Let $E$ and $E'$ metric spaces. If $E$ isomorphic to a subspace of $E'$, then $E'$ is isomorphic to an space that contain $E$.
Extend an isomorphism
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analysis
functional-analysis
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0It's clear that exist $F\subset E'$ with $F$ isomorphic to $E$, I think that the isomorphism $f$ between $E'$ and a space that contain $E$ should coincide with the isomorphism $g:F\to E$, that we obtain for hypothesis, if is restricted to $F$. For me, the difficult is that I don't know nothing about $F$ and $E$, they are arbitrary. For me, the answer is clear if $F$ dense in $E'$ and $E$ complete, for example. – 2011-05-05
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0Yes, thanks @Jonas Meyer. – 2011-05-05
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0What is your notion of isomorphism? – 2011-05-05
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0Is that trivial? I can't see it'. – 2011-05-05
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0In 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
1 Answers
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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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0If someone can TeX up the cases of the metric $d_H(x,y)$ I would be grateful. Thanks :-) – 2011-05-05
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0I tried, please check whether I've messed it up. – 2011-05-05
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0@Theo, it is fine. Many thanks. I did not know they added support for the `cases` environment. Good to know! – 2011-05-05
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0@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 explain why not in your question or comments to the answers. – 2011-05-05