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For example, space $A$ has a metric $\rho$, and its subspace $B\subset A$ has a metric $d$, which happens to have much better properties than $\rho$.

So if $x_{1},x_{2}\in A\setminus B$, but they are very close to $B$ such that $\rho\left(x_{1},B\right)<\varepsilon_{1}$ and $\rho\left(x_{2},B\right)<\varepsilon_{2}$, we prefer to somehow use $d$ to approximately measure the distance between $x_{1}$ and $x_{2}$, for example by defining some sort of composite metric such as, roughly speaking,

$$ \rho_{mod}\left(x_{1},x_{2}\right)=\rho\left(x_{1},y_{1}\right)+d\left(y_{1},y_{2}\right)+\rho\left(y_{2},x_{2}\right) $$

where $y_{j}:=\underset{y\in B}{\mathrm{argmin}}\rho\left(x_{j},y\right)\forall j\in\left\{ 1,2\right\}$

I am a physicist, not a mathematician, so I would appreciate any expert advice on a mathematically rigorous roadmap to do this.

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  • $\begingroup$ That kind of modification makes the topology discrete outside $\bar B$, and you probably don't want that. $\endgroup$ Sep 28, 2014 at 13:48
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    $\begingroup$ One can "glue" metric spaces and your construction of $\rho_{mod}$ is very close to the gluing of $B$ and $A$ along the inclusion $B\to A$. Maybe that is it (?). $\endgroup$ Sep 28, 2014 at 17:42
  • $\begingroup$ Thanks @AntonPetrunin, I will look into it from that perspective. $\endgroup$
    – Jake
    Sep 28, 2014 at 18:02

1 Answer 1

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There exists a theorem--by Hausdorff--about extending subspace metrics, according to which:

Given a metric space $\ (X\ d),\ $ a closed subset $\ A\subseteq X,\ $ and a metrics $\ \rho_0\ $ in $\ A\ $ such that metrics $\ \rho_0\ $ and $\ d\,|\,A\times A\ $ are topologically equivalent (in $\ A$),  there exists a metrics $\ \rho\ $ in $\ X\ $ which is topologically equivalent to $\ d,\ $ and such that $\ \rho\,|\,A\times A = \rho_0$.

An especially elegant proof was obtained by Henryk Toruńczyk, "A short proof of Hausdorff’s theorem on extending metrics", Fund. Math. 77 (1972), no. 2, 191–193. MR 47:9559.

(A search will provide you with more information).

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