T1 is hereditary - Revision history

Vipul at 04:32, 30 January 2014

2014-01-30T04:32:55Z

← Older revision		Revision as of 04:32, 30 January 2014
Line 1:		Line 1:
	{{topospace metaproperty satisfaction}}		{{topospace metaproperty satisfaction\|property = T1 space\|metaproperty = subspace-hereditary property of topological spaces}}
			[[Difficulty level::1\| ]]
	==Statement==		==Statement==

	~~===Property-theoretic statement===~~

	~~The [[property of topological spaces]] of being a [[T1 space]] satisfies the [[metaproperty of topological spaces]] of being [[subspace-hereditary property of topological spaces\|subspace-hereditary]].~~

	~~===Verbal statement===~~

	Any subspace of a [[T1 space]], endowed with the [[subspace topology]], is again a <math>T_1</math>-space.		Any subspace of a [[T1 space]], endowed with the [[subspace topology]], is again a <math>T_1</math>-space.

Vipul: /* Proof */

2008-07-21T15:36:24Z

Proof

← Older revision		Revision as of 15:36, 21 July 2008
Line 35:		Line 35:
	===Proof in terms of two-point definition of T1===		===Proof in terms of two-point definition of T1===

	'''Given''': A T1 space <math>X</math>, a subset <math>A</math>		'''Given''': A <math>T_1</math>-space <math>X</math>, a subset <math>A</math>

	'''To prove''': <math>A</math> is a T1 space when endowed with the subspace topology		'''To prove''': <math>A</math> is a <math>T_1</math>-space when endowed with the subspace topology

	'''Proof''': We need to show that if <math>x \ne y</math> are both points of <math>A</math>, then there exists an open subset of <math>A</math> containing <math>x</math> and not containing <math>y</math>.		'''Proof''': We need to show that if <math>x \ne y</math> are both points of <math>A</math>, then there exists an open subset of <math>A</math> containing <math>x</math> and not containing <math>y</math>.

Vipul: New page: {{topospace metaproperty satisfaction}} ==Statement== ===Property-theoretic statement=== The property of topological spaces of being a T1 space satisfies the [[metaproperty of t...

2008-07-21T15:35:04Z

New page: {{topospace metaproperty satisfaction}} ==Statement== ===Property-theoretic statement=== The property of topological spaces of being a T1 space satisfies the [[metaproperty of t...

New page

{{topospace metaproperty satisfaction}}

==Statement==

===Property-theoretic statement===

The [[property of topological spaces]] of being a [[T1 space]] satisfies the [[metaproperty of topological spaces]] of being [[subspace-hereditary property of topological spaces|subspace-hereditary]].

===Verbal statement===

Any subspace of a [[T1 space]], endowed with the [[subspace topology]], is again a <math>T_1</math>-space.

==Definitions used==

===T1 space===
{{further|[[T1 space]]}}

A topological space <math>X</math> is termed <math>T_1</math> if it satisfies the following equivalent conditions:

# For any two distinct points <math>x,y \in X</math>, there exists an open subset <math>U</math> of <math>X</math> such that <math>x \in U, y \notin U</math>
# For every point <math>x \in X</math>, the singleton subset <math>\{ x \}</math> is closed in <math>X</math>
# For every point <math>x \in X</math>, the intersection of all open subsets of <math>X</math> containing <math>x</math>, is <math>\{ x \}</math>

===Subspace topology===

{{further|[[Subspace topology]]}}

The subspace topology on a subset <math>A</math> of <math>X</math> is defined in the following equivalent ways:

# A subset <math>U</math> of <math>A</math> is open in <math>A</math> iff there exists an open subset <math>V</math> of <math>X</math> such that <math>V \cap A = U</math>.
# A subset <math>C</math> of <math>A</math> is closed in <math>A</math> iff there exists a closed subset <math>D</math> of <math>X</math> such that <math>D \cap A = C</math>.

==Proof==

===Proof in terms of two-point definition of T1===

'''Given''': A T1 space <math>X</math>, a subset <math>A</math>

'''To prove''': <math>A</math> is a T1 space when endowed with the subspace topology

'''Proof''': We need to show that if <math>x \ne y</math> are both points of <math>A</math>, then there exists an open subset of <math>A</math> containing <math>x</math> and not containing <math>y</math>.

Since <math>x,y</math> are distinct points of <math>A</math>, they are also distinct points of <math>X</math>. Since <math>X</math> is a <math>T_1</math>-space, there exists an open subset <math>V</math> of <math>X</math> such that <math>x \in V</math> and <math>y \notin V</math>. Now consider the set <math>U = V \cap A</math>. Then, by definition of subspace topology, <math>U</math> is open in <math>A</math>. Further, since <math>x \in A</math> and <math>x \in V</math>, we have <math>x \in U</math>. Since <math>y \notin V</math>, we have <math>y \notin U</math>. Thus, <math>U</math> is an open subset of <math>X</math> containing <math>x</math> and not containing <math>y</math>.
===Proof in terms of points-are-closed definition of T1===

===Proof in terms of third definition of T1===