
## 5.18 Jacobson spaces

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 $Z \cap X_0$.

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:

1. Open subspaces.

2. Closed subspaces.

3. Locally closed subspaces.

4. Unions of locally closed subspaces.

5. Constructible sets.

6. 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 $X$. $\square$

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

$\{ \text{finite unions loc.\ closed subsets of } X\} \leftrightarrow \{ \text{finite unions loc.\ closed subsets of } X_0\}$

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

$\{ \text{constructible subsets of } X\} \leftrightarrow \{ \text{constructible subsets of } X_0\}$

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. $\square$

Comment #1196 by JuanPablo on

First it may be helpful to remark that the being Jacobson is equivalent to the property: "every nonempty locally closed subset of $X$ has a point $x$ closed in $X$".

So for example in lemma 5.17.2 (tag 005V) can be improved to $Z$ of the form $Z=\overline{\{x\}}$

Comment #1197 by JuanPablo on

My comment was a bit long and wasn't displayed in its entirety. To continue the lemma 5.17.2 (tag 005V) can be improved to $Z=\overline{\{x\}}$ and $X$ arbitrary because a locally closed nonempty in $X$, $K\cap U$ has a point $x\in K\cap U$, so $\overline{\{x\}}\cap U$ is locally closed contained in $K\cap U$ and has a point closed in $X$ by hypohesis.

Comment #1198 by JuanPablo on

Also lemma 5.17.3 (tag 02I7) can be generalized to $X$ Kolmogorov and with a basis of quasi-compact opens, because if $Z\cap U$ is locally closed nonempty withouth points closed in $X$, then restricting $U$ to a smaller open quasi-compact neighborhood of a point of $Z\cap U$ we may assume $U$ is quasi-compact and so there is a point $x\in Z\cap U$ closed in $Z\cap U$ by lemma 5.11.8 (tag 005E), so that $\{x\}$ is locally closed but not closed in $X$.

Comment #1199 by JuanPablo on

Lemma 5.17.5 (4),(6) can be generalized to arbitrary unions of locally closeds in $X$. And the proof becomes a bit clearer if $(\ast)$ is replaced by "Any nonempty locally closed of $T$ has a point closed in $X$". This is true because locally closeds in $T$ are (after restricting to a locally closed in $X$ in (4),(6)) locally closed in $X$.

Comment #1200 by JuanPablo on

Finally I had a problem in lemma 5.17.8 (tag 005Y), there is a biyective correspondence as in lemma 5.17.9 (tag 005Z) by lemma 5.17.5 (tag 005X) but I could not see why if $U$ is retrocompact open in $X$, $U\cap X_0$ is retrocompact in $X_0$.

In case $X$ is Kolmogorov and has a basis of quasi-compact opens, I could see it:

If $V\subset X_0$ is quasi-compact and open in $X_0$ then it can be extended to a quasi-compact $V'$ open in $X$ with $V=V'\cap X_0$. So $U\cap X_0\cap V=U\cap V'\cap X_0=(U\cap V')_0$, the last equality by lemma 5.17.5. So $U\cap V'$ is quasi-compact because $U$ is retrocompact in $X$, and $U\cap X_0\cap V$ is quasi-compact by lemma 5.11.9 (tag 08ZM).

Comment #1201 by JuanPablo on

Ah, please ignore my last comment, If $X$ is Jacobson then $X$ is compact iff $X_0$ is compact with a proof similar to 5.11.9 (tag 08ZM).

Comment #1208 by on

Hi! If you are willing, then maybe you could just send an edited version of this section to our email address? Thanks!

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