Sequence Converges to Point Relative to Metric iff it Converges Relative to Induced Topology

Theorem
Let $M=(S,d)$ be a metric space or a pseudometric space.

Let $T=(S,\tau)$ be the topological space induced by $d$.

Let $\left\langle{ x_n }\right\rangle$ be a infinite sequence in $S$.

Let $l \in S$.

Then $\left\langle{ x_n }\right\rangle$ converges to $l$ relative to $d$ iff $\left\langle{ x_n }\right\rangle$ converges to $l$ relative to $\tau$.

Proof
Suppose that $\left\langle{ x_n }\right\rangle$ converges to $l$ relative to $d$.

When $B_\epsilon \left({l}\right)$ denotes the open $\epsilon$-ball of $l$, this means:


 * $\forall \epsilon \in \R_{>0}: \exists N \in \R: n > N \implies x_n \in B_\epsilon \left({l}\right)$

Let $U \in \tau$ with $l \in U$.

By definition of induced topology, there exists $\epsilon_0 \in \R_{>0}$ such that $B_{\epsilon_0} \left({l}\right) \subseteq U$.

Then there exists $N_0 \in \R$ such that for all $n > N_0$, we have $x_n \in B_{\epsilon_0} \left({l}\right) \subseteq U$.

Hence, $\left\langle{ x_n }\right\rangle$ converges to $l$ in the induced topology $\tau$.

Suppose that $\left\langle{ x_n }\right\rangle$ converges to $l$ in $\tau$.

Let $\epsilon \in \R_{>0}$.

Then $l \in B_\epsilon \left({l}\right)$, and $B_\epsilon \left({l}\right) \in \tau$.

Then there exists $N \in \N$ such that for all $n > N$, we have $x_n \in B_{\epsilon} \left({l}\right)$.

Hence, $\left\langle{ x_n }\right\rangle$ converges to $l$ relative to $d$.