Compact - Sequentially Compact - Totally Bounded

For a metric space , the following are equivalent:

1. is Compact
2. is Sequentially Compact
3. is Totally bounded and complete

Proof(s)

Suppose is compact.
Suppose it is not sequentially compact.
Find a sequence which doesn’t have a convergent subsequence.
Suppose that there is some
s.t. every nbd of
has infinitely many points of the sequence.
Now there must be a subsequence of converging to - contradiction.
Hence, for every
find open nbd of
which contains at most finitely many points in the sequence.
Then are an open cover of .
But is compact,
so find finite s.t. .
Hence, this union contains at most finitely many points of the sequence,
but there are infinitely many of them in - contradiction.

Suppose is sequentially compact.
Take any Cauchy sequence.
It has a convergent subsequence.
But that convergent subsequence will bind our Cauchy sequence,
so the Cauchy sequence converges,
hence is complete.
Now suppose is not totally bounded.
Find an s.t. every -net is infinite.
Pick any point and keep picking
(always possible because finitely many points can’t cover ).
Now find a convergent subsequence of this
but the sequence is not Cauchy!
And a convergent sequence is always Cauchy!
Contradiction.

Suppose is totally bounded and complete.
Pick a sequence in .
Let .
For any let and find a finite -net.
One of the balls in it will contain infinitely many terms of the sequence ,
so let be this infinite set contained in this ball.
Hence define .
Then the sequence is Cauchy
(for any , the sequence is contained in a ball of radius )
hence convergent.

Suppose is sequentially compact.
Because of we have already proved that is totally bounded and complete.
Let be an open cover of .
Use Lebesgue number lemma to find a
Find a finite -net.
By definition of ,
find for each in the -net.
Then cover .
Hence is compact.