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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.

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    $\begingroup$ Let $(y_n)$ be a Cauchy sequence in $T(X)$ ... $\endgroup$ – Daniel Fischer Sep 22 '13 at 13:20
  • $\begingroup$ Let $y_n=T(x_n)$ be a sequence in $T(X)$ converging to some $y\in Y$. Then $(y_n)$ is a Cauchy sequence ... $\endgroup$ – Jochen Sep 23 '13 at 7:12
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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.

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  • $\begingroup$ The last bit of your argument is a bit clumsy. It's better to argue that because $T(X)$ is a complete subspace of $Y$, it must be closed in $Y$. $\endgroup$ – kahen Sep 23 '13 at 10:45
  • $\begingroup$ Sorry, but please point out the clumsy part,then I can explain that in more details. $\endgroup$ – Arindam Nov 4 '13 at 14:13

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