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Let $(A,d)$ be a metric space with completion $(A',d')$. The definition I use here is that

  1. $A'$ is complete
  2. there is an isometry $i:A\rightarrow A'$
  3. 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.

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  • $\begingroup$ What is the $C$? Should probably be $B$? $\endgroup$ – Henno Brandsma Mar 6 '17 at 19:03
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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?

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  • $\begingroup$ So I want to show that that closure is equal to $A'$? $\endgroup$ – Billy Bob Mar 6 '17 at 13:50
  • $\begingroup$ @BillyBob I added a further hint $\endgroup$ – egreg Mar 6 '17 at 15:43
  • $\begingroup$ Do I want to show that $g$ is the identity? $\endgroup$ – Billy Bob Mar 6 '17 at 15:53
  • $\begingroup$ @BillyBob The image of $g$ is a subset of $i(A)$ containing $A$ and is a complete space. So… $\endgroup$ – egreg Mar 6 '17 at 15:56
  • $\begingroup$ Do you mean that it is $A'$? $\endgroup$ – Billy Bob Mar 6 '17 at 16:41
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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 $g_1(x)=(x,1)$.

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