Theorem
Suppose we have a metric space![]()
Define
with
if
and
then if
is a completion of some metric space
then
is ismorphic to![]()
Proof
is a subspace of![]()
Hence for every
there exists a Cauchy sequence
converging to![]()
Define a mapping
by![]()
The mapping
is well defined, since if
converges to
then
so that![]()
Also,
is subjective. Suppose
then
is a Cauchy sequence in
But
is complete hence
converges to
and![]()
Now suppose
then there are sequences
in
such that![]()
Then![]()
Hence
is an isometry between
and![]()