The Timpson Lyceum · General Topology (Point Set Topology)

Interior and Closure of Sets – Lecture 10 (Topology)

Overview

This lecture introduces the two fundamental constructions you can perform on any subset of a topological space: the interior and the closure. The interior is the largest open set sitting inside a set, and the closure is the smallest closed set containing it; every subset of a topological space has both, no matter how badly behaved it is. After defining them precisely as unions and intersections, we prove a family of basic theorems (monotonicity, the open/closed characterizations, the complement duality laws, and behavior under unions and intersections) and establish the two neighborhood-based "membership" criteria that the rest of the course leans on. We close with worked computations across several topologies on $\mathbb{R}$ and the definition of a dense subset.

Key Concepts

Recap — open and closed sets. A topology $\tau$ on $X$ satisfies: $\varnothing, X \in \tau$ ; any finite intersection of open sets is open; any arbitrary union of open sets is open. A set is closed iff its complement is open. By De Morgan's laws ( $(\alpha \cap \beta)^c = \alpha^c \cup \beta^c$ ), complementing flips intersections to unions: a finite union of closed sets is closed, and an arbitrary intersection of closed sets is closed.

Interior. Let $A \subseteq X$ . The interior of $A$ , written $\operatorname{int} A$ or $A^{\circ}$ , is the union of all open sets contained in $A$ : $A^{\circ} = \bigcup \{\, U : U \text{ open},\ U \subseteq A \,\}.$ It is open (an arbitrary union of open sets) and is the largest open set contained in $A$ .

Closure. The closure of $A$ , written $\operatorname{cl} A$ or $\overline{A}$ , is the intersection of all closed sets containing $A$ : $\overline{A} = \bigcap \{\, C : C \text{ closed},\ A \subseteq C \,\}.$ It is closed (an arbitrary intersection of closed sets) and is the smallest closed set containing $A$ .

The nesting relation. For every set, $A^{\circ} \subseteq A \subseteq \overline{A}$ .

Basic theorems.

If $U$ is open and $U \subseteq A$ , then $U \subseteq A^{\circ}$ (open subsets land inside the interior).
If $C$ is closed and $A \subseteq C$ , then $\overline{A} \subseteq C$ (the closure fits inside any closed superset).
Monotonicity: if $A \subseteq B$ then $A^{\circ} \subseteq B^{\circ}$ and $\overline{A} \subseteq \overline{B}$ .
Open characterization: $A$ is open $\iff A = A^{\circ}$ .
Closed characterization: $A$ is closed $\iff A = \overline{A}$ .

Neighborhood criteria. A neighborhood of $y$ is any open set $U_y$ containing $y$ .

$y \in A^{\circ} \iff$ there exists a neighborhood $U_y \subseteq A$ .
Theorem 2.5 (the workhorse): $y \in \overline{A} \iff$ every neighborhood of $y$ meets $A$ , i.e. $U_y \cap A \neq \varnothing$ for all $U_y$ . Equivalently (contrapositive): $y \notin \overline{A} \iff$ some neighborhood $U_y$ satisfies $U_y \cap A = \varnothing$ .

Complement duality laws. For $A \subseteq X$ : $(A^c)^{\circ} = \left(\overline{A}\right)^c, \qquad \overline{A^c} = \left(A^{\circ}\right)^c.$ Pushing a complement into a closure turns it into an interior, and vice versa.

Unions and intersections.

$A^{\circ} \cup B^{\circ} \subseteq (A \cup B)^{\circ}$ (subset, not equality in general).
$A^{\circ} \cap B^{\circ} = (A \cap B)^{\circ}$ (equality).

Dense sets. $A \subseteq X$ is dense in $X$ iff $\overline{A} = X$ . In the standard topology on $\mathbb{R}$ , both $\mathbb{Q}$ and the irrationals are dense: their closures are all of $\mathbb{R}$ .

Sample computations.

Standard topology on $\mathbb{R}$ , $A = [0,1)$ : $A^{\circ} = (0,1)$ , $\overline{A} = [0,1]$ . Neither open nor closed.
$\mathbb{Q}$ in standard $\mathbb{R}$ : $\mathbb{Q}^{\circ} = \varnothing$ (every interval contains irrationals), $\overline{\mathbb{Q}} = \mathbb{R}$ .
Lower-limit topology, basis sets $[a,b)$ : $[a,b)$ is clopen, so $A^{\circ} = A = \overline{A}$ .
Finite-complement topology on $\mathbb{R}$ (open $\iff$ complement finite): for an interval-type set $A$ , $A^{\circ} = \varnothing$ and $\overline{A} = \mathbb{R}$ , since the only closed sets are finite sets, $\varnothing$ , and $\mathbb{R}$ .
Discrete topology (power set): every set is clopen, so $A^{\circ} = A = \overline{A}$ for all $A$ .

Worked Example

Prove the duality law $(A^c)^{\circ} = \left(\overline{A}\right)^c$ by showing each side is a subset of the other (set equality).

\begin{aligned} &(\subseteq)\quad x \in (A^c)^{\circ} \;\Rightarrow\; \exists\,U_x \text{ open},\ U_x \subseteq A^c \;\Rightarrow\; U_x \cap A = \varnothing \\ &\phantom{(\subseteq)\quad} \;\overset{\text{Thm 2.5}}{\Longrightarrow}\; x \notin \overline{A} \;\Rightarrow\; x \in \left(\overline{A}\right)^c \\ &(\supseteq)\quad x \in \left(\overline{A}\right)^c \;\Rightarrow\; x \notin \overline{A} \;\overset{\text{Thm 2.5}}{\Longrightarrow}\; \exists\,U_x \text{ open},\ U_x \cap A = \varnothing \\ &\phantom{(\supseteq)\quad} \;\Rightarrow\; U_x \subseteq A^c \;\Rightarrow\; x \in U_x \subseteq (A^c)^{\circ} \\ &\therefore\quad (A^c)^{\circ} = \left(\overline{A}\right)^c \end{aligned}

The forward direction uses that an open subset of $A^c$ is a witnessing neighborhood disjoint from $A$ , which by Theorem 2.5 keeps $x$ out of $\overline{A}$ . The reverse direction runs the same equivalence backward: a point outside the closure has a neighborhood missing $A$ , hence sitting inside $A^c$ , hence inside its interior. Since each side contains the other, they are equal.

Summary

Every subset $A$ of a topological space has an interior $A^{\circ}$ (largest open set inside $A$ , a union of open subsets) and a closure $\overline{A}$ (smallest closed set containing $A$ , an intersection of closed supersets), with $A^{\circ} \subseteq A \subseteq \overline{A}$ .
$A$ is open $\iff A = A^{\circ}$ ; $A$ is closed $\iff A = \overline{A}$ . Both operations are monotone under $\subseteq$ .
Theorem 2.5 is the key tool: $y \in \overline{A}$ exactly when every neighborhood of $y$ intersects $A$ ; dually, $y \in A^{\circ}$ exactly when some neighborhood of $y$ lies entirely in $A$ .
The duality laws $(A^c)^{\circ} = (\overline{A})^c$ and $\overline{A^c} = (A^{\circ})^c$ let you convert between interior and closure by moving complements.
Interior commutes with finite intersection ( $A^{\circ} \cap B^{\circ} = (A \cap B)^{\circ}$ ) but only gives containment for union ( $A^{\circ} \cup B^{\circ} \subseteq (A \cup B)^{\circ}$ ); e.g. $(1,2) \cup (2,3)$ misses the point $2$ that $(1,3)^{\circ}$ contains.
A set is dense when its closure is the whole space; $\mathbb{Q}$ and the irrationals are both dense in standard $\mathbb{R}$ , while interior/closure depend heavily on the topology (clopen everywhere in the discrete and lower-limit cases, collapsing to $\varnothing$ and $\mathbb{R}$ in the finite-complement topology).