Difference between revisions of "Compactness"
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Then <math>\{A_i\}^n_{i=1}</math> is a subcollection of <math>\mathcal{A}</math> that 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>\impliedby</math>==== |
{{Definition|Topology}} | {{Definition|Topology}} | ||
Revision as of 17:45, 13 February 2015
Contents
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].
