0708-1300/Class notes for Thursday, November 1: Difference between revisions

From Drorbn
Jump to navigationJump to search
No edit summary
No edit summary
Line 12: Line 12:
{{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>,}}
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>.

===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

Revision as of 13:12, 1 November 2007

Announcements go here

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