Compact-to-Hausdorff theorem

From Maths
Revision as of 12:36, 13 August 2015 by Alec (Talk | contribs) (Proof)

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Jump to: navigation, search

Statement

Given a continuous and bijective function between two topological spaces [ilmath]f:X\rightarrow Y[/ilmath] where [ilmath]X[/ilmath] is compact and [ilmath]Y[/ilmath] is Hausdorff

Proof

We wish to show [ilmath](f^{-1})^{-1}(U)[/ilmath] is open (where [ilmath]U[/ilmath] is open in [ilmath]X[/ilmath]), that is that the inverse of [ilmath]f[/ilmath] is continuous.

Proof:

Let [ilmath]U\subseteq X[/ilmath] be a given open set
[ilmath]U[/ilmath] open [ilmath]\implies X-U[/ilmath] is closed [ilmath]\implies X-U[/ilmath] is compact
[ilmath]\implies f(X-U)[/ilmath] is compact
[ilmath]\implies f(X-U)[/ilmath] is closed in [ilmath]Y[/ilmath]
[ilmath]\implies Y-f(X-U)[/ilmath] is open in [ilmath]Y[/ilmath]
But [ilmath]Y-f(X-U)=f(U)[/ilmath]
  • So we conclude [ilmath]f(U)[/ilmath] is open in [ilmath]Y[/ilmath]

As [ilmath]f=(f^{-1})^{-1}[/ilmath] we have shown that a continuous bijective function's inverse is continuous, thus [ilmath]f[/ilmath] is a homeomorphism

References

  1. Introduction to Topology - Nov 2013 - Lecture Notes - David Mond