Section 5.9 (0050): Noetherian topological spaces

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.

Any subset of $X$ with the induced topology is Noetherian.
The space $X$ has finitely many irreducible components.
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 (Lemma 5.8.4), 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.

If $X$ is Noetherian, then $f(X)$ is Noetherian.
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. Let $\{ F_ m\} _{m \in \mathbf{N}}$ a decreasing sequence of closed subsets of $X' = \bigcup _{i = 1, \ldots , n} X_ i$ with the induced topology. Then we can find a decreasing sequence $\{ G_ m\} _{m \in \mathbf{N}}$ of closed subsets of $X$ verifying $F_ m = G_ m \cap X'$ for all $m$ (small detail omitted). As $X_ i$ is noetherian and $\{ G_ m \cap X_ i\} _{m \in \mathbf{N}}$ a decreasing sequence of closed subsets of $X_ i$, there exists $m_ i \in \mathbf{N}$ such that for all $m \geq m_ i$ we have $G_ m \cap X_ i = G_{m_ i} \cap X_ i$. Let $m_0 = \max _{i = 1, \ldots , n} m_ i$. Then clearly

\[ F_ m = G_ m \cap X' = G_ m \cap (X_1 \cup \ldots \cup X_ n) = (G_ m \cap X_1) \cup \ldots (G_ m \cap X_ n) \]

stabilizes for $m \geq m_0$ and the proof is complete. $\square$

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 28.5.9.

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 (8)

Comment #955 by Antoine Chambert-Loir on August 28, 2014 at 09:15

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

Comment #963 by Johan on August 28, 2014 at 17:10

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 September 06, 2014 at 09:57

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

Comment #997 by Johan on September 06, 2014 at 13:15

Yes, indeed, sorry for misunderstanding your point and thanks for persisting. The corresponding change is here. Thanks!

Comment #5560 by Patrick Rabau on October 29, 2020 at 02:54

Lemma 0052, part (2): At the end of the proof of part (2), $Z$ has been written as a finite union of closed irreducible sets, and by removing any redundant member it is claimed that this is the decomposition of $Z$ into its irreducible components. That step is not entirely clear and I think it would be worthwhile to expand it, or even better, make it a lemma in section 5.8 about irreducible components. Something like this maybe:

Lemma: Let $X$ be a topological space and suppose $X=\bigcup_{i=1,\ldots,n}X_i$ where each $X_i$ is an irreducible closed subset of $X$ and no $X_i$ is contained in the union of the other members. Then each $X_i$ is an irreducible component of $X$ and each irreducible component of $X$ is one of the $X_i$ .

Proof: If $Y$ is an irreducible component of $X$ , $Y=\bigcup_{i=1,\ldots,n}(Y\cap X_i)$ and each $Y\cap X_i$ is closed in $Y$ since $X_i$ is closed in $X$ . By irreducibility of $Y$ , $Y$ is equal to one of the $Y\cap X_i$ , that is, $Y\subseteq X_i$ . By maximality of irreducible components, $Y=X_i$ .

Conversely, take one of the $X_i$ and expand it to an irreducible component $Y$ , which we have already shown is one of the $X_j$ . So $X_i\subseteq X_j$ and since the original union does not have redundant members, $X_i=X_j$ , which is an irreducible component.

Comment #5743 by Johan on November 16, 2020 at 14:56

Thanks very much. I added your lemma with your proof. See this commit.

Comment #9885 by Patrick Rabau on December 06, 2024 at 19:55

Lemma 5.9.6 (04MF): At the end of the proof, one can exhibit the open connected neighborhood of $x$ directly without having to mention $E''$ . Let $A=\bigcup\\{Y_i:x\notin Y_i\\}$ , which is closed in $E\cap E'$ . So $V=(E\cap E')\setminus A$ is an open neighborhood of $x$ . Claim: $V$ is connected. Reason: $V$ is the union of the $Y_j\setminus A$ for all the $Y_j$ containing $x$ . Every open subset of an irreducible set is irreducible. So for $Y_j$ containing $x$ , the set $Y_j\setminus A$ is also irreducible, hence connected. And so is their union since they all contain the common point $x$ .

Comment #9886 by Patrick Rabau on December 06, 2024 at 19:57

Sorry the formatting of my previous comment was messed up. $A$ is supposed to be the union of the $Y_i$ that don't contain $x$ .

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5.9 Noetherian topological spaces

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