15 - Limit Points

Limit Point

Let $X$ be a metric space and let $S\subseteq X$ be any subset then

Point $a\in X$ is a limit point of $S$ if

$$\text{any open ball about }a\text{ contains a point of }S\text{ other than }a\text{ itself}$$

Set of Limit Points

Let $X$ be a metric space and let $S\subseteq X$ be any subset then

$$L(s):=\text{set of limit points of S}$$

Isolated Points

Let $X$ be a metric space and let $S\subseteq X$ be any subset then

$a$ is an isolated point of $S$ if there exists ball $B(a,ϵ)$ such that

$$B(a,ϵ)\cap S=\{a\}$$

Set of Limits Points is a Closed Subset lemma

Let $S$ be a subset of a metric space $X$ then

$$L(s)\text{ is a closed subset of }X$$

Proof

Need to show that complement $L(S{)}^c$ is open

Suppose $a\in L(S{)}^c$ then there exists ball $B(a,ϵ)$ such that

$$B(a,ϵ)\cap S=\{a\}\text{ or }\varnothing$$

Claim that $B\left(a,\frac{ϵ}{2}\right)\subseteq L(S{)}^c$
Let $b\in B\left(a,\frac{ϵ}{2}\right)$

If $b=a$ then $b\in L(S{)}^c$
I f $b\ne a$ then there exists ball about $b$ contained in $B(a,ϵ)$ that doesn’t contain $a$
Ball $B(b,\delta )$ where $\delta =\min \left(\frac{ϵ}{2},d(a,b)\right)$ has this property

This ball meets $S$ in the empty set hence $b\in L(S{)}^c$ as well
So $B\left(a,\frac{ϵ}{2}\right)\subseteq L(S{)}^c$ so $L(S)$ is closed

Characterisation of the Closure via Limit Points

Let $S$ be a subset of metric space $X$
Let $L(S)$ be its set of limit points and $\overline{S}$ be it’s closure then

$$\overline{S}=S\cup L(S)$$

Proof

Showing $S\cup L(S)\subseteq \overline{S}$

Naturally $S\subseteq \overline{S}$ so we need $L(S)\subseteq \overline{S}$
Suppose $a\in {\overline{S}}^c$
Since ${\overline{S}}^c$ is open then there exists ball $B(a,ϵ)$ lying in ${\overline{S}}^c$ (hence also in $S^c$)
Hence $a$ cannot be a limit point of $S$
Hence $L(s)\subseteq \overline{S}$ (by contrapositive)

Showing $\overline{S}\subseteq S\cup L(S)$
By Interior points and convergent sequences then
Exists sequence $(x_n{)}_{n=1}^{\infty }$ of elements of $S$ with $x_n\to a$

If $x_n=a$ for some $n$ then $a\in S$
Suppose $x_n\ne a$ for all $n$ then
Let $ϵ>0$ there exists $N$ such that for all $n\ge N$ we have $x_n\in B(a,ϵ)\setminus \{a\}$ so they all in $S$
Hence $a$ is a limit point of $S$ as $B(a,ϵ)$ contains $x_N\ne a$
Thus $a\in L(S)$

Closed Sets contain all Limit Points

Let $S$ be a subset of a metric space $X$ then

$$S\text{ is closed }⟺S\text{ contains all its limit points}$$

Proof

From Property of closed sets and closures then

$$S\text{ is closed}⟺S=\overline{S}$$

Hence result follows