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单词 AlternativeCharacterizationsOfNoetherianTopologicalSpacesProofOf
释义

alternative characterizations of Noetherian topological spaces, proof of


We prove the equivalence of the following five conditions for a topological spaceMathworldPlanetmath X:

  • (DCC) X satisfies the descending chain conditionMathworldPlanetmathPlanetmathPlanetmath (http://planetmath.org/DescendingChainCondition) for closed subsets.

  • (ACC) X satisfies the ascending chain conditionMathworldPlanetmathPlanetmath (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 compactPlanetmathPlanetmath.

Proof.

Let f:P(X)P(X) be the complementPlanetmathPlanetmath map,i.e., f(A)=XA for any subset A of X.Then f induces an order-reversing bijectiveMathworldPlanetmathPlanetmath mapbetween the open subsets of X and the closed subsets of X.Sending arbitrary chains (http://planetmath.org/TotalOrder)/setsof open/closed subsets of X through fthen immediately yields the equivalence of conditions of the theorem, and also the equivalence of (Min) and (Max).

(Min) (DCC) is obvious, since the elements ofan infiniteMathworldPlanetmath strictly descending chain of closed subsets of Xwould form a set of closed subsets of X without minimal element.Likewise, given a non-empty set Sconsisting of closed subsets of X without minimal element,one can construct an infinite strictly descending chain in Ssimply by starting with any A0S and choosing for An+1any proper subsetMathworldPlanetmathPlanetmath of An satisfying An+1S.Hence, we have proven conditions (ACC), (DCC), (Min) and (Max) of the theoremto be equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath,and will be done if we can prove equivalence of statement (HC) to the others.

To this end, first assume statement (HC),and let (Ui)i be an ascending sequencePlanetmathPlanetmath of open subsets of X.Then obviously, the Ui form an open cover of U=iUi,which by assumptionPlanetmathPlanetmath is bound to have a finite subcover.Hence, there exists nsuch that i=0nUi=iUi,so our ascending sequence is in fact stationary.Conversely, assume statement (Max) of the theorem,let AX be any subset of Xand let (Ui)iI be a family of open sets in Xsuch that the UiA form an open cover of Awith respect to the subspace topology.Then, by the assumption,the set of finite unions of the Ui has at least one maximal element,say U,and with any iI we obtain UiU=Ubecause of maximality of U.Hence, we have UiU for all iI,so in fact iIUi=U.But, U was a union of a finite number of Ui by construction;hence, a finite subcovering of U and thereby of A has been found. ∎

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