Difference between revisions of "Compactness"

Revision as of 17:45, 13 February 2015 (view source) Alec (Talk | contribs) m ← Older edit		Revision as of 18:07, 13 February 2015 (view source) Alec (Talk | contribs) m Newer edit →
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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>\{A_i\cap Y\}^n_{i=1}</math> covers <math>Y</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>\{A_i\}^n_{i=1}</math> is a subcollection of <math>\mathcal{A}</math> that covers <math>Y</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>
		+
		+
		+	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|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>
		+
		+	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>
		+	====<math>\impliedby</math>====
	{{Definition|Topology}}		{{Definition|Topology}}

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Revision as of 18:07, 13 February 2015

Definition

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

$$\mathcal{A}$$ ) of $$X$$ contains a finite sub-collection that also covers $$X$$

Lemma for a set being compact

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 $$Y$$

Proof

$$\implies$$

Suppose that the space

$$(Y,\mathcal{J}_\text{subspace})$$ is compact and that $$\mathcal{A}=\{A_\alpha\}_{\alpha\in I}$$ where each $$A_\alpha\in\mathcal{J}$$ (that is each set is open in $$X$$ ).

Then the collection

$$\{A_\alpha\cap Y|\alpha\in I\}$$ is a covering of $$Y$$ by sets open in $$Y$$ (by definition of being a subspace)

By hypothesis

$$Y$$ is compact, hence a finite subcollection $$\{A_{\alpha_i}\cap Y\}^n_{i=1}$$ covers $$Y$$ (as to be compact every open cover must have a finite subcover)

Then

$$\{A_{\alpha_i}\}^n_{i=1}$$ is a subcollection of $$\mathcal{A}$$ that covers $$Y$$ .

Details

As The intersection of sets is a subset of each set and

$$\cup^n_{i=1}(A_{\alpha_i}\cap Y)=Y$$ we see
$$x\in\cup^n_{i=1}(A_{\alpha_i}\cap Y)\implies\exists k\in\mathbb{N}\text{ with }1\le k\le n:x\in A_{\alpha_k}\cap Y\implies x\in A_{\alpha_k}\implies x\in\cup^n_{i=1}A_{\alpha_i}$$
The important part being $$x\in\cup^n_{i=1}(A_{\alpha_i}\cap Y)\implies x\in\cup^n_{i=1}A_{\alpha_i}$$
then by the implies and subset relation we have $$Y=\cup^n_{i=1}(A_{\alpha_i}\cap Y)\subset\cup^n_{i=1}A_{\alpha_i}$$ and conclude $$Y\subset\cup^n_{i=1}A_{\alpha_i}$$

Lastly, as

$$\mathcal{A}$$ was a covering $$\cup_{\alpha\in I}A_\alpha=Y$$ .

It is clear that

$$x\in\cup^n_{i=1}A_{\alpha_i}\implies x\in\cup_{\alpha\in I}A_\alpha$$ so again implies and subset relation we have:
$$\cup^n_{i=1}A_{\alpha_i}\subset\cup_{\alpha\in I}A_\alpha=Y$$ thus concluding $$\cup^n_{i=1}A_{\alpha_i}\subset Y$$

Combining

$$Y\subset\cup^n_{i=1}A_{\alpha_i}$$ and $$\cup^n_{i=1}A_{\alpha_i}\subset Y$$ we see $$\cup^n_{i=1}A_{\alpha_i}=Y$$

Thus

$$\{A_{\alpha_i}\}^n_{i=1}$$ is a finite covering of $$Y$$ consisting of open sets from $$X$$

$$\impliedby$$