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.
Best Answer
Pick an arbitrary Cauchy sequence $\{y_n\}\subset Y$, and let $f\colon X\to Y$ be the isometry. For each $n\geq1$, $y_n=f(x_n)$ for some $x_n\in X$. We have \begin{equation*} \|y_n-y_m\|=\|f(x_n)-f(x_m)\|=\|x_n-x_m\|, \end{equation*} so that $\{x_n\}$ is a Cauchy sequence in $X$. Since $X$ is complete, $x_n\to x$ for some $x\in X$. Therefore, $y=f(x)$ is the limit of $\{y_n\}$, since \begin{equation*} \|y_n-y\|=\|f(x_n)-f(x)\|=\|x_n-x\| \end{equation*}
Therefore, $Y$ is also complete.