Though the following was found in Hartshorne’s *Algebraic Geometry* (read, ‘terror maths’), this is really point-set topology.

**Definition: **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 .

We have the following properties of Noetherian spaces:

**Proposition: **A Noetherian space is compact.

**Proof: **It is easy to show that the Noetherian property is equivalent to every nonempty family of open subsets having a maximal element. (The proof resembles a similar one in ring theory).

Let be the Noetherian space and be an open cover of ; . Define to be the family of finite union of ‘s. Then is a family of open subsets of and hence must have a maximal element, . I claim, because if there is an element , then since is an open cover of , I can find an open set containing . Now, since it is the finite union of elements of the cover of X, and contains properly. This is a contradiction to the maximality of . Hence , a finite union of elements of the open cover.

**Lemma:** (Proof easy) If is any subset of a topological space , then dim dim .

**Theorem:** If is a topological space covered by a family of open subsets , then dim = dim .

**Proof:**

We only prove the case that dim .

Using the above lemma, dim dim is clear, and so taking supremum gives one side of the equality, dim dim . Consider the chain

of closed subsets of , where is a singleton. We show that this chain is preserved when interescted with , for some . This would mean that dim dim . Consider

and assume that for some . Now,

Hence, or, and thus, . In particular, for every . But since is a cover of , some must contain the singleton , thus giving us a contradiction. Hence there exists a such that

thus proving that dim dim .

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March 12, 2010 at 11:21

Partha Solapurkarand this is a blog entry? 😉

March 12, 2010 at 14:10

abhishekparabYes this is indeed a blog post, written by and (exclusively) for me. I write things so that I don’t forget the proofs; why, even Anand took an hour — one lecture — to prove it!

March 3, 2016 at 08:30

AnonymousI have one question: Why do you need X_1 to be a singleton?

If there is U \in {U_i} such that U intersected with the chain will still be strict, we’re done. So there is your assumption that there is no such U:

Your assumption that any U intersected with the chain provides an index i where X_i \cap U = X_{i+1} \cap U already delivers a contradiction: Any U of the open cover has a trivial intersection with X_1 \neq \emptyset, which is false because {U_i} is an open cover of X.

Therefore X_1 could be arbitrary. Am I right?

March 3, 2016 at 09:18

AnonymousIn addition: Not every singleton needs to be closed.