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I'm trying to prove that if the convergent sequences of $(X,d)$ and $(X,\rho)$ are the same, then the metrics $d$ and $\rho$ are equivalent.

Equivalent metrics are those that generate the same open sets.

Say there is one open set generated by $d$ that is not generated by $\rho$. Does this create any contradiction? If $\langle x_i\rangle$ converges in $(X,d)$, it will still converge to the same point in $(X,\rho)$. I don't know where I'm going wrong here. Any help will be greatly appreciated.

user1992
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1 Answers1

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Hints: (a) A set $U$ in a metric space $(X,d)$ is open iff any sequence $(x_i)$ converging to some element $x\in U$ has all but finitely many terms in $U$.

(b) Suppose $U$ is $d$-open, but not $\rho$-open. Then there is some $x \in U$, such that for any $\epsilon > 0$, the $\rho$-$\epsilon$-neighbourhood of $x$ is not contained in $U$. That is for any $i$ there is a $x_i \in X \setminus U$ such that $\rho(x,x_i)< 1/i$.

martini
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