Compactness

From Maths
Revision as of 17:45, 13 February 2015 by Alec (Talk | contribs)

Jump to: navigation, search

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]