Interior, Boundary, Exterior

Interior, Boundary and Exterior

Definition: Isolated Point

Let $(X,\tau )$ be a topological space and let $S$ be a subset of $X$.

A point $p\in S$ is an isolated point of $S$ if there is a neighborhood $N$ of $p$ which contains no points of $S$ other than $p$:

$$N\cap S=\{p\}$$

Definition: Accumulation Point

Let $(X,\tau )$ be a topological space and let $S$ be a subset of $X$.

A point $p\in X$ is an accumulation point / limit point / cluster point of $S$ if every neighborhood of $p$ contains a point of $S$ which is different from $p$:

$$\forall N(p),\exists p^{\prime }\in S:p^{\prime }\ne p$$

Theorem: Accumulation Points and Isolated Points

Let $(X,\tau )$ be a topological space and let $S$ be a subset of $X$.

Every point $p\in S$ is either an isolated point or an accumulation point of $S$.

PROOF

TODO

Theorem: Closedness and Accumulation Points

Let $(X,\tau )$ be a topological space.

A subset $S\subseteq X$ is closed if and only if it contains all of its accumulation points.

PROOF

TODO

Every subset $S$ of a topological space $(X,\tau )$ divides $X$ into three regions such that $X$ is always equal to the union of those regions.

Definition: Interior of a Set

Let $(X,\tau )$ be a topological space $(X,\tau )$ and let $S$ be a subset of $X$.

A point $p\in X$ is an interior point of $S$ if it has a neighborhood $N(p)$ such that $N(p)\subseteq S$.

The interior of $S$ is the set of all interior points of $S$.

$$\{p\in X\mid \exists N(p):N(p)\subseteq S\}$$

NOTATION

$$\mathring{S}{S}^{\circ }\mathrm{int}S{\mathrm{int}}_{X}S$$

Definition: Boundary

Let $(X,\tau )$ be a topological space $(X,\tau )$ and let $S$ be a subset of $X$.

A point $p\in X$ is a boundary point of $S$ if every neighborhood of $p$ contains at least one other point of $S$ and at least one other point of the complement $X\setminus S$.

The boundary of $S$ is the set of all boundary points of $S$.

NOTATION

$$\partial S{\partial }_{X}S{\mathrm{Bd}}_{X}S$$

Definition: Exterior of a Set

Let $(X,\tau )$ be a topological space $(X,\tau )$ and let $S$ be a subset of $X$.

A point $p\in X$ is an exterior point of $S$ if it has a neighborhood which contains no points of $S$.

The exterior of $S$ is the set of all boundary points of $S$.

$$\{p\in X\mid \exists N(p):N(p)\cap S=\varnothing \}$$

NOTATION

$$\mathrm{ext}S\mathrm{Ext}S$$

Theorem: Union of Interior, Boundary and Exterior

Let $(X,\tau )$ be a topological space.

If $S$ is any subset of $X$, then $X$ is the union of $S$‘s interior, boundary and exterior:

$$X=\mathrm{int}S\cup \partial S\cup \mathrm{ext}S$$

PROOF

TODO

Theorem: Interior via Open Sets

Let $(X,\tau )$ be a topological space and let $S$ be a subset of $X$.

The interior of $S$ is the union of all open sets contained in $S$.

$$\mathrm{int}S=\bigcup \{U\in \tau \mid U\subseteq S\}$$

PROOF

TODO

Theorem: Interior is a Subset

Let $(X,\tau )$ be a topological space and let $S$ be a subset of $X$.

The interior of $S$ is a subset of $S$:

$$\mathrm{int}S\subseteq S$$

PROOF

Suppose that $\mathrm{int}S$ is not a subset of $S$. Then there must exist some $s\in \mathrm{int}S$ such that $s\notin S$. Since $s\in \mathrm{int}S$, there must exist some open set $O$ such that $s\in O$ and $O\subseteq S$. However, $s\notin S$ implies that $O$ is not a subset of $S$, which is a contradiction.

Theorem: Openness $⟺$ Interior

Let $(X,\tau )$ be a topological space.

A subset $S\subseteq X$ is open if and only if it is equal to its own interior.

$$S=\mathrm{int}S$$

PROOF

We need to prove two things:

• (I) If $S$ is open, then $S=\mathrm{int}S$.
• (II) If $S=\mathrm{int}S$, then $S$ is open.

Proof of (I):

Suppose $S$ is open. Recall the definition of the interior $\mathrm{int}S$:

$$\mathrm{int}S\stackrel{\text{def}}{=}\bigcup \{O\subseteq S\mid O\text{ is open}\}$$

Since $S\subseteq S$ and $S$ is open, we know that $S\in \{O\subseteq S\mid O\text{ is open}\}$ and thus $S\subseteq \mathrm{int}S$. However, the interior is a subset of $S$. Since $S\subseteq \mathrm{int}S$ and $\mathrm{int}S\subseteq S$, we know deduce that $S=\mathrm{int}S$.

Proof of (II):

Suppose that $S=\mathrm{int}S$. Since the interior $\mathrm{int}S$ is a union of open sets, it is itself open. Therefore, $S$ is open.

Theorem: Exterior via Open Sets

Let $(X,\tau )$ be a topological space and let $S$ be a subset of $X$.

The exterior of $S$ is the union of all open sets which are disjoint from $S$.

$$\mathrm{ext}S=\bigcup \{U\in \tau \mid U\cap S=\varnothing \}$$

PROOF

TODO

Closure

Every subset $S$ of a topological space $(X,\tau )$ may include none, some or all of its boundary points.

Definition: Closure of a Set

Let $(X,\tau )$ be a topological space $(X,\tau )$ and let $S$ be a subset of $X$.

The closure of $S$ is the union of its interior and its boundary.

$$\mathrm{int}S\cup \partial S$$

NOTATION

$$\overline{S}$$

Theorem: Closedness of Closure

Let $(X,\tau )$ be a topological space $(X,\tau )$.

The closure of any subset $S\subseteq X$ is closed.

PROOF

TODO

Theorem: Closure is a Superset

Let $(X,\tau )$ be a topological space.

Every subset $S\subseteq X$ is a subset of its own closure.

$$S\subseteq \overline{S}$$

PROOF

TODO

Theorem: Closedness $⟺$ Closure

Let $(X,\tau )$ be a topological space.

A subset $C\subseteq X$ is closed if and only if it is equal to its own closure.

PROOF

We have to prove two things:

• (I) If $C$ is closed, then $C=\overline{C}$.
• (II) If $C=\overline{S}$, then $C$ is closed.

Proof of (I):

TODO

Proof of (II):

TODO

Theorem: Closure of a Closure

Let $(X,\tau )$ be a topological space and let $S$ be a subset of $X$.

The closure of the closure of $S$ is still the closure of $S$.

$$\overline{\overline{S}}=\overline{S}$$

PROOF

TODO

Theorem: Closure of a Union

Let $(X,\tau )$ be a topological space and let $A$ and $B$ be two arbitrary subsets of $X$.

The closure of the union of $A$ and $B$ is the union of the closures of $A$ and $B$.

$$\overline{A\cup B}=\overline{A}\cup \overline{B}$$

PROOF

TODO

Theorem: Closure of the Empty Set

Let $(X,\tau )$ be a topological space

The closure of the empty set is itself.

$$\overline{\varnothing }=\varnothing$$

PROOF

TODO