Difference between revisions of "Compactness"

From Maths
Jump to: navigation, search
m
m
Line 9: Line 9:
  
 
That is to say that <math>Y</math> is compact if and only if every covering of <math>Y</math> by sets open in <math>X</math> contains a finite subcovering covering <math>Y</math>
 
That is to say that <math>Y</math> is compact if and only if every covering of <math>Y</math> by sets open in <math>X</math> contains a finite subcovering covering <math>Y</math>
{{Todo|Proof}}
+
 
 +
===Proof===
 +
====<math>\implies</math>====
 +
Suppose that the space <math>(Y,\mathcal{J}_\text{subspace})</math> is compact and that <math>\mathcal{A}=\{A_\alpha\}_{\alpha\in I}</math> where each <math>A_\alpha\in\mathcal{J}</math> (that is each set is open in <math>X</math>).
 +
 
 +
Then the collection <math>\{A_\alpha\cap Y|\alpha\in I\}</math> is a covering of <math>Y</math> by sets open in <math>Y</math> (by definition of [[Subspace topology|being a subspace]])
 +
 
 +
By hypothesis <math>Y</math> is compact, hence a finite subcollection <math>\{A_i\cap Y\}^n_{i=1}</math> covers <math>Y</math>
 +
 
 +
Then <math>\{A_i\}^n_{i=1}</math> is a subcollection of <math>\mathcal{A}</math> that covers <math>Y</math>.
 +
====<math>\impliedby<\math>====
  
  
 
{{Definition|Topology}}
 
{{Definition|Topology}}

Revision as of 17:44, 13 February 2015

Definition

A topological space is compact if every open cover (often denoted [math]\mathcal{A}[/math]) of [math]X[/math] contains a finite sub-collection that also covers [math]X[/math]

Lemma for a set being compact

Take a set [math]Y\subset X[/math] in a topological space [math](X,\mathcal{J})[/math].

To say [math]Y[/math] is compact is for [math]Y[/math] to be compact when considered as a subspace of [math](X,\mathcal{J})[/math]

That is to say that [math]Y[/math] is compact if and only if every covering of [math]Y[/math] by sets open in [math]X[/math] contains a finite subcovering covering [math]Y[/math]

Proof

[math]\implies[/math]

Suppose that the space [math](Y,\mathcal{J}_\text{subspace})[/math] is compact and that [math]\mathcal{A}=\{A_\alpha\}_{\alpha\in I}[/math] where each [math]A_\alpha\in\mathcal{J}[/math] (that is each set is open in [math]X[/math]).

Then the collection [math]\{A_\alpha\cap Y|\alpha\in I\}[/math] is a covering of [math]Y[/math] by sets open in [math]Y[/math] (by definition of being a subspace)

By hypothesis [math]Y[/math] is compact, hence a finite subcollection [math]\{A_i\cap Y\}^n_{i=1}[/math] covers [math]Y[/math]

Then [math]\{A_i\}^n_{i=1}[/math] is a subcollection of [math]\mathcal{A}[/math] that covers [math]Y[/math]. ====[math]\impliedby<\math>==== {{Definition|Topology}}[/math]