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Given two metric spaces $(X , \rho_1)$ and $(X, \rho_2)$ are isometrically isomorphic, then is it true that for a sequence $\{ x_n \} \subset X $, it converges in $\rho_1$ if and only if it converges in $\rho_2$?

In other words, is it a true argument that for a sequence $\{ x_n \} \subset X $,
$x_n \to x$ in $\rho_1$ $\iff$ $x_n \to x$ in $\rho_2$?

I thought it is true, because isometry is a distance-preserving transformation.

Could anyone please explain this more clearly and precisely?

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It doesn't have to be true because we don't know if the identity function is the isometric isomorphism. Consider the following example.

$X = [0,1]\cup[2,3]$. For $x,y \in [0,1]$, $\rho_1(x,y) = 1$ if and only if $x\not = y$, and of course $\rho_1(x,x) = 0$. For $x,y \in [2,3]$, we set $\rho_1(x,y) = |x-y|$. And define $\rho_1(x,y) = 2$ for $x \in [0,1], y \in [2,3]$ or vice-versa. So $[0,1]$ is discrete in $[0,1]$, the normal Euclidean metric in $[2,3]$ and just $2$ when $x,y$ are in different intervals.

Define $\rho_2$ as the opposite of $\rho_1$ in the sets $[0,1]$ and $[2,3]$ but still $2$ for $x \in [0,1], y \in [2,3]$ or vice-versa. Namely, $\rho_2$ is discrete on $[2,3]$ and normal on $[0,1]$ and then $2$ for $x,y$ in the different intervals.

Then $\phi: X \to X$ given by $\phi(x) = x+2$ for $x \in [0,1]$ and $\phi(x) = x-2$ for $x \in [2,3]$ is an isometric isomorphism. However, $2+\frac{1}{n} \to 2$ in $\rho_1$ but not in $\rho_2$ since $\rho_2$ is discrete in $[2,3]$.

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  • $\begingroup$ Thanks @mathworker21. But what if the topological space $X$ is connected? Another question is that when the above statement is true for $x_n \to x$ in $\rho_1$ $\iff$ $x_n \to x $ in $\rho_2$? In other words, what kind of sufficient condition we need to add to ensure that the above statement is true? $\endgroup$ – Paradiesvogel Mar 28 '17 at 21:49

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