subset T$ a locally closed nonempty subset in $T$, then there exists a $T_ i$ such that $A\cap T_ i$ is nonempty, it is locally closed in $T_ i$ and so in $X$. As $X$ is Jacobson $A$ has a point closed in $X$. $\square$
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