Irreducible not implies Noetherian: Difference between revisions

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(New page: ==Statement== An irreducible space need not be Noetherian. ==Example== Consider a topological space whose underlying set is uncountable, and where the proper closed subsets are prec...)
 
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{{topospace property non-implication}}
==Statement==
==Statement==


An [[irreducible space]] need not be Noetherian.
An [[irreducible space]] need not be [[Noetherian space|Noetherian]].


==Example==
==Example==


Consider a topological space whose underlying set is uncountable, and where the proper closed subsets are precisely the countable subsets (the [[cocountable topology]]). The topological space is clearly irreducible, because a union of proper closed subsets is countable, and hence again proper. However, it is not Noetherian, because one can have an infinite descending chain of closed subsets.
Consider a topological space whose underlying set is uncountable, and where the proper closed subsets are precisely the countable subsets (the [[cocountable topology]]). The topological space is clearly irreducible, because a union of proper closed subsets is countable, and hence again proper. However, it is not Noetherian, because one can have an infinite descending chain of closed subsets.

Latest revision as of 19:47, 11 May 2008

This article gives the statement and possibly, proof, of a non-implication relation between two topological space properties. That is, it states that every topological space satisfying the first topological space property need not satisfy the second topological space property
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Statement

An irreducible space need not be Noetherian.

Example

Consider a topological space whose underlying set is uncountable, and where the proper closed subsets are precisely the countable subsets (the cocountable topology). The topological space is clearly irreducible, because a union of proper closed subsets is countable, and hence again proper. However, it is not Noetherian, because one can have an infinite descending chain of closed subsets.