Let $(A,d)$ be a metric space with completion $(A',d')$. The definition I use here is that
- $A'$ is complete
- there is an isometry $i:A\rightarrow A'$
- for a complete metric space $(B,d_B)$ and all isometries $f:A\rightarrow B$ there exists a unique isometry $g:A'\rightarrow C$ such that $g\circ i=f$.
How do I prove that $i(A)$ is dense in $A'$?
What I thought:
I want to show that for every $a\in A'$ there is a sequence $(a_n)$ in $A$ such that $i(a_n)\rightarrow a$. So I should define such a sequence. The question is which one?
For the converse of the statement (which I proved already) I had to define $f$ in terms of sequences, so I think I should define $(a_n)$ in terms of $f,g$ and $i$. I just don't know how.