Compactness is weakly hereditary - Revision history

Vipul at 04:26, 30 January 2014

2014-01-30T04:26:55Z

← Older revision		Revision as of 04:26, 30 January 2014
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	property = compact space\|		property = compact space\|
	metaproperty = weakly hereditary property of topological spaces}}		metaproperty = weakly hereditary property of topological spaces}}
			[[Difficulty level::1\| ]]
	==Statement==		==Statement==

Vipul: /* Proof */

2010-09-15T14:57:54Z

Proof

← Older revision		Revision as of 14:57, 15 September 2010
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	! Step no. !! Assertion/construction !! Given data used !! Facts used !! Previous steps used !! Explanation		! Step no. !! Assertion/construction !! Given data used !! Facts used !! Previous steps used !! Explanation
	\|-		\|-
	\| 1 \|\|By the definition of subspace topology, we can find open sets <math>V_i</math> of <math>X</math> such that <math>V_i \cap A = U_i</math>, thus the union of the <math>V_i</math>s contains <math>A</math>. \|\| <math>A</math> is a subspace of <math>X</math> \|\| -- \|\| --		\| 1 \|\|By the definition of subspace topology, we can find open sets <math>V_i</math> of <math>X</math> such that <math>V_i \cap A = U_i</math>, thus the union of the <math>V_i</math>s contains <math>A</math>. \|\| <math>A</math> is a subspace of <math>X</math> \|\| -- \|\| -- \|\|
	\|-		\|-
	\| 2 \|\| The <math>V_i</math>s along with <math>X \setminus A</math> form an open cover of <math>X</math> \|\| <math>A</math> is closed in <math>X</math> \|\| -- \|\| Step (1) \|\| <toggledisplay><math>X \setminus A</math> is open because <math>A</math> is closed. Further, since the union of all the <math>V_i</math>s contains <math>A</math>, that along with <math>X \setminus A</math> covers all of <math>X</math>.</toggledisplay>		\| 2 \|\| The <math>V_i</math>s along with <math>X \setminus A</math> form an open cover of <math>X</math> \|\| <math>A</math> is closed in <math>X</math> \|\| -- \|\| Step (1) \|\| <toggledisplay><math>X \setminus A</math> is open because <math>A</math> is closed. Further, since the union of all the <math>V_i</math>s contains <math>A</math>, that along with <math>X \setminus A</math> covers all of <math>X</math>.</toggledisplay>

Vipul: /* Proof */

2010-09-15T14:55:55Z

Proof

← Older revision		Revision as of 14:55, 15 September 2010
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	'''Proof''':		'''Proof''':

	# By the definition of subspace topology, we can find open sets <math>V_i</math> of <math>X</math> such that <math>V_i \cap A = U_i</math>, thus the union of the <math>V_i</math>s contains <math>A</math>.		{\| class="sortable" border="1"
	~~# Since~~ <math>A</math> is closed~~, we can ''throw~~ in~~'' the open set~~ <math>X \setminus A</math>, ~~and get an open cover~~ of the ~~''whole space''~~ <math>X</math>.		! Step no. !! Assertion/construction !! Given data used !! Facts used !! Previous steps used !! Explanation
	~~# Since the whole space is compact, this~~ open cover has a finite subcover. In other words, there is a finite subcollection of the <math>V_i</math>s, that, ~~possibly~~ along with <math>X \setminus A</math>, covers ~~the whole of~~ <math>X</math>.		\|-
	# By ''throwing out'' <math>X \setminus A</math>, we get a finite collection of <math>V_i</math>s whose union contains <math>A</math>. The corresponding <math>U_i</math> now form a finite subcover of the original cover of <math>A</math>.		\| 1 \|\|By the definition of subspace topology, we can find open sets <math>V_i</math> of <math>X</math> such that <math>V_i \cap A = U_i</math>, thus the union of the <math>V_i</math>s contains <math>A</math>. \|\| <math>A</math> is a subspace of <math>X</math> \|\| -- \|\| --
			\|-
			\| 2 \|\| The <math>V_i</math>s along with <math>X \setminus A</math> form an open cover of <math>X</math> \|\| <math>A</math> is closed in <math>X</math> \|\| -- \|\| Step (1) \|\| <toggledisplay><math>X \setminus A</math> is open because <math>A</math> is closed. Further, since the union of all the <math>V_i</math>s contains <math>A</math>, that along with <math>X \setminus A</math> covers all of <math>X</math>.</toggledisplay>
			\|-
			\| 3 \|\| The open cover from step (2) has a finite subcover. In other words, there is a finite subcollection of the <math>V_i</math>s, that, along with <math>X \setminus A</math>, covers <math>X</math>. \|\| <math>X</math> is compact \|\| -- \|\| Step (2) \|\|
			\|-
			\| 4 \|\| By ''throwing out'' <math>X \setminus A</math>, we get a finite collection of <math>V_i</math>s whose union contains <math>A</math> \|\| -- \|\| -- \|\| Step (3) \|\|
			\|-
			\| 5 \|\| The corresponding <math>U_i</math> now form a finite subcover of the original cover of <math>A</math>. \|\| -- \|\| -- \|\| Steps (1), (4) \|\|
			\|}

	===Proof in terms of finite intersection property===		===Proof in terms of finite intersection property===
	{{fillin}}		{{fillin}}

	==References==		==References==

Vipul at 02:01, 17 July 2009

2009-07-17T02:01:04Z

← Older revision		Revision as of 02:01, 17 July 2009
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	Any [[closed subset]] of a [[compact space]] is compact (when given the [[subspace topology]]).		Any [[closed subset]] of a [[compact space]] is compact (when given the [[subspace topology]]).

			==Related facts==

			* [[Hausdorff implies KC]]: In other words, every compact subset of a Hausdorff space is a [[closed subset]].

			===Weakly hereditary for properties related to compactness===

			* [[Paracompactness is weakly hereditary]]: Every [[closed subset]] of a paracompact space is paracompact.
			* [[Orthocompactness is weakly hereditary]]
			* [[Metacompactness is weakly hereditary]]


	==Proof==		==Proof==
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	'''Given''': <math>X</math> a [[compact space]], <math>A</math> a [[closed subset]] (given the subspace topology)		'''Given''': <math>X</math> a [[compact space]], <math>A</math> a [[closed subset]] (given the subspace topology)

	'''To prove''': ~~<math>A</math> is compact~~		'''To prove''': Consider an open cover of <math>A</math> by open sets <math>U_i</math> with <math>i \in I</math>, an indexing set. The <math>U_i</math> have a finite subcover.

	~~'''Proof''': Let's start with~~ an open cover of <math>A</math> by open sets <math>U_i</math> with <math>i \in I</math>, an indexing set. ~~Our goal is to exhibit~~ a finite subcover.

	By the definition of subspace topology, we can find open sets <math>V_i</math> of <math>X</math> such that <math>V_i \cap A = U_i</math>, thus the union of the <math>V_i</math>s contains <math>A</math>.		'''Proof''':

	Since <math>A</math> is closed, we can ''throw in'' the open set <math>X \setminus A</math>, and get an open cover of the ''whole space''. ~~But since~~ the whole space is compact, this open cover has a finite subcover. In other words, there is a finite subcollection of the <math>V_i</math>s, that, possibly along with <math>X \setminus A</math>, covers the whole of <math>X</math>. By ''throwing out'' <math>X \setminus A</math>, we get a finite collection of <math>V_i</math>s whose union contains <math>A</math>. The corresponding <math>U_i</math> now form a finite subcover of the original cover of <math>A</math>.		# By the definition of subspace topology, we can find open sets <math>V_i</math> of <math>X</math> such that <math>V_i \cap A = U_i</math>, thus the union of the <math>V_i</math>s contains <math>A</math>.
			# Since <math>A</math> is closed, we can ''throw in'' the open set <math>X \setminus A</math>, and get an open cover of the ''whole space'' <math>X</math>.
			# Since the whole space is compact, this open cover has a finite subcover. In other words, there is a finite subcollection of the <math>V_i</math>s, that, possibly along with <math>X \setminus A</math>, covers the whole of <math>X</math>.
			# By ''throwing out'' <math>X \setminus A</math>, we get a finite collection of <math>V_i</math>s whose union contains <math>A</math>. The corresponding <math>U_i</math> now form a finite subcover of the original cover of <math>A</math>.

	===Proof in terms of finite intersection property===		===Proof in terms of finite intersection property===

Vipul at 11:19, 8 August 2008

2008-08-08T11:19:59Z

← Older revision		Revision as of 11:19, 8 August 2008
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	{{topospace metaproperty satisfaction}}		{{topospace metaproperty satisfaction\|
			property = compact space\|
			metaproperty = weakly hereditary property of topological spaces}}

	==Statement==		==Statement==

Vipul: /* References */

2008-07-20T19:55:33Z

References

← Older revision		Revision as of 19:55, 20 July 2008
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	===Textbook references===		===Textbook references===

	* {{booklink\|Munkres}}, Page 165 (Theorem 26.2)		* {{booklink-proved\|Munkres}}, Page 165, Theorem 26.2, Chapter 3, Section 26
	* {{booklink\|SingerThorpe}}, Page 12 (Theorem 4)		* {{booklink-proved\|SingerThorpe}}, Page 12 (Theorem 4)

Vipul at 19:51, 20 July 2008

2008-07-20T19:51:53Z

← Older revision		Revision as of 19:51, 20 July 2008
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	===Proof in terms of open covers===		===Proof in terms of open covers===

	~~Let~~ <math>X</math> be a [[compact space]] ~~and~~ <math>A</math> be a [[closed subset]] (given the subspace topology)~~. We need to prove that <math>A</math> is compact.~~		'''Given''': <math>X</math> a [[compact space]], <math>A</math> a [[closed subset]] (given the subspace topology)

	~~Let~~'~~s start with an open cover of~~ <math>A</math> ~~by open sets <math>U_i</math> with <math>i \in I</math>, an indexing set. Our goal~~ is ~~to exhibit a finite subcover.~~		'''To prove''': <math>A</math> is compact

	By the definition of subspace topology, we can find open sets <math>V_i</math> of <math>X</math> such that <math>V_i \cap A = U_i</math>, thus the union of the <math>V_i</math>s ~~contaiins~~ <math>A</math>.		'''Proof''': Let's start with an open cover of <math>A</math> by open sets <math>U_i</math> with <math>i \in I</math>, an indexing set. Our goal is to exhibit a finite subcover.

			By the definition of subspace topology, we can find open sets <math>V_i</math> of <math>X</math> such that <math>V_i \cap A = U_i</math>, thus the union of the <math>V_i</math>s contains <math>A</math>.

	Since <math>A</math> is closed, we can ''throw in'' the open set <math>X \setminus A</math>, and get an open cover of the ''whole space''. But since the whole space is compact, this open cover has a finite subcover. In other words, there is a finite subcollection of the <math>V_i</math>s, that, possibly along with <math>X \setminus A</math>, covers the whole of <math>X</math>. By ''throwing out'' <math>X \setminus A</math>, we get a finite collection of <math>V_i</math>s whose union contains <math>A</math>. The corresponding <math>U_i</math> now form a finite subcover of the original cover of <math>A</math>.		Since <math>A</math> is closed, we can ''throw in'' the open set <math>X \setminus A</math>, and get an open cover of the ''whole space''. But since the whole space is compact, this open cover has a finite subcover. In other words, there is a finite subcollection of the <math>V_i</math>s, that, possibly along with <math>X \setminus A</math>, covers the whole of <math>X</math>. By ''throwing out'' <math>X \setminus A</math>, we get a finite collection of <math>V_i</math>s whose union contains <math>A</math>. The corresponding <math>U_i</math> now form a finite subcover of the original cover of <math>A</math>.

			===Proof in terms of finite intersection property===
			{{fillin}}
	==References==		==References==

Vipul: 2 revisions

2008-05-11T19:41:57Z

2 revisions

← Older revision	Revision as of 19:41, 11 May 2008
(No difference)

Vipul at 20:48, 21 April 2008

2008-04-21T20:48:14Z

← Older revision		Revision as of 20:48, 21 April 2008
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	Since <math>A</math> is closed, we can ''throw in'' the open set <math>X \setminus A</math>, and get an open cover of the ''whole space''. But since the whole space is compact, this open cover has a finite subcover. In other words, there is a finite subcollection of the <math>V_i</math>s, that, possibly along with <math>X \setminus A</math>, covers the whole of <math>X</math>. By ''throwing out'' <math>X \setminus A</math>, we get a finite collection of <math>V_i</math>s whose union contains <math>A</math>. The corresponding <math>U_i</math> now form a finite subcover of the original cover of <math>A</math>.		Since <math>A</math> is closed, we can ''throw in'' the open set <math>X \setminus A</math>, and get an open cover of the ''whole space''. But since the whole space is compact, this open cover has a finite subcover. In other words, there is a finite subcollection of the <math>V_i</math>s, that, possibly along with <math>X \setminus A</math>, covers the whole of <math>X</math>. By ''throwing out'' <math>X \setminus A</math>, we get a finite collection of <math>V_i</math>s whose union contains <math>A</math>. The corresponding <math>U_i</math> now form a finite subcover of the original cover of <math>A</math>.

			==References==

			===Textbook references===

			* {{booklink\|Munkres}}, Page 165 (Theorem 26.2)
			* {{booklink\|SingerThorpe}}, Page 12 (Theorem 4)

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

2008-01-22T17:07:12Z

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

New page

{{topospace metaproperty satisfaction}}

==Statement==

===Property-theoretic statement===

The [[property of topological spaces]] of being [[compact space|compact]] satisfies the [[metaproperty of topological spaces|metaproperty]] of being [[weakly hereditary property of topological spaces|weakly hereditary]]: in other words, it is inherited by [[closed subset]]s.

===Verbal statement===

Any [[closed subset]] of a [[compact space]] is compact (when given the [[subspace topology]]).

==Proof==

===Proof in terms of open covers===

Let <math>X</math> be a [[compact space]] and <math>A</math> be a [[closed subset]] (given the subspace topology). We need to prove that <math>A</math> is compact.

Let's start with an open cover of <math>A</math> by open sets <math>U_i</math> with <math>i \in I</math>, an indexing set. Our goal is to exhibit a finite subcover.

By the definition of subspace topology, we can find open sets <math>V_i</math> of <math>X</math> such that <math>V_i \cap A = U_i</math>, thus the union of the <math>V_i</math>s contaiins <math>A</math>.

Since <math>A</math> is closed, we can ''throw in'' the open set <math>X \setminus A</math>, and get an open cover of the ''whole space''. But since the whole space is compact, this open cover has a finite subcover. In other words, there is a finite subcollection of the <math>V_i</math>s, that, possibly along with <math>X \setminus A</math>, covers the whole of <math>X</math>. By ''throwing out'' <math>X \setminus A</math>, we get a finite collection of <math>V_i</math>s whose union contains <math>A</math>. The corresponding <math>U_i</math> now form a finite subcover of the original cover of <math>A</math>.