Are these two metrics equivalent?

Question

Let $d_1$ and $d_2$ be metrics on the space $X$. Assume that for any sequence $\{x_n\}_{n=1}^\infty \subset X$ and point $x_0 \in X$ we have that $$ \lim_{n \to \infty}d_1(x_n,x_0)=0 \iff \lim_{n \to \infty}d_2(x_n,x_0)=0. $$ Can we conclude that the metrics $d_1$ and $d_2$ are equivalent, i.e. that they induce the same (metric) topology? I would be tempted to trivially say "yes", since the spaces $(X,d_1)$ and $(X,d_2)$ are homeomorphic, an isomorphism being given by the identity function. Am I missing something?

Answer 1

The metrics $d_1,d_2$ on $X$ are equivalent iff $\textrm{id}_X: (X,d_1) \to (X,d_2)$ is a homeomorphism.

And the sequential continuity criterion for $\textrm{id}_X$ applies by the given property, in both directions. So the idea you have is good; just state it more accurately.

Another observation: in any metric topology $O$ is open iff

$$\forall x \in O: \text{ for all sequences } (x_n)_n \text{ in } X: (x_n \to x) \implies (\exists N \in \Bbb N: \forall n \ge N: x_n \in O)$$

and so as $d_1$ and $d_2$ have the same convergent seequences, they have the same open sets too, so are equivalent. A variation of this for closed sets can also be made.

Answer 2

You are not missing anything. Closed sets are same for the two metrics since a set is closed iff the limit of any sequence from it belongs to it. Hence the two metrics have the same closed sets (and therefore the same open sets).