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?


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.

  • $\begingroup$ 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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