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Noetherian topological space
A topological space is called Noetherian if it satisfies the descending chain condition for closed subsets: for any sequence
of closed subsets of , there is an integer such that .
As a first example, note that all finite topological spaces are Noetherian.
There is a lot of interplay between the Noetherian condition and compactness:
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Every Noetherian topological space is quasi-compact.
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A Hausdorff topological space is Noetherian if and only if every subspace of is compact. (i.e. is hereditarily compact)
Note that if is a Noetherian ring, then , the prime spectrum of , is a Noetherian topological space.
Example of a Noetherian topological space:
The space (affine -space over a field ) under the Zariski topology is an example of a Noetherian topological space. By properties of the ideal of a subset of , we know that if
is a descending chain of Zariski-closed subsets, then is an ascending chain of ideals of .
Since is a Noetherian ring, there exists an integer such that . But because we have a one-to-one correspondence between radical ideals of and Zariski-closed sets in , we have for all . Hence as required.
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Comments
Relationship to noetherian rings
Suppose $R$ is a commutative unital ring. Are the following two concepts equivalent?
* Spec R is a noetherian topological space
* R is noetherian
Re: Relationship to noetherian rings
[Sent this via PM mail as well; sorry about the duplication.]
Apparently not. According to Problem 2.13 (p. 80) from Hartshorne, the latter condition implies the former, but the implication is one-way: Spec(R) can be noetherian even when R is not. I can't supply proofs of these results offhand.