## Equivalence of Compactness, Countable Compactness and Sequential Compactness

Theorem

Ifis a metric space then the following statements are equivalent.

1.is compact

2.is countably compact

3.is sequentially compact

Proof

1. Supposeis compact, then letrepresent a subset ofwith no accumulation points inEach pointbelongs to an open setcontaining at most one point of Consider the familyThis is a family of open sets inandhenceis an open cover of

Sinceis compact we can find a finite subcoverand

Since eachcontains at most one point ofis finite and every infinite subset of contains an accumulation point ofsois countably compact.

2. We prove that ifis not bounded thenis not sequentially compact. Ifexists such that no finite– net ofexists . Letthenexists such that

Otherwisewould– net.

Similarly a pointexists such thatotherwise would be– net.

This procedure gives a sequence of pointssuch thatfor

This sequence does not have a convergent subsequence henceis not sequentially compact.

3. Letbe sequentially compact and letbe an open cover ofSethas a Lebesgue numbersince every compact subset of a metric space has a Lebesgue number. Sinceis totally bounded there is a decomposition ofinto a finite number of subsetswithbutis a Lebesgue number ofhence open setsexist such that

Hence andis a finite subcover of

Henceis compact.