alternative characterizations of Noetherian topological spaces, proof of
We prove the equivalence of the following five conditions for a topological space^{} $X$:

•
(DCC) $X$ satisfies the descending chain condition^{} (http://planetmath.org/DescendingChainCondition) for closed subsets.

•
(ACC) $X$ satisfies the ascending chain condition^{} (http://planetmath.org/AscendingChainCondition) for open subsets.

•
(Min) Every nonempty family of closed subsets has a minimal element.

•
(Max) Every nonempty family of open subsets has a maximal element.

•
(HC) Every subset of $X$ is compact^{}.
Proof.
Let $f:P(X)\to P(X)$ be the complement^{} map, i.e., $f(A)=X\setminus A$ for any subset $A$ of $X$. Then $f$ induces an orderreversing bijective^{} map between the open subsets of $X$ and the closed subsets of $X$. Sending arbitrary chains (http://planetmath.org/TotalOrder)/sets of open/closed subsets of $X$ through $f$ then immediately yields the equivalence of conditions of the theorem, and also the equivalence of (Min) and (Max).
(Min) $\Rightarrow $ (DCC) is obvious, since the elements of an infinite^{} strictly descending chain of closed subsets of $X$ would form a set of closed subsets of $X$ without minimal element. Likewise, given a nonempty set $S$ consisting of closed subsets of $X$ without minimal element, one can construct an infinite strictly descending chain in $S$ simply by starting with any ${A}_{0}\in S$ and choosing for ${A}_{n+1}$ any proper subset^{} of ${A}_{n}$ satisfying ${A}_{n+1}\in S$. Hence, we have proven conditions (ACC), (DCC), (Min) and (Max) of the theorem to be equivalent^{}, and will be done if we can prove equivalence of statement (HC) to the others.
To this end, first assume statement (HC), and let ${({U}_{i})}_{i\in \mathbb{N}}$ be an ascending sequence^{} of open subsets of $X$. Then obviously, the ${U}_{i}$ form an open cover of $U={\bigcup}_{i\in \mathbb{N}}{U}_{i}$, which by assumption^{} is bound to have a finite subcover. Hence, there exists $n\in \mathbb{N}$ such that ${\bigcup}_{i=0}^{n}{U}_{i}={\bigcup}_{i\in \mathbb{N}}{U}_{i}$, so our ascending sequence is in fact stationary. Conversely, assume statement (Max) of the theorem, let $A\subseteq X$ be any subset of $X$ and let ${({U}_{i})}_{i\in I}$ be a family of open sets in $X$ such that the ${U}_{i}\cap A$ form an open cover of $A$ with respect to the subspace topology. Then, by the assumption, the set of finite unions of the ${U}_{i}$ has at least one maximal element, say $U$, and with any $i\in I$ we obtain ${U}_{i}\cup U=U$ because of maximality of $U$. Hence, we have ${U}_{i}\subseteq U$ for all $i\in I$, so in fact ${\bigcup}_{i\in I}{U}_{i}=U$. But, $U$ was a union of a finite number of ${U}_{i}$ by construction; hence, a finite subcovering of $U$ and thereby of $A$ has been found. ∎
Title  alternative characterizations of Noetherian topological spaces, proof of 

Canonical name  AlternativeCharacterizationsOfNoetherianTopologicalSpacesProofOf 
Date of creation  20130322 15:25:39 
Last modified on  20130322 15:25:39 
Owner  yark (2760) 
Last modified by  yark (2760) 
Numerical id  12 
Author  yark (2760) 
Entry type  Proof 
Classification  msc 14A10 