Complete regularity is hereditary - Revision history

Vipul at 04:27, 30 January 2014

2014-01-30T04:27:27Z

← Older revision		Revision as of 04:27, 30 January 2014
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	property = completely regular space\|		property = completely regular space\|
	metaproperty = subspace-hereditary property of topological spaces}}		metaproperty = subspace-hereditary property of topological spaces}}
			[[Difficulty level::1\| ]]
	{{basic fact}}		{{basic fact}}

Vipul: /* Proof */

2012-01-25T01:03:31Z

Proof

← Older revision		Revision as of 01:03, 25 January 2012
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	! Step no. !! Assertion/construction !! Facts used !! Given data used !! Previous steps used !! Explanation		! Step no. !! Assertion/construction !! Facts used !! Given data used !! Previous steps used !! Explanation
	\|-		\|-
	\| 1 \|\| There exists a closed subset <matH>D</math> of <math>X</math> such that <math>C= D \cap A</math>. \|\| definition of subspace topology \|\| <math>C</math> is closed in <math>A</math>. \|\| \|\|		\| 1 \|\| There exists a closed subset <matH>D</math> of <math>X</math> such that <math>C= D \cap A</math> \|\| definition of subspace topology \|\| <math>C</math> is closed in <math>A</math>. \|\| \|\| Direct from definition. Note: We can, for concreteness, take <math>D = \overline{C}</math>, the closure of <math>C</math> in <math>X</math> -- this is the smallest <math>D</math> that works.
	\|-		\|-
	\| 2 \|\| <math>x</math> is not in <math>D</math> \|\| \|\| <math>x \in A, x \notin C</math> \|\| Step (1) \|\| <toggledisplay>We are given that <math>x \in A</math>, so if<math>x \in D</math>, then <math>x \in D \cap A</math>. By Step (1), <math>D \cap A = C</math>. Thus, if <math>x \in D</math>, then <math>x \in C</math>, which we know is not true. Hence, <math>x \notin D</math>.</toggledisplay>		\| 2 \|\| <math>x</math> is not in <math>D</math> \|\| \|\| <math>x \in A, x \notin C</math> \|\| Step (1) \|\| <toggledisplay>We are given that <math>x \in A</math>, so if<math>x \in D</math>, then <math>x \in D \cap A</math>. By Step (1), <math>D \cap A = C</math>. Thus, if <math>x \in D</math>, then <math>x \in C</math>, which we know is not true. Hence, <math>x \notin D</math>.</toggledisplay>

Vipul: /* Proof */

2012-01-25T00:07:33Z

Proof

← Older revision		Revision as of 00:07, 25 January 2012
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	\| 5 \|\| The function <math>f</math> defined in Step (4) has the property that <math>f(x) = 0</math> and <math>f(a) = 1</math> for all <math>a \in C</math>. \|\| \|\| \|\| Steps (1), (3), (4) \|\| The condition on <math>f</matH> follows directly from the condition on <math>g</math> in Step (3) and the fact that <math>C = D \cap A</math>, so <matH>C \subseteq D</math>.		\| 5 \|\| The function <math>f</math> defined in Step (4) has the property that <math>f(x) = 0</math> and <math>f(a) = 1</math> for all <math>a \in C</math>. \|\| \|\| \|\| Steps (1), (3), (4) \|\| The condition on <math>f</matH> follows directly from the condition on <math>g</math> in Step (3) and the fact that <math>C = D \cap A</math>, so <matH>C \subseteq D</math>.
	\|}		\|}

			{{tabular proof format}}

	==References==		==References==

Vipul: /* Related facts */

2012-01-24T23:46:20Z

Related facts

← Older revision		Revision as of 23:46, 24 January 2012
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	\| [[regular space]] \|\| [[regularity is hereditary]]		\| [[regular space]] \|\| [[regularity is hereditary]]
	\|-		\|-
	\| [[Urysohn space]] \|\| [[Urysohn is hereditary]		\| [[Urysohn space]] \|\| [[Urysohn is hereditary]]
	\|-		\|-
	\| [[T1 space]] \|\| [[T1 is hereditary]]		\| [[T1 space]] \|\| [[T1 is hereditary]]

Vipul: /* Definitions used */

2012-01-24T23:46:04Z

Definitions used

@@ Line 18: / Line 18: @@
 | [[subspace topology]] || For a subset <math>A</math> of the space <math>X</math>, the subspace topology on <math>A</math> is defined as follows: a subset of <math>A</math> is open in <math>A</math> iff it can be expressed as the intersection with <math>A</math> of an open subset of <math>X</math>.<br>Also, a subset of <math>A</math> is closed in <math>A</math> iff it can be expressed as the intersection with <math>A</math> of a closed subset of <math>X</math>.
 |}
 ==Facts used==

Vipul: /* Proof */

2012-01-24T23:43:51Z

Proof

← Older revision		Revision as of 23:43, 24 January 2012
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	\| [[subspace topology]] \|\| For a subset <math>A</math> of the space <math>X</math>, the subspace topology on <math>A</math> is defined as follows: a subset of <math>A</math> is open in <math>A</math> iff it can be expressed as the intersection with <math>A</math> of an open subset of <math>X</math>.<br>Also, a subset of <math>A</math> is closed in <math>A</math> iff it can be expressed as the intersection with <math>A</math> of a closed subset of <math>X</math>.		\| [[subspace topology]] \|\| For a subset <math>A</math> of the space <math>X</math>, the subspace topology on <math>A</math> is defined as follows: a subset of <math>A</math> is open in <math>A</math> iff it can be expressed as the intersection with <math>A</math> of an open subset of <math>X</math>.<br>Also, a subset of <math>A</math> is closed in <math>A</math> iff it can be expressed as the intersection with <math>A</math> of a closed subset of <math>X</math>.
	\|}		\|}

			==Facts used==

			# [[uses::T1 is hereditary]]

	==Proof==		==Proof==

	===Proof ~~outline===~~		'''Given''': A topological space <math>X</math>, a subset <math>A</math> of <math>X</math>. <math>X</math> is completely regular.

			'''To prove''': <math>A</math> is completely regular.

			'''Proof''': The T1 property for <math>A</math> follows from Fact (1). It thus suffices to show the separation property for <math>A</math>, i.e., that any point and disjoint closed subset in <math>A</math> can be separated by a continuous function. In other words, we want to prove the following.

			'''To prove (specific)''': For any point <math>x \in A</math> and any closed subset <math>C</math> of <math>A</math> such that <math>x \notin A</math>, there exists a continuous function from <math>A</math> to <math>[0,1]</math> such that <math>f(x) = 0</math> and <math>f(a) = 1</math> for all <math>a \in C</math>.

			'''Proof (specific)''':

	* Pick a point and a closed subset of ~~the~~ subspace		{\| class="sortable" border="1"
	* Find a closed subset of the whole space, ~~whose intersection with the subspace~~ is ~~the given subset~~		! Step no. !! Assertion/construction !! Facts used !! Given data used !! Previous steps used !! Explanation
	* Find a continuous function ~~separating the point~~, and the ~~bigger closed subset~~, in the ~~whole space~~		\|-
	* Restrict this continuous function to the subspace, and ~~observe~~ that ~~this works~~		\| 1 \|\| There exists a closed subset <matH>D</math> of <math>X</math> such that <math>C= D \cap A</math>. \|\| definition of subspace topology \|\| <math>C</math> is closed in <math>A</math>. \|\| \|\|
			\|-
			\| 2 \|\| <math>x</math> is not in <math>D</math> \|\| \|\| <math>x \in A, x \notin C</math> \|\| Step (1) \|\| <toggledisplay>We are given that <math>x \in A</math>, so if<math>x \in D</math>, then <math>x \in D \cap A</math>. By Step (1), <math>D \cap A = C</math>. Thus, if <math>x \in D</math>, then <math>x \in C</math>, which we know is not true. Hence, <math>x \notin D</math>.</toggledisplay>
			\|-
			\| 3 \|\| There exists a continuous function <math>g:X \to [0,1]</math> such that <math>g(x) = 0</math> and <math>g(a) = 1</math> for all <math>a \in D</math>. \|\| \|\| <math>X</math> is completely regular \|\| Steps (1), (2) \|\| Step-given combination direct
			\|-
			\| 4 \|\| Define <math>f</math> as the restriction of <math>g</math> to <math>A</math>. In other words, if <math>i: A \to X</math> is the inclusion mapping, then <math>f = g \circ i</math>. Then, <math>f</math> is a continuous function from <math>A</math> to <math>[0,1]</math>. \|\| definition of subspace topology (the inclusion mapping from a subset with the subspace topology is continuous), a composite of continuous maps is continuous \|\| \|\| Step (3) \|\| Step-given combination direct
			\|-
			\| 5 \|\| The function <math>f</math> defined in Step (4) has the property that <math>f(x) = 0</math> and <math>f(a) = 1</math> for all <math>a \in C</math>. \|\| \|\| \|\| Steps (1), (3), (4) \|\| The condition on <math>f</matH> follows directly from the condition on <math>g</math> in Step (3) and the fact that <math>C = D \cap A</math>, so <matH>C \subseteq D</math>.
			\|}

	==References==		==References==

Vipul: /* Definitions used */

2012-01-24T23:33:31Z

Definitions used

← Older revision		Revision as of 23:33, 24 January 2012
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	! Term !! Definition used		! Term !! Definition used
	\|-		\|-
	\| [[completely regular space]] \|\| A space <math>X</math> is completely regular if it is a [[T1 space]] and given any point <math>x \in X</math> and closed subset <math>A \subseteq X</math> such that <math>x \notin A</math>, there exists a [[continuous map]]<math>f:X \to [0,1]</math> such that <math>f(x) = 0</math> and <math>f(a) = 1</math> for all <math>a \in A</math>.		\| [[completely regular space]] \|\| A space <math>X</math> is completely regular if it is a [[T1 space]] and given any point <math>x \in X</math> and closed subset <math>C \subseteq X</math> such that <math>x \notin C</math>, there exists a [[continuous map]]<math>f:X \to [0,1]</math> such that <math>f(x) = 0</math> and <math>f(a) = 1</math> for all <math>a \in C</math>.
	\|-		\|-
	\| [[subspace topology]] \|\| For a subset <math>A</math> of the space <math>X</math>, the subspace topology on <math>A</math> is defined as follows: a subset of <math>A</math> is open in <math>A</math> iff it can be expressed as the intersection with <math>A</math> of an open subset of <math>X</math>.<br>Also, a subset of <math>A</math> is closed in <math>A</math> iff it can be expressed as the intersection with <math>A</math> of a closed subset of <math>X</math>.		\| [[subspace topology]] \|\| For a subset <math>A</math> of the space <math>X</math>, the subspace topology on <math>A</math> is defined as follows: a subset of <math>A</math> is open in <math>A</math> iff it can be expressed as the intersection with <math>A</math> of an open subset of <math>X</math>.<br>Also, a subset of <math>A</math> is closed in <math>A</math> iff it can be expressed as the intersection with <math>A</math> of a closed subset of <math>X</math>.

Vipul: /* Definitions used */

2012-01-24T23:32:50Z

Definitions used

← Older revision		Revision as of 23:32, 24 January 2012
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	\| [[completely regular space]] \|\| A space <math>X</math> is completely regular if it is a [[T1 space]] and given any point <math>x \in X</math> and closed subset <math>A \subseteq X</math> such that <math>x \notin A</math>, there exists a [[continuous map]]<math>f:X \to [0,1]</math> such that <math>f(x) = 0</math> and <math>f(a) = 1</math> for all <math>a \in A</math>.		\| [[completely regular space]] \|\| A space <math>X</math> is completely regular if it is a [[T1 space]] and given any point <math>x \in X</math> and closed subset <math>A \subseteq X</math> such that <math>x \notin A</math>, there exists a [[continuous map]]<math>f:X \to [0,1]</math> such that <math>f(x) = 0</math> and <math>f(a) = 1</math> for all <math>a \in A</math>.
	\|-		\|-
	\| [[subspace topology]] \|\| For a subset <math>A</math> of the space <math>X</math>, the subspace topology on <math>A</math> is defined as follows: a subset of <math>A</~~matH~~> is open in <math>A</math> iff it can be expressed as the intersection with <math>A</math> of an open subset of <math>X</math>.		\| [[subspace topology]] \|\| For a subset <math>A</math> of the space <math>X</math>, the subspace topology on <math>A</math> is defined as follows: a subset of <math>A</math> is open in <math>A</math> iff it can be expressed as the intersection with <math>A</math> of an open subset of <math>X</math>.<br>Also, a subset of <math>A</math> is closed in <math>A</math> iff it can be expressed as the intersection with <math>A</math> of a closed subset of <math>X</math>.
	\|}		\|}

Vipul at 23:20, 24 January 2012

2012-01-24T23:20:16Z

← Older revision		Revision as of 23:20, 24 January 2012
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	{{topospace metaproperty satisfaction\|		{{topospace metaproperty satisfaction\|
	property = completely regular space\|		property = completely regular space\|
	metaproperty = hereditary property of topological spaces}}		metaproperty = subspace-hereditary property of topological spaces}}

	{{basic fact}}		{{basic fact}}

	==Statement==		==Statement==

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

	~~The [[property of topological spaces]] of being [[completely regular space\|completely regular]] is a [[hereditary property of topological spaces]].~~

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

	Any subset of a [[completely regular space]] is completely regular in the [[subspace topology]].		Any subset of a [[completely regular space]] is completely regular in the [[subspace topology]].
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	==Definitions used==		==Definitions used==

	==~~=Completely regular space===~~		{\| class="sortable" border="1"
			! Term !! Definition used
	~~{{further~~\|[[completely regular space]]}}		\|-
			\| [[completely regular space]] \|\| A space <math>X</math> is completely regular if it is a [[T1 space]] and given any point <math>x \in X</math> and closed subset <math>A \subseteq X</math> such that <math>x \notin A</math>, there exists a [[continuous map]]<math>f:X \to [0,1]</math> such that <math>f(x) = 0</math> and <math>f(a) = 1</math> for all <math>a \in A</math>.
	~~===Subspace topology=~~==		\|-
			\| [[subspace topology]] \|\| For a subset <math>A</math> of the space <math>X</math>, the subspace topology on <math>A</math> is defined as follows: a subset of <math>A</matH> is open in <math>A</math> iff it can be expressed as the intersection with <math>A</math> of an open subset of <math>X</math>.
	~~{{further~~\|[[subspace topology]]}}		\|}

	==Proof==		==Proof==

Vipul at 11:21, 8 August 2008

2008-08-08T11:21:08Z

← Older revision		Revision as of 11:21, 8 August 2008
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	{{topospace metaproperty satisfaction}}		{{topospace metaproperty satisfaction\|
			property = completely regular space\|
			metaproperty = hereditary property of topological spaces}}

	{{basic fact}}		{{basic fact}}