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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 (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.

  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 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 (6)

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!

Comment #5560 by Patrick Rabau on

Lemma 0052, part (2): At the end of the proof of part (2), 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 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 be a topological space and suppose where each is an irreducible closed subset of and no is contained in the union of the other members. Then each is an irreducible component of and each irreducible component of is one of the .

Proof: If is an irreducible component of , and each is closed in since is closed in . By irreducibility of , is equal to one of the , that is, . By maximality of irreducible components, .

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

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