Difference between revisions of "Compactness"
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==Lemma for a set being compact== | ==Lemma for a set being compact== | ||
| − | Take a set <math>Y\subset X</math> in a [[Topological space|topological space]] <math>(X,\mathcal{J})</math> | + | Take a set <math>Y\subset X</math> in a [[Topological space|topological space]] <math>(X,\mathcal{J})</math>. |
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| + | To say <math>Y</math> is compact is for <math>Y</math> to be compact when considered as a [[Subspace topology|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> | 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> | ||
Revision as of 17:37, 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]
TODO: Proof
