Interior, Boundary and Exterior#

Every subset \(S\) of a topological space \((X, \tau)\) divides \(X\) into three pairwise disjoint subsets whose union is \(X\).

Definition: Interior Point

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

We say that \(p \in X\) is an interior point of \(S\) if it has a neighborhood contained in \(S\).

Definition: Topological Interior

The (topological) interior of \(S\) is the set of all its interior points:

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

Notation

\[\mathring S \qquad S^\circ \qquad \operatorname{int} S \qquad \operatorname{int}_X S\]

Definition: Boundary Point

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

We say that \(p \in X\) is a boundary point of \(S\) if every neighborhood of \(p\) intersects both \(S\) and and its complement \(X \setminus S\).

Definition: Topological Boundary

The (topological) boundary of \(S\) is the set of all its boundary points.

Notation

\[\partial S \qquad \partial_X S \qquad \operatorname{Bd}_X S\]

Definition: Exterior Point

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

We say that \(p \in X\) is an exterior point of \(S\) if it has a neighborhood which contains no points of \(S\).

Definition: Topological Exterior

The (topological) exterior of \(S\) is the set of all its exterior points:

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

Notation

\[\operatorname{ext} S \qquad \operatorname{Ext} S\]

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:

\[X = \mathop{\operatorname{int}} S \cup \partial S \cup \mathop{\operatorname{ext}} S\]

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 \(S\).

\[\operatorname{int} S = \bigcup\{U \in \tau \mid U \subseteq S \}\]

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 \(S\):

\[\operatorname{int} S \subseteq S\]

Proof

Suppose that \(\operatorname{int} S\) is not a subset of \(S\). Then there must exist some \(s \in \operatorname{int} S\) such that \(s \notin S\). Since \(s \in \operatorname{int} S\), 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 \(\iff\) 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.

\[S = \operatorname{int} S\]

Proof

We need to prove two things:

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

Proof of (I):

Suppose \(S\) is open. Recall the definition of the interior \(\operatorname{int} S\):

\[\operatorname{int} S \overset{\text{def}}{=} \bigcup \{O \subseteq S \mid O \text{ is open}\}\]

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

Proof of (II):

Suppose that \(S = \operatorname{int} S\). Since the interior \(\operatorname{int} S\) 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 \(S\):

\[\operatorname{ext} S = \bigcup\{U \in \tau \mid U \cap S = \varnothing\}\]

Proof

TODO