Equivalence of Definitions of Limit Point in Metric Space/Definition 2 implies Definition 3

Theorem

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

Let $\tau$ be the topology induced by the metric $d$.

Let $A \subseteq S$ be a subset of $S$.

Let $\alpha \in S$.

Let there exist a sequence $\sequence{\alpha_n}$ in $A \setminus \set \alpha$ such that $\sequence{\alpha_n}$ converges to $\alpha$ in $S$.

Then:

$\alpha$ is a limit point in the topological space $\struct{S, \tau}$.

Proof

Let $U \in \tau$ be an arbitrary open set:

$\alpha \in U$

By Definition of Topology Induced by Metric:

$\exists \epsilon \in \R_{>0} : \map {B_\epsilon} \alpha \subseteq U$

By Definition of Convergent Sequence (Metric Space):

$\exists N \in \N : \forall n \ge N : d(\alpha, \alpha_n) < \epsilon$

By Definition of Open Ball:

$\alpha_N \in \map {B_\epsilon} \alpha$

By hypothesis:

$\alpha_N \in A \setminus \set \alpha$

We have:

\(\ds \alpha_N \in \paren{A \setminus \set \alpha} \cap \map {B_\epsilon} \alpha\)

\(=\)

\(\ds \paren{A \cap \map {B_\epsilon} \alpha} \setminus \set \alpha\)

Intersection with Set Difference is Set Difference with Intersection

\(\ds \)

\(\subseteq\)

\(\ds \paren{A \cap U} \setminus \set \alpha\)

Set Intersection Preserves Subsets and Set Difference over Subset

\(\ds \)

\(=\)

\(\ds A \cap \paren{U \setminus \set \alpha}\)

Intersection with Set Difference is Set Difference with Intersection

Hence:

$A \cap \paren{U \setminus \set \alpha} \ne \O$

Since $U$ was arbitrary, it follows that $\alpha$ is a limit point in the topological space $\struct{S, \tau}$ by definition.

$\blacksquare$