I want to know if the claim is true and someone know a proof for it.
Claim:
Let $(F,d_F)$ be any metric space. Then $F$ is complete if and only if, for any space $(X,d_X)$ with $F\subset X$ and ${d_{X_{|F\times F}}}=d_F$, $F$ is closed in $X$.
I also found this, which is pretty much the same. I would like to know whether we always have a set that is the actual completion of $F$, where $F$ is a subset and not the image under an isometry is a subset?
Show: $(X,d_X)$ is complete $\Leftrightarrow $ $f(X)$ is closed in $(Y, d_Y)$ ($f: X \to Y$ is an isometric embedding)
My proof (sketch):
$"\Rightarrow"$ Let $(F,d_F)$ be a complet metric space and $(X,d_X)$ as above. Let $(x_n)_n\subset F$ convegent sequence with $x_n\rightarrow x \in X$, then $(x_n)_n$ is a cauchy sequence in $d_X$ and so as well in $d_F$. Since $F$ is complete its limit exist in $F$, so $x\in F$.
$"\Leftarrow"$ Assume $(F,d_F)$ is not complete. Then there exist a cauchy sequence $(x_n)_n\subset F$ that dose not convege. We define $\hat X:=F\cup \{\hat x\}$, where $\hat x$ is some element not in $F$ (for example the set $F$ itself) and $$d^*(x,y)= \left\{\begin{array}{l,l} d(x,y) & \text{ if }x,y\in F\\ \lim_{n \to \infty}d(x_n,y)& \text{ if } x=\hat{x}\end{array}\right.$$ (Similar if $y=\hat x$ The limit exists as $(x_n)$ is a cauchy sequence). Now $(\hat X, d^*)$ is a metric space. $(x_n)$ is a towards $\hat x$ converging sequence. $F$ is not closed in $\hat X$.
I have skipped some details and may have mad some mistakes. If someone think so I can work them out. But I wanted to ask first whether there is a better way to show the second direction or even if the statement as I have written it down is false.
Note: For normed spaces every isomorphism, that is a bijective linear bounded map who's inverse is bounded as well, preserve completeness. For a normed space the canonical embedding is $\Phi:E\rightarrow E^{**},\; x\mapsto \hat x$, where $\hat x:E^{*}\rightarrow\mathbb{K},\; \ell\mapsto \ell(x)$. $E^{**}$ is complete and the completion of $E$ is $\overline{\Phi(E)}$