# alternative characterizations of Noetherian topological spaces, proof of

Proof.

Let $f\colon P(X)\to P(X)$ be the complement  map, i.e., $f(A)=X\setminus A$ for any subset $A$ of $X$. Then $f$ induces an order-reversing 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 non-empty 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 AlternativeCharacterizationsOfNoetherianTopologicalSpacesProofOf 2013-03-22 15:25:39 2013-03-22 15:25:39 yark (2760) yark (2760) 12 yark (2760) Proof msc 14A10