TITLE: Isometry of a complete normed space is also complete.
QUESTION [5 upvotes]: Let $X$ be a complete normed space and assume the normed space $Y$ is isometric to $X$. Show that $Y$ is complete.
I tried:
Since X is complete $||x_n-x_m||<\epsilon, \forall n,m>N$ and since $Y$ is isometric to $X$ there exists an isometry $f:X\to Y$ such that 
$$||f(x_n)-f(x_m)||=||x_n-x_m||<\epsilon, \forall n,m>N.$$ I stuck at this step.

REPLY [2 votes]: There is nothing to prove here: $Y$ is a complete metric space even if it wouldn't  be a vector space. It's all in the word "isometric". This word says that $Y$ is a bijective copy of $X$ whereby the distance between points is preserved. This implies that the notions of "convergence" or "Cauchy sequence" in $X$ and in $Y$ are the same.