# Compactness in Metric Spaces: Is It Possible?

• Bleys
In summary, in metric spaces, compactness is often proven after introducing the notions of closed and bounded or sequential compactness. However, there is a way to prove that a closed interval [a,b] is compact in the usual topology on the real numbers just from definition. This can be done by considering an arbitrary open cover and using the fact that this topology has a natural correspondence with the metric space. This is shown in the proof of the Heine-Borel Theorem, which explicitly proves the compactness of a closed interval using covers. The proof involves considering the supremum of all numbers c' such that [a,c'] can be covered by finitely many elements of the cover, and ultimately shows that the supremum is

#### Bleys

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?

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...

Last edited by a moderator:
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.