# Isometry from Banach Space to a Normed linear space maps

Let $X$ be a Banach Space and $Y$ be a normed linear space. Show that if $T$ is an isometry then $T(X)$ is closed in $Y$.

Let me have some idea to solve this. Thank you for your help.

#### Solutions Collecting From Web of "Isometry from Banach Space to a Normed linear space maps"

I had problem in showing every sequence is cauchy in T(X).Now it is done.

Let $y_{n}$ be sequence in $T(X)$. Since $y_{n}=T(x_{n})$ $\forall n$. So we have got a sequence $(x_{n})_{{n \in \mathbf{N}}}$ in X.
Now X is a Banach Space therefore it is complete and hence $x_{n} \longrightarrow x$ for some $x$ in $X$.
$\|T(x_{n})-T(x_{m})\|=\|T(x_{n}-x_{m})\|=\|x_{n}-x_{m}\|$ as $T$ is isometry.

Since $(x_{n})_{n \in \mathbf{N}}$ is convergent it is Cauchy. This tells that $(T(x_{n}))_{n \in \mathbf{N}}$ is cauchy. But T is continious so $T(x_{n}) \longrightarrow T(x)$ and uniqueness of limit point gives $y_{n} \longrightarrow T(x)$. It shows that every sequence in $T(X)$ converges in $T(x)$. Therefore T(X) is closed in Y.