TITLE: Question on problem: Equivalence of two metrics $\iff$ same convergent sequences
QUESTION [8 upvotes]: 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!

REPLY [5 votes]: 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.