I have the following task:
Show that a metric Space $(X,d_X)$ is complete if and only if for every isometric embedding $f: X \to Y$ in another metric Space $(Y, d_Y)$ it holds true that $f(X)$ is closed in $Y$
Any ideas or hints on how to prove this? Thanks in advance.
(I know that since $f$ is isometric it preserves the distance between two points and that since it's an embedding it's injective and continous. I further know that for every metric space there exists a completion i. e. there exists a metric space $(Z, d_Z)$ and an isometry $\phi : X \to Z$ so that $\phi (X) $ is dense in $Z$. But i don't really know how to pack all this together and get the prove)
The more difficult part is, in fact "$ \Leftarrow $", i.e the fact that $(X,d_x)$ is complete, but according to what you have written you have all the necessary ingredients.
Consider the obvious inclusion $i$ of $(X,d_x)$ into it's completion $(X_c, d_{X_c})$ (which, also obviously, is an isometry) and consider the image of a Cauchy sequence $x_n$ under this inclusion. This is, of course, also a Cauchy sequence since $i$ is an isometry. Since the target $X_c$ is complete, $i(x_n) $ converges to $x_c\in X_c$, and since the image is closed, $x_c=i(x)$ for some $x$. Since $i$ is an isometry, $d(x_n, x) = d(i(x_n), x_c)$ which tends to $0$, so $x_n\rightarrow x$ and $X$ is complete w.r.t $d_X$
I leave the other direction, which is a bit simpler, to you (or someone else).