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

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m
m (Proof)
 
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: Let {{M|U\subseteq X}} be a given open set
 
: Let {{M|U\subseteq X}} be a given open set
 
:: {{M|U}} open {{M|\implies X-U}} is closed {{M|\implies X-U}} is [[Compactness|compact]]
 
:: {{M|U}} open {{M|\implies X-U}} is closed {{M|\implies X-U}} is [[Compactness|compact]]
::* (Using the compactness of {{M|X}}) - a [[Closed set in compact space is compact]])
+
::* (Using the compactness of {{M|X}}) - a [[Closed set in a compact space is compact]])
 
:: {{M|\implies f(X-U)}} is compact
 
:: {{M|\implies f(X-U)}} is compact
 
::* (Using [[Image of a compact set is compact]])
 
::* (Using [[Image of a compact set is compact]])
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As {{M|1=f=(f^{-1})^{-1} }} we have shown that a continuous bijective function's inverse is continuous, '''thus {{M|f}} is a homeomorphism'''
 
As {{M|1=f=(f^{-1})^{-1} }} we have shown that a continuous bijective function's inverse is continuous, '''thus {{M|f}} is a homeomorphism'''
 +
 
==References==
 
==References==
 
<references/>
 
<references/>

Latest revision as of 12:36, 13 August 2015

Statement

Given a continuous and bijective function between two topological spaces f:XY where X is compact and Y is Hausdorff

Proof

We wish to show (f1)1(U) is open (where U is open in X), that is that the inverse of f is continuous.

Proof:

Let UX be a given open set
U open XU is closed XU is compact
f(XU) is compact
f(XU) is closed in Y
Yf(XU) is open in Y
But Yf(XU)=f(U)
  • So we conclude f(U) is open in Y

As f=(f1)1 we have shown that a continuous bijective function's inverse is continuous, thus f is a homeomorphism

References

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