Theorem
Any countably compact subset
of a metric space
is also sequentially compact.
Proof
Let
be a metric space and let
be a sequence in
If the set
is finite then one of the elements, say
appears infinitely often, hence the subsequence
converges to![]()
Suppose the set
is infinite.
is countably compact. The infinite subset of![]()
has an accumulation point
Since
is a metric space, we can choose a subsequence
which converges to
Hence
is sequentially compact.