[Math] Question on problem: Equivalence of two metrics $\iff$ same convergent sequences

general-topologymetric-spaces

Community!

I'm working on the following problem:
Let $X$ be a non-empty set and $d_1,d_2$ metrics on X. Show that the following conditions are equivalent:

1) $d_1$ and $d_2$ are equivalent, i.e.
$\forall x \in X, \epsilon < 0 \exists \delta >0: U_\delta^1 \subset U_\epsilon^2 \land U_\delta^2 \subset U_\epsilon^1 $

2) The convergent sequences in $(X,d_1)$ and $(X,d_2)$ coincide

This question has been asked and answered multiple times, but I still have one particular question. I am not really sure what condition 2 means.
The two interpretations I see:

a) $\{(a_n)_{n \ \in \mathbb{N}} : (a_n)_{n \ \in \mathbb{N}} \text{ converges in } (X,d_1) \}$ =
$\{(a_n)_{n \ \in \mathbb{N}} : (a_n)_{n \ \in \mathbb{N}} \text{ converges in } (X,d_2) \}$, which would allow a sequence to converge to different points for the different metrics.

b) $\forall a \in X, (a_n)_{n \ \in \mathbb{N}} \subset X: a_n \rightarrow_{d_1} a \Leftrightarrow a_n \rightarrow_{d_2} a$, where a convergent sequence must have the same limit point in both metrics.

I have already done the proof for b), but I could not find a way to prove a). I think that b) is actually wrong, but I would be glad for a reassurance

Best regards!

Best Answer

Conditions (a) and (b) are equivalent. Obviously (b) implies (a). One way to see the reverse would be by proving the following:

Lemma. Let $(a_n)$ be a sequence in a metric space. Then $a_n \to a$ if and only if the sequence $(a_1, a, a_2, a, a_3, a, \dots)$ is convergent.