Vipul at 21:39, 3 March 2010

2010-03-03T21:39:56Z

← Older revision		Revision as of 21:39, 3 March 2010
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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]].		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.

Vipul: Created page with '{{topospace property non-implication| stronger = US-space| weaker = KC-space}} ==Statement== It is possible to have a US-space (i.e., a topological space in which every con…'

2009-10-26T20:28:52Z

Created page with '{{topospace property non-implication| stronger = US-space| weaker = KC-space}} ==Statement== It is possible to have a US-space (i.e., a topological space in which every con…'

New page

{{topospace property non-implication|
stronger = US-space|
weaker = KC-space}}

==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]].

==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.

US not implies KC - Revision history

Vipul at 21:39, 3 March 2010

Vipul: Created page with '{{topospace property non-implication| stronger = US-space| weaker = KC-space}} ==Statement== It is possible to have a US-space (i.e., a topological space in which every con…'