Compact spaces

  • Thread starter Bleys
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  • #1
Bleys
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I've never actually seen a proof that a space is compact just from the definition. In metric spaces it was usually after the notion of closed and bounded or sequential compactness was introduced.
For example is there a way to prove [a,b] is compact (with the usual topology on the real numbers) just from definition? It seems almost impossible... you have to consider an arbitrary open cover!
Is it possible to use the fact that this topology has a natural correspondence with the metric space? Because open sets in the metric space are the same as the open sets in the topology?
 

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  • #2
micromass
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See http://planetmath.org/encyclopedia/ProofOfHeineBorelTheorem.html [Broken]
There they explicitely prove that a closed interval is compact using covers. The proof is somewhat involved though...
 
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  • #3
Tinyboss
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It's not too involved--let c be the supremum of all numbers c' such that [a,c'] can be covered by finitely many elements of the cover you're given. Well, c is contained in some element of the cover, so add that element to your finite subcover to get another finite subcover that includes a neighborhood of c, so that it covers [a,c+epsilon] (you don't end up with two disconnected components because of the definition of supremum). So c was not the supremum after all, unless c=b.
 

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