US not implies KC: Difference between revisions

Latest revision as of 21:39, 3 March 2010

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 (i.e., US-space (?)) need not satisfy the second topological space property (i.e., KC-space (?))
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Statement

It is possible to have a US-space (i.e., a topological space in which every convergent sequence has at most one limit) that is not a KC-space.

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 It is possible to have a [[US-space]] (i.e., a topological space in which every convergent sequence has at most one [[limit]]) that is not a [[KC-space]].
-==Proof==
-===Example of cofinite topology===
-Consider a countable set, say <math>X = \{ 1,2,3,\dots \}</math>, equipped with the [[cofinite topology]]. With this topology, the set is a US-space, because by definition, the only convergent sequence are those that are eventually constant, with the unique limit being the eventual constant value.
-On the other hand, <math>X</math> is not a KC-space, because, in fact, ''every'' subset of <math>X</math> is compact, including the infinite proper subsets, which are not closed. To see why every subset of <math>X</math> is compact, note that if <math>Y</math> is a subset of <math>X</math>, any nonempty open subset of <math>Y</math> is cofinite in <math>Y</math>. Hence, an open cover of <math>Y</math> must have a finite subcover.