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Suppose $X_1$ and $X_2$ are isometric and $X_1$ is a complete space; show that $X_2$ is a complete space. Here I need somebody to help me or to give me ideas.

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Use the definition of a complete space, show that because $X_2$ is isometric to $X_1$ every Cauchy sequence in $X_2$ can be pulled to a Cauchy sequence in $X_1$, and therefore it must have a limit. – Asaf Karagila Apr 14 '12 at 9:41

Let $f:X_1\to X_2$ be an isometry. To show that $X_2$ is complete, you must show that every Cauchy sequence in $X_2$ converges, so let $\langle x_k:k\in\mathbb N\rangle$ be a Cauchy sequence in $X_2$. What do you know about $\langle f^{-1}(x_k):k\in\mathbb N\rangle$?

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let $f: X_1 \to X_2$ be a isometric isomorphism, $(x_n)$ a Cauchy-Seqence in $X_2$. Then $((f^{-1}(x_n))$ is Cauchy in $X_1$ (can you show why?). As $X_1$ is complete, $f^{-1}(x_n) \to y$, but now ...

Hope this helps,

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