Difference between revisions of "Compactness"
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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> | ||
| − | {{ | + | |
| + | ===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>). | ||
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| + | 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]]) | ||
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| + | 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]
