Vipul: /* Related facts */

2009-10-26T18:46:17Z

Related facts

← Older revision		Revision as of 18:46, 26 October 2009
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	* [[Hausdorff implies US]]		* [[Hausdorff implies US]]
	* [[Hausdorff implies KC]]		* [[Hausdorff implies KC]]
	* [[US not implies ~~Huasdorff~~]]		* [[US not implies Hausdorff]]

	==Proof==		==Proof==

Vipul: Created page with '{{topospace property implication| stronger = US-space| weaker = T1 space}} ==Statement== Any US-space (i.e., a space where every convergent sequence has a unique limit)…'

2009-10-26T18:45:50Z

Created page with '{{topospace property implication| stronger = US-space| weaker = T1 space}} ==Statement== Any US-space (i.e., a space where every convergent sequence has a unique limit)…'

New page

{{topospace property implication|
stronger = US-space|
weaker = T1 space}}

==Statement==

Any [[US-space]] (i.e., a space where every convergent sequence has a unique [[limit]]) is a [[T1 space]] (i.e., every point is closed).

==Related facts==

* [[T1 not implies US]]
* [[Hausdorff implies US]]
* [[Hausdorff implies KC]]
* [[US not implies Huasdorff]]

==Proof==

'''Given''': A US-space <math>X</math>.

'''To prove''': <math>X</math> is a <math>T_1</math>-space.

'''Proof''': We need to show that for every point <math>x \in X</math>, the singleton subset <math>\{ x \}</math> is closed. For this, take the sequence <math>\{ x,x,x, \dots \}</math>. By the definition of limit, <math>x</math> is a limit of the sequence, and hence, since the space is US, it is the only limit. Thus, for any point <math>y \ne x</math>, there exists an open neighborhood of <math>y</math> not containing <math>x</math>. Thus, <math>\{ x \}</math> is closed.

US implies T1 - Revision history

Vipul: /* Related facts */

Vipul: Created page with '{{topospace property implication| stronger = US-space| weaker = T1 space}} ==Statement== Any US-space (i.e., a space where every convergent sequence has a unique limit)…'