0708-1300/Class notes for Thursday, November 1: Difference between revisions
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{{Equation*|<math>f(B)\cap U=f(B\cap f^{-1}(U))\cap U\subset f(B\cap f^{-1}(\bar U))\cap U\subset f(B)\cap U</math>,}} |
{{Equation*|<math>f(B)\cap U=f(B\cap f^{-1}(U))\cap U\subset f(B\cap f^{-1}(\bar U))\cap U\subset f(B)\cap U</math>,}} |
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so that <math>f(B)\cap U=f(B\cap f^{-1}(\bar U))\cap U</math>. But <math>\bar U</math> is compact by choice, so <math>f^{-1}(\bar U)</math> is compact as <math>f</math> is proper, so <math>B\cap f^{-1}(\bar U)</math> is compact as <math>B</math> is closed, so <math>f(B\cap f^{-1}(\bar U))</math> is compact (and hence closed) as a continuous image of a compact set, so <math>f(B)\cap U</math> is the intersection <math>f(B\cap f^{-1}(\bar U))\cap U</math> of a closed set with <math>U</math>, hence it is closed in <math>U</math>. |
so that <math>f(B)\cap U=f(B\cap f^{-1}(\bar U))\cap U</math>. But <math>\bar U</math> is compact by choice, so <math>f^{-1}(\bar U)</math> is compact as <math>f</math> is proper, so <math>B\cap f^{-1}(\bar U)</math> is compact as <math>B</math> is closed, so <math>f(B\cap f^{-1}(\bar U))</math> is compact (and hence closed) as a continuous image of a compact set, so <math>f(B)\cap U</math> is the intersection <math>f(B\cap f^{-1}(\bar U))\cap U</math> of a closed set with <math>U</math>, hence it is closed in <math>U</math>. |
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===Note=== |
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The example of a non-contractible "comb" seen today is, in fact, "Cantor's comb". See, for example, page 25 of www.karlin.mff.cuni.cz/~pyrih/e/e2000v0/c/ect.ps |
Revision as of 12:12, 1 November 2007
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Today's Agenda
- HW4 and TE1.
- Continue with Tuesday's agenda:
- Debt on proper functions.
- Prove that "the sphere is not contractible".
- Complete the proof of the "tubular neighborhood theorem".
Proper Implies Closed
Theorem. A proper function from a topological space to a locally compact (Hausdorff) topological space is closed.
Proof. Let be closed in , we need to show that is closed in . Since closedness is a local property, it is enough to show that every point has a neighbourhood such that is closed in . Fix , and by local compactness, choose a neighbourhood of whose close is compact. Then
so that . But is compact by choice, so is compact as is proper, so is compact as is closed, so is compact (and hence closed) as a continuous image of a compact set, so is the intersection of a closed set with , hence it is closed in .
Note
The example of a non-contractible "comb" seen today is, in fact, "Cantor's comb". See, for example, page 25 of www.karlin.mff.cuni.cz/~pyrih/e/e2000v0/c/ect.ps