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?


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 $[2,3]$.

  • $\begingroup$ 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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