Equivalence of Definitions of Limit Point in Metric Space

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 $\alpha \in S$.

The following definitions of the concept of limit point in metric space are equivalent:

Definition 1

$\alpha$ is a limit point of $A$ if and only if every deleted $\epsilon$-neighborhood $\map {B_\epsilon} \alpha \setminus \set \alpha$ of $\alpha$ contains a point in $A$:

$\forall \epsilon \in \R_{>0}: \paren {\map {B_\epsilon} \alpha \setminus \set \alpha} \cap A \ne \O$

that is:

$\forall \epsilon \in \R_{>0}: \set {x \in A: 0 < \map d {x, \alpha} < \epsilon} \ne \O$

Note that $\alpha$ does not have to be an element of $A$ to be a limit point.

Definition 2

$\alpha$ is a limit point of $A$ if and only if there is a sequence $\sequence{\alpha_n}$ in $A \setminus \set \alpha$ such that $\sequence{\alpha_n}$ converges to $\alpha$, that is, $\alpha$ is a limit of the sequence $\sequence{\alpha_n}$ in $S$.

Definition 3

$\alpha$ is a limit point of $A$ if and only if $\alpha$ is a limit point in the topological space $\struct{S, \tau}$.

Proof

$(1) \implies (2)$

Let:

$\forall \epsilon \in \R_{>0}: \paren {\map {B_\epsilon} \alpha \setminus \set \alpha} \cap A \ne \O$

A sequence $\sequence {\alpha_n}$ is constructed using finite recursion.

Let $\alpha_0$ be some arbitrary point in $A \cap \paren{\map {B_1} \alpha \setminus \set \alpha}$.

$\alpha_{n + 1}$ is some arbitrary point in $A \cap \paren{\map {B_{\frac {\map d {\alpha, \alpha_n} } 2} } \alpha \setminus \set \alpha}$ where $\map {B_{\frac {\map d {\alpha, \alpha_n} } 2} } \alpha$ denotes the open $\dfrac {\map d {\alpha, \alpha_n} } 2$-ball of $\alpha$.

We can be assured that such a point exists by hypothesis.

Then:

$\map d {\alpha_n, \alpha} < 2^{-n} \map d {\alpha_0, \alpha}$

Since $\map d {\alpha_0, \alpha}$ is fixed:

$\map d {\alpha_n, \alpha} < 2^{-n} \map d {\alpha_0, \alpha} < r$

for some $n$ because the natural numbers are Archimedean.

Therefore $\alpha$ is the limit of the sequence $\sequence {\alpha_n}$ by definition.

$\Box$

$(2) \implies (3)$

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

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.

$\Box$

$(3) \implies (1)$

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

From Open Ball is Open Set:

$\forall \epsilon \in \R_{>0}: \map {B_\epsilon} \alpha$ is open set in $M$

By Definition of Topology Induced by Metric:

$\forall \epsilon \in \R_{>0}: \map {B_\epsilon} \alpha$ is open set in $\struct{S,\tau}$

By Definition of Limit Point of Set:

$\forall \epsilon \in \R_{>0}: \paren {\map {B_\epsilon} \alpha \setminus \set \alpha} \cap A \ne \O$

$\blacksquare$