## Equivalence of Compactness, Countable Compactness and Sequential Compactness

Theorem

If is a metric space then the following statements are equivalent.

1. is compact

2. is countably compact

3. is sequentially compact

Proof

1. Suppose is compact, then let represent a subset of with no accumulation points in Each point belongs to an open set containing at most one point of Consider the family This is a family of open sets in and hence is an open cover of Since is compact we can find a finite subcover and Since each contains at most one point of is finite and every infinite subset of contains an accumulation point of so is countably compact.

2. We prove that if is not bounded then is not sequentially compact. If exists such that no finite – net of exists . Let then exists such that Otherwise would – net.

Similarly a point exists such that otherwise would be – net.

This procedure gives a sequence of points such that for This sequence does not have a convergent subsequence hence is not sequentially compact.

3. Let be sequentially compact and let be an open cover of Set has a Lebesgue number since every compact subset of a metric space has a Lebesgue number. Since is totally bounded there is a decomposition of into a finite number of subsets with but is a Lebesgue number of hence open sets exist such that Hence and is a finite subcover of Hence is compact. 