Difference between revisions of "Compactness"

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	Then <math>\{A_i\}^n_{i=1}</math> is a subcollection of <math>\mathcal{A}</math> that 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>====	+	====<math>\impliedby</math>====
	{{Definition|Topology}}		{{Definition|Topology}}

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Revision as of 17:45, 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 $$\{A_i\cap Y\}^n_{i=1}$$ covers $$Y$$

Then

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

$$\impliedby$$