The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

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. $\square$

slogan

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. $\square$


Comments (8)

Comment #1196 by JuanPablo on

Some comments on this section:

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

So for example in lemma 5.17.2 (tag 005V) can be improved to of the form

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 and arbitrary because a locally closed nonempty in , has a point , so is locally closed contained in and has a point closed in by hypohesis.

Comment #1198 by JuanPablo on

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

Comment #1199 by JuanPablo on

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

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 is retrocompact open in , is retrocompact in .

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

If is quasi-compact and open in then it can be extended to a quasi-compact open in with . So , the last equality by lemma 5.17.5. So is quasi-compact because is retrocompact in , and is quasi-compact by lemma 5.11.9 (tag 08ZM).

Comment #1201 by JuanPablo on

Ah, please ignore my last comment, If is Jacobson then is compact iff 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!


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.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 005T. Beware of the difference between the letter 'O' and the digit '0'.