Closed Sets, Closure, Interior, and Boundary

Definition. A subset $F \subseteq X$ is closed if its complement $X \setminus F$ is open.

Proposition 3.1. Closed sets satisfy:

1. $\emptyset$ and $X$ are closed.
2. Any intersection of closed sets is closed.
3. Any finite union of closed sets is closed.

Example 3.1. In $\mathbb{R}$ with the standard topology, $[a, b]$ is closed since $\mathbb{R} \setminus [a, b] = (-\infty, a) \cup (b, \infty)$ is open.

Example 3.2. Sets can be both open and closed (clopen). In the discrete topology every set is clopen. In any topological space, $\emptyset$ and $X$ are clopen.

Example 3.3. Sets can be neither open nor closed. In $\mathbb{R}$, $[0, 1)$ is neither open nor closed.

Definition. The closure of $A \subseteq X$, denoted $\overline{A}$, is the smallest closed set containing $A$:

$\overline{A} = \bigcap \{F : F \text{ is closed and } A \subseteq F\}.$

Equivalently, $x \in \overline{A}$ if and only if every open set containing $x$ intersects $A$.

Example 3.4. In $\mathbb{R}$: $\overline{(0, 1)} = [0, 1]$, $\overline{\mathbb{Q}} = \mathbb{R}$.

Example 3.5. In $\mathbb{R}^2$ with the standard topology, the closure of the open unit disc $\{(x, y) : x^2 + y^2 < 1\}$ is the closed unit disc $\{(x, y) : x^2 + y^2 \leq 1\}$.

Definition. The interior of $A \subseteq X$, denoted $\mathring{A}$ or $\operatorname{int}(A)$, is the largest open set contained in $A$:

$\operatorname{int}(A) = \bigcup \{U : U \text{ is open and } U \subseteq A\}.$

Equivalently, $x \in \operatorname{int}(A)$ if and only if there exists an open set $U$ with $x \in U \subseteq A$.

Example 3.6. In $\mathbb{R}$: $\operatorname{int}([0, 1]) = (0, 1)$, $\operatorname{int}(\mathbb{Q}) = \emptyset$.

Definition. The boundary of $A \subseteq X$, denoted $\partial A$, is:

$\partial A = \overline{A} \cap \overline{X \setminus A} = \overline{A} \setminus \operatorname{int}(A).$

Example 3.7. In $\mathbb{R}$: $\partial(0, 1) = \{0, 1\}$, $\partial\mathbb{Q} = \mathbb{R}$, $\partial\emptyset = \emptyset$.

Definition. A subset $A \subseteq X$ is dense in $X$ if $\overline{A} = X$.

Example 3.8. $\mathbb{Q}$ is dense in $\mathbb{R}$ with the standard topology.

Example 3.9. In the cofinite topology on an infinite set $X$, every infinite subset is dense.

Proposition 3.2. $A$ is dense in $X$ if and only if every nonempty open set intersects $A$.

1. Closed is not the opposite of open: A set can be both open and closed (clopen), or neither. In $\mathbb{R}$ with the standard topology, $\emptyset$ and $\mathbb{R}$ are clopen; $[0,1)$ is neither.
2. Closure depends on the ambient space: $(0,1)$ has closure $[0,1]$ in $\mathbb{R}$, but in $\mathbb{R}$ with the lower-limit topology, the closure includes additional limit points. The closure operation is relative to the topology.
3. Boundary points need not belong to the set: $\partial(0,1) = \{0,1\}$ in $\mathbb{R}$, but neither 0 nor 1 is in $(0,1)$. The boundary of an open set is always contained in its complement.
4. Dense sets can have empty interior: $\mathbb{Q}$ is dense in $\mathbb{R}$ yet has empty interior. A dense set has full closure but may be “full of holes” topologically.
5. Finite union vs infinite intersection: Closed sets are only guaranteed closed under finite unions. An infinite union of closed sets (e.g., $\bigcup_{n=1}^\infty [1/n, 1]$) may not be closed.

Problem. Let $A = (0,1] \times \{0\} \subset \mathbb{R}^2$. Find $\overline{A}$, $\operatorname{int}(A)$, and $\partial A$ in $\mathbb{R}^2$ with the standard topology.

Solution. $A$ is a line segment on the $x$-axis, open at 0 and closed at 1.

• $\overline{A} = [0,1] \times \{0\}$: The point $(0,0)$ is a limit point because every neighbourhood contains points of $A$.
• $\operatorname{int}(A) = \emptyset$: No open ball in $\mathbb{R}^2$ is contained in a line segment.
• $\partial A = \overline{A} = [0,1] \times \{0\}$: Every point of $A$ is also a boundary point since every neighbourhood contains points both in $A$ and $A^c$.

This illustrates that a set with empty interior is entirely contained in its boundary.

• Continuous functions on dense subsets: If two continuous functions $f, g : X \to Y$ agree on a dense subset $D \subset X$, then $f = g$ everywhere. This is fundamental in analysis for extending functions.
• Baire category theorem: Complete metric spaces cannot be expressed as a countable union of nowhere dense sets. This theorem underlies proofs of existence of continuous nowhere differentiable functions.
• Closure in function spaces: The Stone—Weierstrass theorem characterises dense subalgebras of $C(X)$, enabling polynomial approximation of continuous functions.
• Manifold boundaries: The topological boundary of a manifold with boundary is distinct from its manifold boundary. For an $n$-manifold with boundary, $\partial M$ as a topological space is an $(n-1)$-manifold without boundary.

• Analysis: The closure operation is essential for defining compactness (every open cover has a finite subcover) and the Heine—Borel theorem.
• Algebraic topology: The boundary operator $\partial$ in singular homology is defined using topological boundaries of simplices.
• Functional analysis: The closure of a subspace in a normed space is central to the Hahn—Banach theorem and the definition of the dual space.
• Differential geometry: The interior and boundary of a manifold with boundary are defined analogously, using charts to $\mathbb{R}^n_+$.

Problem. Show that the set of polynomials $\mathcal{P}[0,1]$ is dense in $C([0,1], \mathbb{R})$ (continuous functions on $[0,1]$) under the supremum norm.

Solution. By the Stone—Weierstrass theorem, any subalgebra of $C([0,1])$ that separates points and contains constant functions is dense. The polynomials clearly form a subalgebra, contain constants, and separate points (the polynomial $p(x) = x$ distinguishes $x_1 \neq x_2$). Therefore $\overline{\mathcal{P}[0,1]} = C([0,1])$.

In the language of closure: the closure of the polynomials in the sup-norm topology is the entire space of continuous functions. Every continuous function on $[0,1]$ can be approximated uniformly by a sequence of polynomials (Weierstrass approximation theorem).