Difference between revisions of "Compact-to-Hausdorff theorem"

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==Statement==
 
==Statement==
 
Given a [[Continuous map|continuous]] and [[Bijection|bijective]] function between two [[Topological space|topological spaces]] {{M|f:X\rightarrow Y}} where
 
Given a [[Continuous map|continuous]] and [[Bijection|bijective]] function between two [[Topological space|topological spaces]] {{M|f:X\rightarrow Y}} where
{{M|X}} is [[Compactness|compact]] and {{M|Y}} is [[Hausdorff]]
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{{M|X}} is [[Compactness|compact]] and {{M|Y}} is [[Hausdorff space|Hausdorff]]
 
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* '''Then {{M|f}} is a [[Homeomorphism|homeomorphism]]'''<ref>Introduction to Topology - Nov 2013 - Lecture Notes - David Mond</ref>
 
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'''Then {{M|f}} is a [[Homeomorphism|homeomorphism]]'''<ref>Introduction to Topology - Nov 2013 - Lecture Notes - David Mond</ref>
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==Proof==
 
==Proof==
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{{Theorem Of|Topology}}
 
{{Theorem Of|Topology}}
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[[Category:Theorems involving compactness]]

Revision as of 05:06, 9 June 2015

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