Difference between revisions of "Compactness"

Revision as of 17:37, 13 February 2015

A topological space is compact if every open cover (often denoted

$\mathcal{A}$ ) of

$X$ contains a finite sub-collection that also covers

$X$

Take a set

$Y\subset X$ in a topological space

$(X,\mathcal{J})$ .

To say

$Y$ is compact is for

$Y$ to be compact when considered as a subspace of

$(X,\mathcal{J})$

That is to say that

$Y$ is compact if and only if every covering of

$Y$ by sets open in

$X$ contains a finite subcovering covering

$Y$

TODO: Proof

@@ Line 4: / Line 4: @@
 ==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>, <math>Y</math> is compact considered as a [[Subspace topology|subspace]] of <math>(X,\mathcal{J})</math>
+Take a set <math>Y\subset X</math> in a [[Topological space|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 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>