The Stacks project

Proof. Let $Z$ be closed subset of $X$ and $U$ be and open subset of $X$ such that $U\cap Z$ is nonempty. Then for $x\in U\cap Z$ we have that $\overline{\{ x\} }\cap U$ is a nonempty subset of $Z\cap U$, and by hypothesis it contains a point closed in $X$ as required. $\square$

Proof. As $X$ is not Jacobson there exists a closed set $Z$ and an open set $U$ in $X$ such that $Z \cap U$ is nonempty and does not contain points closed in $X$. As $X$ has a basis of quasi-compact open sets we may replace $U$ by an open quasi-compact neighborhood of a point in $Z\cap U$ and so we may assume that $U$ is quasi-compact open. By Lemma 5.12.8, there exists a point $x \in Z \cap U$ closed in $Z \cap U$, and so $\{ x\} $ is locally closed but not closed in $X$. $\square$

Proof. Let $X$ be a topological space. Let $X_0$ be the set of closed points of $X$. Let $U_{i, 0}$ be the set of closed points of $U_ i$. Then $X_0 \cap U_ i \subset U_{i, 0}$ but equality may not hold in general.

First, assume that each $U_ i$ is Jacobson. We claim that in this case $X_0 \cap U_ i = U_{i, 0}$. Namely, suppose that $x \in U_{i, 0}$, i.e., $x$ is closed in $U_ i$. Let $\overline{\{ x\} }$ be the closure in $X$. Consider $\overline{\{ x\} } \cap U_ j$. If $x \not\in U_ j$, then $\overline{\{ x\} } \cap U_ j = \emptyset $. If $x \in U_ j$, then $U_ i \cap U_ j \subset U_ j$ is an open subset of $U_ j$ containing $x$. Let $T' = U_ j \setminus U_ i \cap U_ j$ and $T = \{ x\} \amalg T'$. Then $T$, $T'$ are closed subsets of $U_ j$ and $T$ contains $x$. As $U_ j$ is Jacobson we see that the closed points of $U_ j$ are dense in $T$. Because $T = \{ x\} \amalg T'$ this can only be the case if $x$ is closed in $U_ j$. Hence $\overline{\{ x\} } \cap U_ j = \{ x\} $. We conclude that $\overline{\{ x\} } = \{ x \} $ as desired.

Let $Z \subset X$ be a closed subset (still assuming each $U_ i$ is Jacobson). Since now we know that $X_0 \cap Z \cap U_ i = U_{i, 0} \cap Z$ are dense in $Z \cap U_ i$ it follows immediately that $X_0 \cap Z$ is dense in $Z$.

Conversely, assume that $X$ is Jacobson. Let $Z \subset U_ i$ be closed. Then $X_0 \cap \overline{Z}$ is dense in $\overline{Z}$. Hence also $X_0 \cap Z$ is dense in $Z$, because $\overline{Z} \setminus Z$ is closed. As $X_0 \cap U_ i \subset U_{i, 0}$ we see that $U_{i, 0} \cap Z$ is dense in $Z$. Thus $U_ i$ is Jacobson as desired. $\square$

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 $X$. $\square$

Proof. If $X$ is finite then the set $X_0 \subset X$ of closed points is finite. Assume $X_0$ is is finite and $X$ is Jacobson. Then $\overline{X_0} = X$ by the Jacobson property. Now $X_0 =\{ x_1, \ldots , x_ n\} = \bigcup _{i = 1, \ldots , n} \{ x_ i\} $ is a finite union of closed sets, hence closed, so $X = \overline{X_0} = X_0$. Every point is closed, and by finiteness, every point is open. $\square$

Proof. We just prove that the correspondence $E \mapsto E \cap X_0$ is injective. Indeed if $E\neq E'$ then without loss of generality $E\setminus E'$ is nonempty, and it is a finite union of locally closed sets (details omitted). As $X$ is Jacobson, we see that $(E \setminus E') \cap X_0 = E \cap X_0 \setminus E' \cap X_0$ is not empty. $\square$

Proof. From Lemma 5.18.7 above, we just have to see that if $U$ is open in $X$ then $U\cap X_0$ is retrocompact in $X_0$ if and only if $U$ is retrocompact in $X$. This follows if we prove that for $U$ open in $X$ then $U\cap X_0$ is quasi-compact if and only if $U$ is quasi-compact. From Lemma 5.18.5 it follows that we may replace $X$ by $U$ and assume that $U = X$. Finally notice that any collection of opens $\mathcal{U}$ of $X$ cover $X$ if and only if they cover $X_0$, using the Jacobson property of $X$ in the closed $X\setminus \bigcup \mathcal{U}$ to find a point in $X_0$ if it were nonempty. $\square$