Difference between revisions of "Compact-to-Hausdorff theorem"
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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''' | ||
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==References== | ==References== | ||
<references/> | <references/> | ||
Latest revision as of 12:36, 13 August 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
- Then [ilmath]f[/ilmath] is a homeomorphism[1]
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
- (Using the compactness of [ilmath]X[/ilmath]) - a Closed set in a compact space 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]
- [ilmath]U[/ilmath] open [ilmath]\implies X-U[/ilmath] is closed [ilmath]\implies X-U[/ilmath] is compact
- 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
- ↑ Introduction to Topology - Nov 2013 - Lecture Notes - David Mond
