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Suppose $d_1$, $d_2$ are metrics on $X$ and whenever $x_n \rightarrow x$ using $d_1$ we have that $x_n \rightarrow x$ using $d_2$.

Let $\tau_1$ be the collection of open sets of $(X,d_1)$ and $\tau_2$ be the collection of open sets of $(X,d_2)$.

Find a relationship between $\tau_1$ and $\tau_2$.

My attempt:

From the information given, the map $f:(X,d_1) \rightarrow (X,d_2)$ by $f(x) = x$ is continuous. Hence if $G$ is open $(X,d_2)$ then $f^{-1}(G) = G$ is open in $(X,d_1)$ and so $\tau_2 \subseteq \tau_1$.

Is there any better relation than this? Or is this as much as we can say?

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The next thing that one would try is to prove that these two topologies are equivalent. But they are not. If they were, we would also have that $x_n \to x$ using $d_2$ implies $x_n \to x$ using $d_1$, but we don't have this.

Hence, I believe that $\tau_2 \subseteq \tau_1$ is the best you can get.

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  • $\begingroup$ E.g. $d_1$ could be the discrete metric, and $d_2$ the usual metric on the reals. $\endgroup$ – Henno Brandsma Apr 27 '16 at 17:39

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