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Let $E$ and E' metric spaces. If $E$ isomorphic to a subspace of E', then E' is isomorphic to an space that contain $E$.

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    In that case, let $A$ and $A'$ metric spaces, $f:A\to A'$ is an isomorphism if is surjective and if $x,y\in A$, $d(x,y)=d'(f(x),f(y)),$ where $d$ and $d'$ are the respective metrics of each space.2011-05-05

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Let the f\colon E\to E' be an embedding of metric spaces (that is isomorphism of $E$ with a subspace of $E'$). Define a metric space $H$ as following:

H=E'\setminus \text{Rng}(f)\cup E, and the metric of $H$ defined as:

d_{H}(x,y) = \begin{cases} d_E(x,y) & \text{ if } x,y \in E \\d_{E'}(x,y) & \text{ if } x,y\in E' \setminus E \\d_{E'}(x,f(y)) & \text{ if } x \in E', y \in E \end{cases}

This is a metric spaces, and it is isomorphic to E' by taking $f$ on $E$ and the identity on $H\setminus E$.

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    @leo: If an answer (not only this, but to other questions you have asked, and will ask) has helped you to solve your question it is usual to accept it by the check sign below the voting arrows, if several were given choose the one most fitting to you, if no answer satisfied your problem please e$x$plain why not in your question or comments to the answers.2011-05-05