Hammar experiment - Revision history

en>Colonies Chris: /* Overview */sp, date & link fixes; unlinking common words, replaced: Michelson-Morley → Michelson–Morley, Trouton-Noble → Trouton–Noble using AWB

2014-03-12T18:41:07Z

Overview: sp, date & link fixes; unlinking common words, replaced: Michelson-Morley → Michelson–Morley, Trouton-Noble → Trouton–Noble using AWB

← Older revision		Revision as of 20:41, 12 March 2014
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	~~In [[mathematics]], particularly [[topology]], the '''tube lemma''~~' ~~is a useful tool in order to prove that the finite product of [[compact space]]~~s is ~~compact~~. It is ~~in general, a concept of [[point-set topology]].~~		The writer's title is Christy. My day job is an information officer but I've currently applied for an additional 1. North Carolina is where we've been living for years and will never transfer. The preferred hobby for him and his children is to perform lacross and he would never give it up.<br><br>Also visit my site - [http://www.weddingwall.com.au/groups/easy-advice-for-successful-personal-development-today/ authentic psychic readings]

	~~==Tube lemma==~~

	~~Before giving the lemma, one notes the following terminology:~~

	* If ''X'' and ''Y'' are [[topological spaces]] and ''X'' × ''Y'' is the product space, a slice in ''X'' × ''Y'' is a set of the form {''x''} × ''Y'' for ~~''x'' ∈ ''X''~~

	* A tube in ''X'' × ''Y'' is just a [[Base (topology)\|basis element]], ''K'' × ''Y'', in ''X'' × ''Y'' containing a slice in ''X'' × ''Y''

	~~'''Tube Lemma:''' Let ''X'' and ''Y'' be topological spaces with ''Y'' compact, and consider the [[Product topology\|product space]] ''X'' × ''Y''. If ''N'' is~~ an ~~open set containing a slice in ''X'' × ''Y'', then there exists a tube in ''X'' × ''Y'' containing this slice and contained in ''N''.~~

	Using the concept of [[closed map]]s, this can be rephrased concisely as follows: if ''X'' is any topological space and ''Y'' a compact space, then the projection map ''X'' × ''Y'' → ''X'' is closed.

	'''Generalized Tube Lemma:''' Let ''X'' and ''Y'' be topological spaces and consider the product space ''X'' × ''Y''. Let ''A'' be a compact subset of ''X'' and ''B'' be a compact subset of ''Y''. If ''N'' is an open set containing ''A'' × ''B'', then there exists ''U'' open in ''X'' and ''V'' open in ''Y'' such that <math>A\times B\subset U\times V\subset N</math>.

	~~==Examples and properties==~~

	1. Consider ''R'' × ''R'' in the product topology, that is the [[Euclidean plane]], and the open set ''N'' = { (''x'', ''y'') : \|''x''·''y''\| < 1 }. The open set ''N'' contains {0} × ''R'', but contains no tube, so in this case the tube lemma fails. Indeed, if ''W'' × ''R'' is a tube containing {0} × ''R'' and contained in ''N'', ''W'' must be a subset of (−1/''x'', +1/''x'') for all positive integers ''x'' which means ''W'' = {0} contradicting the fact that ''W'' is open in ''R'' (because ''W'' × ''R'' is a tube). This shows that the compactness assumption is essential.

	~~2. The tube lemma can be used to prove that if ''X'' and ''Y'' are compact topological spaces, then ''X'' × ''Y''~~ is ~~compact as follows:~~

	~~Let {''G''<sub>''a''</sub>} be an open cover of ''X'' × ''Y'~~'; for each ''x'' belonging to ''X'', cover the slice {''x''} × ''Y'' by finitely many elements of {''G''<sub>''a''</sub>} (this is possible since {''x''} × ''Y'' is compact being [[homeomorphism\|homeomorphic]] to ''Y''). Call the union of these finitely many elements ''N''<sub>''x''</sub>. By the tube lemma, there is an open set of the form ''W''<sub>x</sub> × ''Y'' containing {''x''} × ''Y'' and ~~contained in ''N''<sub>''x''</sub>~~. The ~~collection of all ''W''<sub>''x''</sub>~~ for ~~''x'' belonging to ''X'' is an open cover of ''X''~~ and hence has a finite subcover ''W''<sub>''x''<sub>1</sub></sub>  ∪ ... ∪ ''W''<sub>''x''<sub>''n''</sub></sub>. Then for each ''x''<sub>''i''</sub>, ''W''<sub>''x''<sub>''i''</sub></sub> × ''Y'' is ~~contained in ''N''<sub>''x''<sub>''i''</sub></sub>. Using the fact that each ''N''<sub>''x''<sub>''i''</sub></sub> is the finite union of elements of ''G''<sub>''a''</sub>~~ and ~~that the finite collection (''W''<sub>''x''<sub>1</sub></sub> × ''Y'') ∪ ..~~.~~ ∪ (''W''~~<~~sub~~>~~''x''~~<~~sub~~>''n''</sub></sub> × ''Y'') covers ''X'' × ''Y'', the collection ''N''<sub>''x''<sub>1</sub></sub> ∪ ... ∪ ''N''<sub>''x''<sub>''n''</sub></sub> is a finite subcover of ''X'' × ''Y''.

	~~3. By example 2 and induction, one can show that the finite product of compact spaces is compact.~~

	~~4. The tube lemma cannot be used to prove the [~~[~~Tychonoff theorem]], which generalizes the above to infinite products.~~

	~~==Proof==~~
	The tube lemma follows from the generalized tube lemma by taking <math>A=\{x\}</math> and <math>B=Y</math>. It therefore suffices to prove the generalized tube lemma. By the definition of the product topology, for each <math>(a,b)\in A\times B</math> there are open sets <math>U_{a,b}\subset X</math> and <math>V_{a,b}\subset Y</math> such that <math>(a,b)\in U_{a,b}\times V_{a,b}\subset N</math>. Fix some <math>a\in A</math>. Then <math>(V_{a,b}:b\in B)</math> is an open cover of <math>B</math>. Since <math>B</math> is compact, this cover has a finite subcover; namely, there is a finite <math>B_0(a)\subset B</math> such that <math>V'_{a}:=\bigcup_{b\in B_0(a)} V_{a,b}\supset B</math>. Set <math>U'_a:~~=\bigcap_{b\in B_0(a)} U_{a,b}</math>. Since <math>B_0(a)<~~/~~math> is finite, <math>U'_a<~~/~~math> is open~~. ~~Also <math>V'_{a}</math> is open~~. ~~Moreover, the construction of <math>U'_a</math> and <math>V'_a</math> implies that <math>\{a\}\times B\subset U'_a\times V'_a\subset N</math>~~. ~~We now essentially repeat the argument to drop the dependence on <math>a<~~/~~math>. Let <math>A_0\subset A<~~/math> be a finite subset such that <math>U'':=\bigcup_{a\in A_0}U'_a\supset A</math> and set <math>V'':=\bigcap_{a\in A_0}V'_a</math>. It then follows by the above reasoning that <math>A\times B\subset U''\times V''\subset N</math> and <math>U''\subset X</math> and <math>V''\subset Y</math> are open, which completes the proof.

	~~==See also==~~

	*[[Tychonoff theorem]]
	*[[Compact space]]
	*[[Product topology]]

	~~==References==~~

	* {{cite book
	~~\| author = [[James Munkres]]~~
	~~\| year = 1999~~
	~~\| title = Topology~~
	~~\| edition = 2nd edition~~
	~~\| publisher = [[Prentice Hall]]~~
	~~\| isbn = 0~~-13-~~181629~~-2
	}}

	~~[[Category:Topology]]~~
	~~[[Category:Lemmas]]~~
	~~[[Category:Articles containing proofs]~~]

en>Bibcode Bot: Adding 0 arxiv eprint(s), 1 bibcode(s) and 1 doi(s). Did it miss something? Report bugs, errors, and suggestions at User talk:Bibcode Bot

2013-10-31T19:50:07Z

Adding 0 arxiv eprint(s), 1 bibcode(s) and 1 doi(s). Did it miss something? Report bugs, errors, and suggestions at User talk:Bibcode Bot

← Older revision		Revision as of 21:50, 31 October 2013
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	~~The name~~ of the ~~writer~~ is ~~Figures but it~~'s ~~not~~ the ~~most masucline name out~~ there. ~~Body developing~~ is ~~one~~ of the ~~things I adore most~~. ~~Years ago we moved~~ to ~~Puerto Rico~~ and ~~my family members enjoys it~~. ~~Since she was 18 she~~'~~s been working as~~ a ~~meter reader but she~~'~~s always wanted her personal business~~.<br><br>~~My web blog~~: ~~home std test kit~~ (~~[http~~://~~www~~.~~pinaydiaries~~.~~com~~/~~blog~~/~~104745 mouse click~~ the ~~up coming internet site~~])		In [[mathematics]], particularly [[topology]], the '''tube lemma''' is a useful tool in order to prove that the finite product of [[compact space]]s is compact. It is in general, a concept of [[point-set topology]].

			==Tube lemma==

			Before giving the lemma, one notes the following terminology:

			* If ''X'' and ''Y'' are [[topological spaces]] and ''X'' × ''Y'' is the product space, a slice in ''X'' × ''Y'' is a set of the form {''x''} × ''Y'' for ''x'' ∈ ''X''

			* A tube in ''X'' × ''Y'' is just a [[Base (topology)\|basis element]], ''K'' × ''Y'', in ''X'' × ''Y'' containing a slice in ''X'' × ''Y''

			'''Tube Lemma:''' Let ''X'' and ''Y'' be topological spaces with ''Y'' compact, and consider the [[Product topology\|product space]] ''X'' × ''Y''. If ''N'' is an open set containing a slice in ''X'' × ''Y'', then there exists a tube in ''X'' × ''Y'' containing this slice and contained in ''N''.

			Using the concept of [[closed map]]s, this can be rephrased concisely as follows: if ''X'' is any topological space and ''Y'' a compact space, then the projection map ''X'' × ''Y'' → ''X'' is closed.

			'''Generalized Tube Lemma:''' Let ''X'' and ''Y'' be topological spaces and consider the product space ''X'' × ''Y''. Let ''A'' be a compact subset of ''X'' and ''B'' be a compact subset of ''Y''. If ''N'' is an open set containing ''A'' × ''B'', then there exists ''U'' open in ''X'' and ''V'' open in ''Y'' such that <math>A\times B\subset U\times V\subset N</math>.

			==Examples and properties==

			1. Consider ''R'' × ''R'' in the product topology, that is the [[Euclidean plane]], and the open set ''N'' = { (''x'', ''y'') : \|''x''·''y''\| < 1 }. The open set ''N'' contains {0} × ''R'', but contains no tube, so in this case the tube lemma fails. Indeed, if ''W'' × ''R'' is a tube containing {0} × ''R'' and contained in ''N'', ''W'' must be a subset of (−1/''x'', +1/''x'') for all positive integers ''x'' which means ''W'' = {0} contradicting the fact that ''W'' is open in ''R'' (because ''W'' × ''R'' is a tube). This shows that the compactness assumption is essential.

			2. The tube lemma can be used to prove that if ''X'' and ''Y'' are compact topological spaces, then ''X'' × ''Y'' is compact as follows:

			Let {''G''<sub>''a''</sub>} be an open cover of ''X'' × ''Y''; for each ''x'' belonging to ''X'', cover the slice {''x''} × ''Y'' by finitely many elements of {''G''<sub>''a''</sub>} (this is possible since {''x''} × ''Y'' is compact being [[homeomorphism\|homeomorphic]] to ''Y''). Call the union of these finitely many elements ''N''<sub>''x''</sub>. By the tube lemma, there is an open set of the form ''W''<sub>x</sub> × ''Y'' containing {''x''} × ''Y'' and contained in ''N''<sub>''x''</sub>. The collection of all ''W''<sub>''x''</sub> for ''x'' belonging to ''X'' is an open cover of ''X'' and hence has a finite subcover ''W''<sub>''x''<sub>1</sub></sub>  ∪ ... ∪ ''W''<sub>''x''<sub>''n''</sub></sub>. Then for each ''x''<sub>''i''</sub>, ''W''<sub>''x''<sub>''i''</sub></sub> × ''Y'' is contained in ''N''<sub>''x''<sub>''i''</sub></sub>. Using the fact that each ''N''<sub>''x''<sub>''i''</sub></sub> is the finite union of elements of ''G''<sub>''a''</sub> and that the finite collection (''W''<sub>''x''<sub>1</sub></sub> × ''Y'') ∪ ... ∪ (''W''<sub>''x''<sub>''n''</sub></sub> × ''Y'') covers ''X'' × ''Y'', the collection ''N''<sub>''x''<sub>1</sub></sub> ∪ ... ∪ ''N''<sub>''x''<sub>''n''</sub></sub> is a finite subcover of ''X'' × ''Y''.

			3. By example 2 and induction, one can show that the finite product of compact spaces is compact.

			4. The tube lemma cannot be used to prove the [[Tychonoff theorem]], which generalizes the above to infinite products.

			==Proof==
			The tube lemma follows from the generalized tube lemma by taking <math>A=\{x\}</math> and <math>B=Y</math>. It therefore suffices to prove the generalized tube lemma. By the definition of the product topology, for each <math>(a,b)\in A\times B</math> there are open sets <math>U_{a,b}\subset X</math> and <math>V_{a,b}\subset Y</math> such that <math>(a,b)\in U_{a,b}\times V_{a,b}\subset N</math>. Fix some <math>a\in A</math>. Then <math>(V_{a,b}:b\in B)</math> is an open cover of <math>B</math>. Since <math>B</math> is compact, this cover has a finite subcover; namely, there is a finite <math>B_0(a)\subset B</math> such that <math>V'_{a}:=\bigcup_{b\in B_0(a)} V_{a,b}\supset B</math>. Set <math>U'_a:=\bigcap_{b\in B_0(a)} U_{a,b}</math>. Since <math>B_0(a)</math> is finite, <math>U'_a</math> is open. Also <math>V'_{a}</math> is open. Moreover, the construction of <math>U'_a</math> and <math>V'_a</math> implies that <math>\{a\}\times B\subset U'_a\times V'_a\subset N</math>. We now essentially repeat the argument to drop the dependence on <math>a</math>. Let <math>A_0\subset A</math> be a finite subset such that <math>U'':=\bigcup_{a\in A_0}U'_a\supset A</math> and set <math>V'':=\bigcap_{a\in A_0}V'_a</math>. It then follows by the above reasoning that <math>A\times B\subset U''\times V''\subset N</math> and <math>U''\subset X</math> and <math>V''\subset Y</math> are open, which completes the proof.

			==See also==

			*[[Tychonoff theorem]]
			*[[Compact space]]
			*[[Product topology]]

			==References==

			* {{cite book
			\| author = [[James Munkres]]
			\| year = 1999
			\| title = Topology
			\| edition = 2nd edition
			\| publisher = [[Prentice Hall]]
			\| isbn = 0-13-181629-2
			}}

			[[Category:Topology]]
			[[Category:Lemmas]]
			[[Category:Articles containing proofs]]

en>Stigmatella aurantiaca: /* The experiment */ center image

2012-08-09T09:40:43Z

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The name of the writer is Figures but it's not the most masucline name out there. Body developing is one of the things I adore most. Years ago we moved to Puerto Rico and my family members enjoys it. Since she was 18 she's been working as a meter reader but she's always wanted her personal business.<br><br>My web blog: home std test kit ([http://www.pinaydiaries.com/blog/104745 mouse click the up coming internet site])