This question is going to sound very trivial, but given a complete metric space $(X,d)$, how does one show that its completion via Cauchy sequences $(X^*,d^*)$ is itself?
I know that $X$ is dense in its completion $X^*$. So how can I make use of this fact to prove the above claim?
Any help/hint will be appreciated. Thanks in advance.
Let $i:X \to X^*$ be the canonical inclusion. As stated in the comments $i$ is an isometry, in particular it is injective. $X$ is complete by assumption, so $\overline{i(X)}=i(X)=X^*$. (Any element in the closure of $i(X)$ is a limit $i(x_n)_n$. Since $i$ is an isometry, the sequence $(x_n)_n$ is Cauchy in $X$. Thus, converges.) Thus, $i$ is bijective. One can show that $i$ is closed (not hard). Hence, $i$ is a homeomorphism.