Denseness of completion 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.
 A: Hint: consider the closure of $i(A)$ in $A'$. This is a closed subspace of a complete space, hence complete as well.
So, let $B$ be the closure of $i(A)$ in $A'$ and $f$ be the corestriction of $i$. Then there exists a unique isometry $g\colon A'\to B$ such that $g\circ i=f$.
Can you finish?
A: Assuming you mean $C=B$ .(Apparently a typo).
[1]. When $A$ is not empty:
For brevity let $D= Cl_{A'} (i(A)).$ 
By contradiction, suppose $p\in A'$ \ $D.$ 
Let $B=\{p\}\cup D.$ Let $d''(u,v)=d'(u,v)$  when $u,v \in D.$ Let $d''(u,p)=1+d'(u,p)$ when $u\in D.$ Confirm that $d''$ satisfies  the triangle inequality so $d''$ is a metric on $B$. 
Now $(D,d'')=(D,d')$ is a closed subspace of the complete metric space $(A',d')$ so $(D,d'')$ is complete. Confirm that $(B,d'')$ is complete. Note that $B\subset A'.$
Let $f=i.$ Then $f:A\to B$ is an isometric embedding.  There does NOT exist  an isometric embedding $g:A'\to B$ such that $gi=f=i.$ 
PROOF: Assume such a $g$ exists.
(i). If $g(p)\in D,$ then $\inf \{d''(g(p),i(x)):x\in A\}= \inf \;\{d'(g(p),i(x)): x\in A\}=0$ by def'n of D  and of $d''$, and because $A\ne \phi$. But  then $$0=\inf \;\{d''(g(p),i(x)):x\in A\}=$$ $$=\inf \;\{d''(g(p),g(i(x)):x\in A\}=$$ $$=\inf \; \{d'(p,i(x)):x\in A\}$$ contradicting $p\not \in Cl_{X'}(i(A)).$
(ii). If $g(p)=p$ then there exists $x \in A$ and we have $$1+d'(i(x),p)=d''(i(x),p)=d''(g(i(x)),g(p))=d'(i(x),p)$$ which is absurd... (End of proof of Part [1]).
[2]. If $A$ is empty and $A'$ is not empty, let $B=A'\times \{0,1\}$ with $d_B((u,i),(v,j))=d_{A'}(u,v)+|i-j|.$ There are at least $2$ isometric embeddings of $A'$ into $B$. E.g. $g_0(x)=(x,0)$ and $g_1(x)=(x,1)$.
