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

Theorem
Let $M = \struct {S, d}$ be a metric space or a pseudometric space.

Let $T = \struct {S, \tau}$ be the topological space induced by $d$.

Let $\sequence {x_n}$ be a infinite sequence in $S$.

Let $l \in S$.

Then $\sequence {x_n}$ converges to $l$ relative to $d$ $\sequence {x_n}$ converges to $l$ relative to $\tau$.

Necessary Condition
Suppose that $\sequence {x_n}$ converges to $l$ relative to $d$.

When $\map {B_\epsilon} l$ denotes the open $\epsilon$-ball of $l$, this means:


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

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

By definition of induced topology, there exists $\epsilon_0 \in \R_{>0}$ such that:
 * $\map {B_{\epsilon_0} } l \subseteq U$

Then there exists $N_0 \in \R$ such that for all $n > N_0$:
 * $x_n \in \map {B_{\epsilon_0} } l \subseteq U$

Hence, $\sequence {x_n}$ converges to $l$ in the induced topology $\tau$.

Sufficient Condition
Suppose that $\sequence {x_n}$ converges to $l$ in $\tau$.

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

Then $l \in \map {B_\epsilon} l$, and $\map {B_\epsilon} l \in \tau$.

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

Hence, $\sequence {x_n}$ converges to $l$ relative to $d$.