Difference between revisions of "Compactness"
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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]]) | 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]]) | ||
| − | By hypothesis <math>Y</math> is compact, hence a finite subcollection <math>\{ | + | By hypothesis <math>Y</math> is compact, hence a finite subcollection <math>\{A_{\alpha_i}\cap Y\}^n_{i=1}</math> covers <math>Y</math> '''(as to be compact ''every'' open cover must have a finite subcover)''' |
| − | Then <math>\{ | + | Then <math>\{A_{\alpha_i}\}^n_{i=1}</math> is a subcollection of <math>\mathcal{A}</math> that covers <math>Y</math>. |
| − | ====<math>\impliedby</math>==== | + | |
| + | =====Details===== | ||
| + | As [[The intersection of sets is a subset of each set]] and <math>\cup^n_{i=1}(A_{\alpha_i}\cap Y)=Y</math> we see <br /> | ||
| + | <math>x\in\cup^n_{i=1}(A_{\alpha_i}\cap Y)\implies\exists k\in\mathbb{N}\text{ with }1\le k\le n:x\in A_{\alpha_k}\cap Y\implies x\in A_{\alpha_k}\implies x\in\cup^n_{i=1}A_{\alpha_i}</math><br /> | ||
| + | The important part being <math>x\in\cup^n_{i=1}(A_{\alpha_i}\cap Y)\implies x\in\cup^n_{i=1}A_{\alpha_i}</math><br /> | ||
| + | then by the [[Implies and subset relation|implies and subset relation]] we have <math>Y=\cup^n_{i=1}(A_{\alpha_i}\cap Y)\subset\cup^n_{i=1}A_{\alpha_i}</math> and conclude <math>Y\subset\cup^n_{i=1}A_{\alpha_i}</math> | ||
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| + | Lastly, as <math>\mathcal{A}</math> was a covering <math>\cup_{\alpha\in I}A_\alpha=Y</math>. | ||
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| + | It is clear that <math>x\in\cup^n_{i=1}A_{\alpha_i}\implies x\in\cup_{\alpha\in I}A_\alpha</math> so again [[Implies and subset relation|implies and subset relation]] we have:<br /> | ||
| + | <math>\cup^n_{i=1}A_{\alpha_i}\subset\cup_{\alpha\in I}A_\alpha=Y</math> thus concluding <math>\cup^n_{i=1}A_{\alpha_i}\subset Y</math> | ||
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| + | Combining <math>Y\subset\cup^n_{i=1}A_{\alpha_i}</math> and <math>\cup^n_{i=1}A_{\alpha_i}\subset Y</math> we see <math>\cup^n_{i=1}A_{\alpha_i}=Y</math> | ||
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| + | Thus <math>\{A_{\alpha_i}\}^n_{i=1}</math> is a finite covering of <math>Y</math> consisting of '''open''' sets from <math>X</math> | ||
| + | ====<math>\impliedby</math>==== | ||
{{Definition|Topology}} | {{Definition|Topology}} | ||
Revision as of 18:07, 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_{\alpha_i}\cap Y\}^n_{i=1}[/math] covers [math]Y[/math] (as to be compact every open cover must have a finite subcover)
Then [math]\{A_{\alpha_i}\}^n_{i=1}[/math] is a subcollection of [math]\mathcal{A}[/math] that covers [math]Y[/math].
Details
As The intersection of sets is a subset of each set and [math]\cup^n_{i=1}(A_{\alpha_i}\cap Y)=Y[/math] we see
[math]x\in\cup^n_{i=1}(A_{\alpha_i}\cap Y)\implies\exists k\in\mathbb{N}\text{ with }1\le k\le n:x\in A_{\alpha_k}\cap Y\implies x\in A_{\alpha_k}\implies x\in\cup^n_{i=1}A_{\alpha_i}[/math]
The important part being [math]x\in\cup^n_{i=1}(A_{\alpha_i}\cap Y)\implies x\in\cup^n_{i=1}A_{\alpha_i}[/math]
then by the implies and subset relation we have [math]Y=\cup^n_{i=1}(A_{\alpha_i}\cap Y)\subset\cup^n_{i=1}A_{\alpha_i}[/math] and conclude [math]Y\subset\cup^n_{i=1}A_{\alpha_i}[/math]
Lastly, as [math]\mathcal{A}[/math] was a covering [math]\cup_{\alpha\in I}A_\alpha=Y[/math].
It is clear that [math]x\in\cup^n_{i=1}A_{\alpha_i}\implies x\in\cup_{\alpha\in I}A_\alpha[/math] so again implies and subset relation we have:
[math]\cup^n_{i=1}A_{\alpha_i}\subset\cup_{\alpha\in I}A_\alpha=Y[/math] thus concluding [math]\cup^n_{i=1}A_{\alpha_i}\subset Y[/math]
Combining [math]Y\subset\cup^n_{i=1}A_{\alpha_i}[/math] and [math]\cup^n_{i=1}A_{\alpha_i}\subset Y[/math] we see [math]\cup^n_{i=1}A_{\alpha_i}=Y[/math]
Thus [math]\{A_{\alpha_i}\}^n_{i=1}[/math] is a finite covering of [math]Y[/math] consisting of open sets from [math]X[/math]
