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*}

5.9 Noetherian topological spaces

Definition 5.9.1. A topological space is called Noetherian if the descending chain condition holds for closed subsets of $X$. A topological space is called locally Noetherian if every point has a neighbourhood which is Noetherian.

Lemma 5.9.2. Let $X$ be a Noetherian topological space.

  1. Any subset of $X$ with the induced topology is Noetherian.

  2. The space $X$ has finitely many irreducible components.

  3. Each irreducible component of $X$ contains a nonempty open of $X$.

Proof. Let $T \subset X$ be a subset of $X$. Let $T_1 \supset T_2 \supset \ldots $ be a descending chain of closed subsets of $T$. Write $T_ i = T \cap Z_ i$ with $Z_ i \subset X$ closed. Consider the descending chain of closed subsets $Z_1 \supset Z_1\cap Z_2 \supset Z_1 \cap Z_2 \cap Z_3 \ldots $ This stabilizes by assumption and hence the original sequence of $T_ i$ stabilizes. Thus $T$ is Noetherian.

Let $A$ be the set of closed subsets of $X$ which do not have finitely many irreducible components. Assume that $A$ is not empty to arrive at a contradiction. The set $A$ is partially ordered by inclusion: $\alpha \leq \alpha ' \Leftrightarrow Z_{\alpha } \subset Z_{\alpha '}$. By the descending chain condition we may find a smallest element of $A$, say $Z$. As $Z$ is not a finite union of irreducible components, it is not irreducible. Hence we can write $Z = Z' \cup Z''$ and both are strictly smaller closed subsets. By construction $Z' = \bigcup Z'_ i$ and $Z'' = \bigcup Z''_ j$ are finite unions of their irreducible components. Hence $Z = \bigcup Z'_ i \cup \bigcup Z''_ j$ is a finite union of irreducible closed subsets. After removing redundant members of this expression, this will be the decomposition of $Z$ into its irreducible components, a contradiction.

Let $Z \subset X$ be an irreducible component of $X$. Let $Z_1, \ldots , Z_ n$ be the other irreducible components of $X$. Consider $U = Z \setminus (Z_1\cup \ldots \cup Z_ n)$. This is not empty since otherwise the irreducible space $Z$ would be contained in one of the other $Z_ i$. Because $X = Z \cup Z_1 \cup \ldots Z_ n$ (see Lemma 5.8.3), also $U = X \setminus (Z_1\cup \ldots \cup Z_ n)$ and hence open in $X$. Thus $Z$ contains a nonempty open of $X$. $\square$

Lemma 5.9.3. Let $f : X \to Y$ be a continuous map of topological spaces.

  1. If $X$ is Noetherian, then $f(X)$ is Noetherian.

  2. If $X$ is locally Noetherian and $f$ open, then $f(X)$ is locally Noetherian.

Proof. In case (1), suppose that $Z_1 \supset Z_2 \supset Z_3 \supset \ldots $ is a descending chain of closed subsets of $f(X)$ (as usual with the induced topology as a subset of $Y$). Then $f^{-1}(Z_1) \supset f^{-1}(Z_2) \supset f^{-1}(Z_3) \supset \ldots $ is a descending chain of closed subsets of $X$. Hence this chain stabilizes. Since $f(f^{-1}(Z_ i)) = Z_ i$ we conclude that $Z_1 \supset Z_2 \supset Z_3 \supset \ldots $ stabilizes also. In case (2), let $y \in f(X)$. Choose $x \in X$ with $f(x) = y$. By assumption there exists a neighbourhood $E \subset X$ of $x$ which is Noetherian. Then $f(E) \subset f(X)$ is a neighbourhood which is Noetherian by part (1). $\square$

Lemma 5.9.4. Let $X$ be a topological space. Let $X_ i \subset X$, $i = 1, \ldots , n$ be a finite collection of subsets. If each $X_ i$ is Noetherian (with the induced topology), then $\bigcup _{i = 1, \ldots , n} X_ i$ is Noetherian (with the induced topology).

Proof. Omitted. $\square$

Example 5.9.5. Any nonempty, Kolmogorov Noetherian topological space has a closed point (combine Lemmas 5.12.8 and 5.12.13). Let $X = \{ 1, 2, 3, \ldots \} $. Define a topology on $X$ with opens $\emptyset $, $\{ 1, 2, \ldots , n\} $, $n \geq 1$ and $X$. Thus $X$ is a locally Noetherian topological space, without any closed points. This space cannot be the underlying topological space of a locally Noetherian scheme, see Properties, Lemma 27.5.9.

Lemma 5.9.6. Let $X$ be a locally Noetherian topological space. Then $X$ is locally connected.

Proof. Let $x \in X$. Let $E$ be a neighbourhood of $x$. We have to find a connected neighbourhood of $x$ contained in $E$. By assumption there exists a neighbourhood $E'$ of $x$ which is Noetherian. Then $E \cap E'$ is Noetherian, see Lemma 5.9.2. Let $E \cap E' = Y_1 \cup \ldots \cup Y_ n$ be the decomposition into irreducible components, see Lemma 5.9.2. Let $E'' = \bigcup _{x \in Y_ i} Y_ i$. This is a connected subset of $E \cap E'$ containing $x$. It contains the open $E \cap E' \setminus (\bigcup _{x \not\in Y_ i} Y_ i)$ of $E \cap E'$ and hence it is a neighbourhood of $x$ in $X$. This proves the lemma. $\square$


Comments (4)

Comment #955 by Antoine Chambert-Loir on

Lemma 5.8.2 should state that every irreducible subset of a noetherian topological space is contained in an irreducible component, equivalently that is the union of its irreducible components. This is more or less proved: Define as the set of closed subspaces of which are not the union of finitely many irreducible subsets.

Comment #963 by on

By our definition of irreducible components (as maximal irreducible subsets) this is true for every topological space. See Lemma 5.8.3.

Comment #996 by Antoine Chambert-Loir on

Sorry to have overlooked that Lemma 004W. But you will agree that it should state: ``every irreducible subset of is contained in an irreducible component''. (With the same proof, starting from the given irreducible subset instead of a singleton.)

Comment #997 by on

Yes, indeed, sorry for misunderstanding your point and thanks for persisting. The corresponding change is here. 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 0050. Beware of the difference between the letter 'O' and the digit '0'.