Lemma 5.18.4. Let $X$ be a topological space. Let $X = \bigcup U_ i$ be an open covering. Then $X$ is Jacobson if and only if each $U_ i$ is Jacobson. Moreover, in this case $X_0 = \bigcup U_{i, 0}$.

**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. $\square$

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