Definition 5.18.1. Let $X$ be a topological space. Let $X_0$ be the set of closed points of $X$. We say that $X$ is *Jacobson* if every closed subset $Z \subset X$ is the closure of $Z \cap X_0$.

## 5.18 Jacobson spaces

Note that a topological space $X$ is Jacobson if and only if every nonempty locally closed subset of $X$ has a point closed in $X$.

Let $X$ be a Jacobson space and let $X_0$ be the set of closed points of $X$ with the induced topology. Clearly, the definition implies that the morphism $X_0 \to X$ induces a bijection between the closed subsets of $X_0$ and the closed subsets of $X$. Thus many properties of $X$ are inherited by $X_0$. For example, the Krull dimensions of $X$ and $X_0$ are the same.

Lemma 5.18.2. Let $X$ be a topological space. Let $X_0$ be the set of closed points of $X$. Suppose that for every point $x\in X$ the intersection $X_0 \cap \overline{\{ x\} }$ is dense in $\overline{\{ x\} }$. Then $X$ is Jacobson.

**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.
$\square$

Lemma 5.18.3. Let $X$ be a Kolmogorov topological space with a basis of quasi-compact open sets. If $X$ is not Jacobson, then there exists a non-closed point $x \in X$ such that $\{ x\} $ is locally closed.

**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 $X$.
$\square$

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. $\square$

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 $X$. $\square$

Lemma 5.18.6. A finite Jacobson space is discrete.

**Proof.**
If $X$ is finite Jacobson, $X_0 \subset X$ the subset of closed points, then, on the one hand, $\overline{X_0} = X$. On the other hand, $X$, and hence $X_0$ is finite, so $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.
$\square$

Lemma 5.18.7. Suppose $X$ is a Jacobson topological space. Let $X_0$ be the set of closed points of $X$. There is a bijective, inclusion preserving correspondence

given by $E \mapsto E \cap X_0$. This correspondence preserves the subsets of locally closed, of open and of closed subsets.

**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.
$\square$

Lemma 5.18.8. Suppose $X$ is a Jacobson topological space. Let $X_0$ be the set of closed points of $X$. There is a bijective, inclusion preserving correspondence

given by $E \mapsto E \cap X_0$. This correspondence preserves the subset of retrocompact open subsets, as well as complements of these.

**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.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (8)

Comment #1196 by JuanPablo on

Comment #1197 by JuanPablo on

Comment #1198 by JuanPablo on

Comment #1199 by JuanPablo on

Comment #1200 by JuanPablo on

Comment #1201 by JuanPablo on

Comment #1208 by Johan on

Comment #1240 by Johan on