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]] | + | {{M|X}} is [[Compactness|compact]] and {{M|Y}} is [[Hausdorff space|Hausdorff]] |
− | + | * '''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> | + | |
==Proof== | ==Proof== | ||
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{{Theorem Of|Topology}} | {{Theorem Of|Topology}} | ||
+ | [[Category:Theorems involving compactness]] |
Revision as of 05:06, 9 June 2015
Statement
Given a continuous and bijective function between two topological spaces f:X→Y where X is compact and Y is Hausdorff
- Then f is a homeomorphism[1]
Proof
We wish to show (f^{-1})^{-1}(U) is open (where U is open in X), that is that the inverse of f is continuous.
Proof:
- Let U\subseteq X be a given open set
- U open \implies X-U is closed \implies X-U is compact
- (Using the compactness of X) - a Closed set in compact space is compact)
- \implies f(X-U) is compact
- \implies f(X-U) is closed in Y
- \implies Y-f(X-U) is open in Y
- But Y-f(X-U)=f(U)
- U open \implies X-U is closed \implies X-U is compact
- So we conclude f(U) is open in Y
As f=(f^{-1})^{-1} we have shown that a continuous bijective function's inverse is continuous, thus f is a homeomorphism
References
- Jump up ↑ Introduction to Topology - Nov 2013 - Lecture Notes - David Mond