06. Interiors, Closures, Limit Points

6.1 Interiors and Closures

Interior

Let $X$ be a metric space and let $S\subseteq X$ then
Interior $\mathrm{int}(S)$ of $S$ is defined as

$$\text{ union of all open subsets of }X\text{ contained in S}$$

Alternate Definition

Let ${\tau }_X$ be the collection of all open subsets of $X$ (Topology of $X$) then

$$\mathrm{int}(S)=\bigcup \{U\in {\tau }_X:U\subseteq S\}$$

Alternate Definition 2

$$\text{unique largest open subset of }X\text{ contained in }S$$

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Closure

Let $X$ be a metric space and let $S\subseteq X$ then
Closure $\overline{S}$ of $S$ is defined as

$$\text{ intersection of all closed subsets of }X\text{ containing S}$$

Alternate Definition

Let $X_C$ be the collection of all closed subsets of $X$ then

$$\overline{S}=\bigcap \{U\in X_C:S\subseteq U\}$$

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Boundary

Let $X$ be a metric space and let $S\subseteq X$ then
Boundary $\partial S$ of $S$ is defined as

$$\overline{S}\setminus \mathrm{int}(S)$$

In other words the “edge / border” of the set

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Dense

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

$S$ is dense if

$$\overline{S}=X$$

Alternate Definition

$$S\text{ is dense }⟺S\text{ meets every open set in }X$$

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Property of Open Sets and Interiors

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

$$S\text{ is open}⟹\mathrm{int}(S)=S$$

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Property of Closed Sets and Closures

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

$$S\text{ is closed}⟹S=\overline{S}$$

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Interior Points

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

$$x\in \mathrm{int}(S)⟹x\text{ is an interior point of }S$$

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Interior Points and Balls lemma

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

$$a\in \overline{S}⟺\text{ every open ball }B(a,ϵ)\text{ contains a point of }S$$

Proof

Proof $⟹$

Suppose $a\in \overline{S}$ and let $ϵ>0$
If $B(a,ϵ)\cap S=\varnothing$ (i.e. $B(a,ϵ)$ does not meet $S$) then $B(a,ϵ{)}^c$ is a closed set containing $S$
Hence $B(a,ϵ{)}^c$ contains $\overline{S}$ so $a\in B(a,ϵ{)}^c$ which is a contradiction
Thus $B(a,ϵ{)}^c\cap S\ne \varnothing$ so $B(a,ϵ)$ contains a point of $S$

Proof $⟸$

Suppose $a\in X$ and suppose every ball $B(a,ϵ)$ meets $S$
If $a\notin \overline{S}$ then as ${\overline{S}}^c$ is open then there is ball $B(a,ϵ)$ contained in ${\overline{S}}^c$ as $a\in {\overline{S}}^c$
So we have

$$B(a,ϵ)\subseteq {\overline{S}}^c\subseteq S^c$$

as $S\subseteq \overline{S}$ then ${\overline{S}}^c\subseteq S^c$
This contradicts assumption hence $a\in \overline{S}$

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Interior Points and Convergent Sequences corollary

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

$$a\in \overline{S}⟺\exists \text{ sequence }(x_n{)}_{n=1}^{\infty }\text{ of elements of }S\text{ with }\underset{n\to \infty }{\lim }{x}_n=a$$

In particular

$$S\text{ is closed }⟺\text{ limit of every convergent sequence }(x_n{)}_{n=1}^{\infty }\text{ of elements of }S\text{ lies in }S$$

Proof

Suppose $a\in \overline{S}$
By Interior points and balls then

$$\text{every ball }B\left(a,\frac{1}{n}\right)\text{ contains a point of }S$$

So pick sequence $(x_n{)}_{n=1}^{\infty }$ with $x_n\in B\left(a,\frac{1}{n}\right)\cap S$
Clearly also having

$$\underset{n\to \infty }{\lim }{x}_n=a$$

Conversely suppose that ${\lim }_{n\to \infty }{x}_n=a$ with $x_n\in S$
If $a\notin \overline{S}$ then by Interior points and balls there exists ball $B(a,ϵ)$ not meeting $S$
For large $n$ then

$$d(x_n,a)<ϵ$$

So $x_n\in S\cap B(a,ϵ)$ is a contradiction hence $a\in \overline{S}$

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6.2 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

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

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Set of Limit Points

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

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

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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

$$B(a,ϵ)\cap S=\{a\}$$

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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

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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 $a\in L(S)$

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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

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