14 - Interiors, Closures, Boundary, Dense

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

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

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

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

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

Dense

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

$S$ is dense if

$$\overline{S}=X$$

Alternate Definition

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

Property of Open Sets and Interiors

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

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

Property of Closed Sets and Closures

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

$$S\text{ is closed}⟹S=\overline{S}$$

Interior Points

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

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

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 $a\in \overline{S}$

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 $a\in \overline{S}$