In mathematics, particularly topology, the tube lemma, also called Wallace's theorem, is a useful tool in order to prove that the finite product of compact spaces is compact.
A tube in is a subset of the form where is an open subset of . It contains all the slices for .
Tube Lemma — Let and be topological spaces with compact, and consider the product space If is an open set containing a slice in then there exists a tube in containing this slice and contained in
Using the concept of closed maps, this can be rephrased concisely as follows: if is any topological space and a compact space, then the projection map is closed.
Generalized Tube Lemma 1 — Let and be topological spaces and consider the product space Let be a compact subset of and be a compact subset of If is an open set containing then there exists open in and open in such that
Generalized Tube Lemma 2 — Let be topological spaces and consider the product space For each , let be a compact subset of If is an open set containing then there exists open in with for all but finite amount of , such that
Examples and propertiesedit
1. Consider in the product topology, that is the Euclidean plane, and the open set The open set contains but contains no tube, so in this case the tube lemma fails. Indeed, if is a tube containing and contained in must be a subset of for all which means contradicting the fact that is open in (because is a tube). This shows that the compactness assumption is essential.
2. The tube lemma can be used to prove that if and are compact spaces, then is compact as follows:
Let be an open cover of . For each , cover the slice by finitely many elements of (this is possible since is compact, being homeomorphic to ). Call the union of these finitely many elements By the tube lemma, there is an open set of the form containing and contained in The collection of all for is an open cover of and hence has a finite subcover . Thus the finite collection covers . Using the fact that each is contained in and each is the finite union of elements of , one gets a finite subcollection of that covers .
3. By part 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.
Proofedit
The tube lemma follows from the generalized tube lemma by taking and It therefore suffices to prove the generalized tube lemma. By the definition of the product topology, for each there are open sets and such that For any is an open cover of the compact set so this cover has a finite subcover; namely, there is a finite set such that contains where observe that is open in For every let which is an open in set since is finite. Moreover, the construction of and implies that We now essentially repeat the argument to drop the dependence on Let be a finite subset such that contains and set It then follows by the above reasoning that and and are open, which completes the proof.
See alsoedit
Alexander's sub-base theorem – Collection of subsets that generate a topologyPages displaying short descriptions of redirect targets
Tubular neighborhood – neighborhood of a submanifold homeomorphic to that submanifold’s normal bundlePages displaying wikidata descriptions as a fallback
Tychonoff theorem – Product of any collection of compact topological spaces is compactPages displaying short descriptions of redirect targets
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In mathematics particularly topology the tube lemma also called Wallace s theorem is a useful tool in order to prove that the finite product of compact spaces is compact Contents 1 Statement 2 Examples and properties 3 Proof 4 See also 5 ReferencesStatement editThe lemma uses the following terminology If X displaystyle X nbsp and Y displaystyle Y nbsp are topological spaces and X Y displaystyle X times Y nbsp is the product space endowed with the product topology a slice in X Y displaystyle X times Y nbsp is a set of the form x Y displaystyle x times Y nbsp for x X displaystyle x in X nbsp A tube in X Y displaystyle X times Y nbsp is a subset of the form U Y displaystyle U times Y nbsp where U displaystyle U nbsp is an open subset of X displaystyle X nbsp It contains all the slices x Y displaystyle x times Y nbsp for x U displaystyle x in U nbsp Tube Lemma Let X displaystyle X nbsp and Y displaystyle Y nbsp be topological spaces with Y displaystyle Y nbsp compact and consider the product space X Y displaystyle X times Y nbsp If N displaystyle N nbsp is an open set containing a slice in X Y displaystyle X times Y nbsp then there exists a tube in X Y displaystyle X times Y nbsp containing this slice and contained in N displaystyle N nbsp Using the concept of closed maps this can be rephrased concisely as follows if X displaystyle X nbsp is any topological space and Y displaystyle Y nbsp a compact space then the projection map X Y X displaystyle X times Y to X nbsp is closed Generalized Tube Lemma 1 Let X displaystyle X nbsp and Y displaystyle Y nbsp be topological spaces and consider the product space X Y displaystyle X times Y nbsp Let A displaystyle A nbsp be a compact subset of X displaystyle X nbsp and B displaystyle B nbsp be a compact subset of Y displaystyle Y nbsp If N displaystyle N nbsp is an open set containing A B displaystyle A times B nbsp then there exists U displaystyle U nbsp open in X displaystyle X nbsp and V displaystyle V nbsp open in Y displaystyle Y nbsp such that A B U V N displaystyle A times B subseteq U times V subseteq N nbsp Generalized Tube Lemma 2 Let X i i I displaystyle X i i in I nbsp be topological spaces and consider the product space i I X i displaystyle prod i in I X i nbsp For each i I displaystyle i in I nbsp let A i displaystyle A i nbsp be a compact subset of X i displaystyle X i nbsp If N displaystyle N nbsp is an open set containing i I A i displaystyle prod i in I A i nbsp then there exists U i displaystyle U i nbsp open in X i displaystyle X i nbsp with U i X i displaystyle U i X i nbsp for all but finite amount of i I displaystyle i in I nbsp such that i I A i i I U i N displaystyle prod i in I A i subseteq prod i in I U i subseteq N nbsp Examples and properties edit1 Consider R R displaystyle mathbb R times mathbb R nbsp in the product topology that is the Euclidean plane and the open set N x y R R x y lt 1 displaystyle N x y in mathbb R times mathbb R xy lt 1 nbsp The open set N displaystyle N nbsp contains 0 R displaystyle 0 times mathbb R nbsp but contains no tube so in this case the tube lemma fails Indeed if W R displaystyle W times mathbb R nbsp is a tube containing 0 R displaystyle 0 times mathbb R nbsp and contained in N displaystyle N nbsp W displaystyle W nbsp must be a subset of 1 x 1 x displaystyle left 1 x 1 x right nbsp for all x gt 0 displaystyle x gt 0 nbsp which means W 0 displaystyle W 0 nbsp contradicting the fact that W displaystyle W nbsp is open in R displaystyle mathbb R nbsp because W R displaystyle W times mathbb R nbsp is a tube This shows that the compactness assumption is essential 2 The tube lemma can be used to prove that if X displaystyle X nbsp and Y displaystyle Y nbsp are compact spaces then X Y displaystyle X times Y nbsp is compact as follows Let G a displaystyle G a nbsp be an open cover of X Y displaystyle X times Y nbsp For each x X displaystyle x in X nbsp cover the slice x Y displaystyle x times Y nbsp by finitely many elements of G a displaystyle G a nbsp this is possible since x Y displaystyle x times Y nbsp is compact being homeomorphic to Y displaystyle Y nbsp Call the union of these finitely many elements N x displaystyle N x nbsp By the tube lemma there is an open set of the form W x Y displaystyle W x times Y nbsp containing x Y displaystyle x times Y nbsp and contained in N x displaystyle N x nbsp The collection of all W x displaystyle W x nbsp for x X displaystyle x in X nbsp is an open cover of X displaystyle X nbsp and hence has a finite subcover W x 1 W x n displaystyle W x 1 dots W x n nbsp Thus the finite collection W x 1 Y W x n Y displaystyle W x 1 times Y dots W x n times Y nbsp covers X Y displaystyle X times Y nbsp Using the fact that each W x i Y displaystyle W x i times Y nbsp is contained in N x i displaystyle N x i nbsp and each N x i displaystyle N x i nbsp is the finite union of elements of G a displaystyle G a nbsp one gets a finite subcollection of G a displaystyle G a nbsp that covers X Y displaystyle X times Y nbsp 3 By part 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 editThe tube lemma follows from the generalized tube lemma by taking A x displaystyle A x nbsp and B Y displaystyle B Y nbsp It therefore suffices to prove the generalized tube lemma By the definition of the product topology for each a b A B displaystyle a b in A times B nbsp there are open sets U a b X displaystyle U a b subseteq X nbsp and V a b Y displaystyle V a b subseteq Y nbsp such that a b U a b V a b N displaystyle a b in U a b times V a b subseteq N nbsp For any a A displaystyle a in A nbsp V a b b B displaystyle left V a b b in B right nbsp is an open cover of the compact set B displaystyle B nbsp so this cover has a finite subcover namely there is a finite set B 0 a B displaystyle B 0 a subseteq B nbsp such that V a b B 0 a V a b displaystyle V a bigcup b in B 0 a V a b nbsp contains B displaystyle B nbsp where observe that V a displaystyle V a nbsp is open in Y displaystyle Y nbsp For every a A displaystyle a in A nbsp let U a b B 0 a U a b displaystyle U a bigcap b in B 0 a U a b nbsp which is an open in X displaystyle X nbsp set since B 0 a displaystyle B 0 a nbsp is finite Moreover the construction of U a displaystyle U a nbsp and V a displaystyle V a nbsp implies that a B U a V a N displaystyle a times B subseteq U a times V a subseteq N nbsp We now essentially repeat the argument to drop the dependence on a displaystyle a nbsp Let A 0 A displaystyle A 0 subseteq A nbsp be a finite subset such that U a A 0 U a displaystyle U bigcup a in A 0 U a nbsp contains A displaystyle A nbsp and set V a A 0 V a displaystyle V bigcap a in A 0 V a nbsp It then follows by the above reasoning that A B U V N displaystyle A times B subseteq U times V subseteq N nbsp and U X displaystyle U subseteq X nbsp and V Y displaystyle V subseteq Y nbsp are open which completes the proof See also editAlexander s sub base theorem Collection of subsets that generate a topologyPages displaying short descriptions of redirect targets Tubular neighborhood neighborhood of a submanifold homeomorphic to that submanifold s normal bundlePages displaying wikidata descriptions as a fallback Tychonoff theorem Product of any collection of compact topological spaces is compactPages displaying short descriptions of redirect targetsReferences editJames Munkres 1999 Topology 2nd ed Prentice Hall ISBN 0 13 181629 2 Joseph J Rotman 1988 An Introduction to Algebraic Topology Springer ISBN 0 387 96678 1 See Chapter 8 Lemma 8 9 Retrieved from https en wikipedia org w index php title Tube lemma amp oldid 1162085659, wikipedia, wiki, book, books, library,