Given a subset $A$ of a [topological space](/page/Topology) $X$, which points of $A$ are "safely inside" — surrounded by other points of $A$ in every direction — and which sit precariously on the edge? This question is not merely taxonomic. Across analysis, the distinction between interior points and boundary points governs what operations are permissible. A [continuous function](/page/Continuity) on a [closed set](/page/Closed%20Set) $F$ can be extended to the ambient space (Tietze), but the extension depends on the geometry of $F$ near its boundary, not on its interior. A [Sobolev function](/page/Sobolev%20Space) satisfies a PDE in the classical sense at interior points where it is smooth, but its behaviour at the boundary requires a separate trace theory. The maximum principle for harmonic functions asserts that nonconstant solutions achieve their extrema on the boundary — making the interior the region where the solution is controlled by the data prescribed elsewhere. The **interior** of a set $A$ collects precisely those points that have "room to move" — each interior point is surrounded by an [open set](/page/Open%20Set) contained entirely within $A$. The interior is the largest open set contained in $A$, the complement of the [closure](/page/Closed%20Set) of the complement, and the set $A$ with its [boundary](/page/Boundary) removed. These three characterisations are equivalent, and each illuminates a different aspect of the concept. [example: Intervals in $\mathbb{R}$] Consider the half-open interval $A = [0, 1) \subset \mathbb{R}$ with the standard topology. Which points of $A$ are interior? For any $x \in (0, 1)$, set $\varepsilon := \min(x, 1 - x) > 0$. Then the open interval $(x - \varepsilon, x + \varepsilon)$ satisfies $(x - \varepsilon, x + \varepsilon) \subset (0, 1) \subset A$, so $x$ is an interior point. The endpoint $x = 0$ is different. Every open interval containing $0$ has the form $(-\delta, \delta)$ for some $\delta > 0$, and $(-\delta, \delta)$ contains the point $-\delta/2 \notin A$. No open set containing $0$ is contained in $A$, so $0$ is not an interior point. Therefore, the interior of $[0, 1)$ is the open interval $(0, 1)$. The left endpoint has been stripped away — it belongs to the boundary, not the interior. [/example] This example reveals a recurring pattern: the interior of a set "opens it up" by removing the boundary points. The formal definition makes this precise for arbitrary topological spaces. ## Definition In practice, one encounters the interior through two different lenses. When verifying that a specific point lies in the interior, one works locally: find an open neighbourhood of the point contained in $A$. When manipulating the interior as a set — taking complements, intersections, or images under maps — one needs a global characterisation: the interior as the largest open subset of $A$. Both perspectives are essential, and the fact that they define the same set is what makes the interior a robust concept. [definition: Interior] Let $(X, \tau)$ be a [topological space](/page/Topology) and let $A \subset X$. The **interior** of $A$, denoted $A^\circ$ or $\operatorname{int}(A)$, is the union of all open sets contained in $A$: \begin{align*} A^\circ := \bigcup \{ U \in \tau : U \subset A \}. \end{align*} A point $x \in A$ is called an **interior point** of $A$ if there exists an open set $U \in \tau$ with $x \in U \subset A$. [/definition] Since arbitrary unions of open sets are open, $A^\circ$ is itself open — it is the **largest open set contained in $A$**. The interior can also be characterised as the set of all interior points: $x \in A^\circ$ if and only if $x$ is an interior point of $A$. To see this, note that if $x \in A^\circ$, then $x$ belongs to some open $U \subset A$, making $x$ an interior point; conversely, if $x$ is an interior point, then the witnessing open set $U$ is one of the sets in the union defining $A^\circ$, so $x \in A^\circ$. [remark: Notation] The notation $A^\circ$ is standard in topology and analysis. Some authors write $\operatorname{int}(A)$, which avoids potential confusion with the connected component notation in some contexts. On Androma, we use $A^\circ$ as the primary notation for the interior. [/remark] ## The Interior in Metric Spaces In a general [topological space](/page/Topology), the definition of the interior relies on the abstract notion of open sets. In a [metric space](/page/Metric%20Space), open sets are built from open balls, and this gives a concrete, computational characterisation of interior points: a point is interior to $A$ if and only if some ball around it fits inside $A$. [definition: Interior Point in a Metric Space] Let $(X, d)$ be a [metric space](/page/Metric%20Space) and let $A \subset X$. A point $x \in A$ is an **interior point** of $A$ if there exists $\varepsilon > 0$ such that the open ball $B(x, \varepsilon) := \{y \in X : d(x, y) < \varepsilon\}$ satisfies $B(x, \varepsilon) \subset A$. [/definition] This characterisation is equivalent to the topological one because the open balls form a basis for the metric topology: every open set is a union of open balls, so the existence of an open set $U$ with $x \in U \subset A$ is equivalent to the existence of a ball $B(x, \varepsilon) \subset A$. The metric characterisation transforms the question "is $x$ interior to $A$?" into a quantitative problem: find an explicit $\varepsilon > 0$, or prove that no such $\varepsilon$ exists. The following examples illustrate this technique. [example: Interior of a Closed Interval] Let $A = [0, 1] \subset \mathbb{R}$ with the standard metric $d(x, y) = |x - y|$. **Interior points:** For $x \in (0, 1)$, set $\varepsilon := \min(x, 1 - x) > 0$. Then for any $y$ with $|y - x| < \varepsilon$, we have $y > x - \varepsilon \geq 0$ and $y < x + \varepsilon \leq 1$, so $y \in [0, 1]$. Thus $B(x, \varepsilon) \subset [0, 1]$, confirming $x \in [0, 1]^\circ$. **Boundary points:** For $x = 0$, any ball $B(0, \varepsilon) = (-\varepsilon, \varepsilon)$ contains the point $-\varepsilon/2 \notin [0, 1]$. Similarly, $B(1, \varepsilon)$ contains $1 + \varepsilon/2 \notin [0, 1]$. Neither endpoint is an interior point. Therefore $[0, 1]^\circ = (0, 1)$. [/example] The closed interval $[0, 1]$ has a large interior — removing two boundary points leaves the full open interval $(0, 1)$. The next example shows that a set can be far more pervasive in $\mathbb{R}$ yet have no interior at all, demonstrating that density and having nonempty interior are independent properties. [example: Interior of the Rationals] Let $A = \mathbb{Q} \subset \mathbb{R}$ with the standard metric. We claim $\mathbb{Q}^\circ = \varnothing$. Let $x \in \mathbb{Q}$ and let $\varepsilon > 0$ be arbitrary. By the density of the irrationals (a consequence of the uncountability of $\mathbb{R}$ and the countability of $\mathbb{Q}$), the interval $(x - \varepsilon, x + \varepsilon)$ contains an irrational number $y$. Since $y \notin \mathbb{Q}$, the ball $B(x, \varepsilon)$ is not contained in $\mathbb{Q}$. Since $\varepsilon > 0$ was arbitrary, $x$ is not an interior point. Since $x \in \mathbb{Q}$ was arbitrary, $\mathbb{Q}$ has no interior points at all: $\mathbb{Q}^\circ = \varnothing$. This is a striking result: the rationals are [dense](/page/Dense%20Subset) in $\mathbb{R}$ — every real number can be approximated by rationals — yet no rational number is an interior point of $\mathbb{Q}$. Density describes how $\mathbb{Q}$ sits relative to $\mathbb{R}$; the empty interior describes how $\mathbb{Q}$ fails to contain any open set. The two properties are independent. [/example] The previous two examples live in $\mathbb{R}$, where sets are intervals or scattered subsets. Moving to higher dimensions, the interior computation becomes a geometric argument involving the triangle inequality rather than simple interval arithmetic. [example: Interior of a Closed Ball in $\mathbb{R}^n$] Let $A = \overline{B}(x_0, r) := \{x \in \mathbb{R}^n : |x - x_0| \leq r\}$ be the closed ball of radius $r > 0$ centred at $x_0$ in $\mathbb{R}^n$ with the Euclidean metric. **Interior points:** Let $x \in A$ with $|x - x_0| < r$. Set $\varepsilon := r - |x - x_0| > 0$. For any $y \in B(x, \varepsilon)$, the triangle inequality gives: \begin{align*} |y - x_0| \leq |y - x| + |x - x_0| < \varepsilon + |x - x_0| = r. \end{align*} So $y \in B(x_0, r) \subset \overline{B}(x_0, r) = A$, and hence $B(x, \varepsilon) \subset A$. **Boundary points:** Let $x \in A$ with $|x - x_0| = r$ (i.e., $x$ lies on the sphere $\partial B(x_0, r)$). For any $\varepsilon > 0$, consider the point $y := x + \frac{\varepsilon}{2} \cdot \frac{x - x_0}{|x - x_0|}$. This point satisfies $|y - x| = \varepsilon/2 < \varepsilon$, so $y \in B(x, \varepsilon)$. But: \begin{align*} |y - x_0| = \left| x - x_0 + \frac{\varepsilon}{2} \cdot \frac{x - x_0}{|x - x_0|} \right| = |x - x_0| + \frac{\varepsilon}{2} = r + \frac{\varepsilon}{2} > r, \end{align*} so $y \notin A$. The ball $B(x, \varepsilon)$ is not contained in $A$. Therefore $\overline{B}(x_0, r)^\circ = B(x_0, r)$: the interior of the closed ball is the open ball. This confirms the geometric intuition that the "skin" of the ball — the sphere $\{x : |x - x_0| = r\}$ — is the boundary, and the "inside" is the interior. [/example] The identity $\overline{B}(x_0, r)^\circ = B(x_0, r)$ established above relies on the geometry of $\mathbb{R}^n$ — specifically, on the ability to "push" a boundary point radially outward. This identity is not a general fact about metric spaces. [remark: Closed Ball Interior in General Metric Spaces] The identity $\overline{B}(x_0, r)^\circ = B(x_0, r)$ holds in $\mathbb{R}^n$ but fails in general metric spaces. In a discrete metric space $(X, d)$ where $d(x, y) = 1$ for all $x \neq y$, the closed ball $\overline{B}(x_0, 1) = X$ (every point has distance $\leq 1$) is the entire space, while the open ball $B(x_0, 1) = \{x_0\}$ is a single point. The interior of $\overline{B}(x_0, 1) = X$ is $X$ itself (since $X$ is open), which is strictly larger than $B(x_0, 1) = \{x_0\}$ when $X$ has more than one element. [/remark] ## The Interior Operator and Its Properties The passage from a set $A$ to its interior $A^\circ$ defines an operation on the power set of $X$. This **interior operator** has four characteristic properties that mirror — in a dual sense — the four [Kuratowski closure axioms](/page/Closed%20Set) for the closure operator. Understanding these properties is important because they allow us to manipulate interiors algebraically, without returning to the definition each time. The following theorem collects the fundamental properties of the interior operator. [quotetheorem:1013] Property (1) says the interior never adds points — it can only remove them. Property (2) says that once we have extracted the interior, repeating the operation changes nothing; $A^\circ$ is already open, and the interior of an open set is itself. Property (3) is the convention that the whole space is open, so every point of $X$ is interior to $X$. Property (4) states that the interior distributes over finite intersections. Property (5) is the monotonicity that follows from (4): if $A \subset B$, then $A = A \cap B$, so $A^\circ = (A \cap B)^\circ = A^\circ \cap B^\circ \subset B^\circ$. Properties (1)--(4) are sometimes called the **dual Kuratowski axioms**. Just as a topology can be axiomatised through its closure operator (a map $A \mapsto \overline{A}$ satisfying the Kuratowski axioms), it can equivalently be axiomatised through an interior operator satisfying (1)--(4), with $\tau := \{A \subset X : A^\circ = A\}$. [explanation: Failure of Infinite Intersection] The interior operator preserves finite intersections but **not** arbitrary intersections. This is a direct reflection of the fact that finite intersections of open sets are open, while infinite intersections need not be. Consider the family of open intervals $A_n := (-1/n, 1/n)$ in $\mathbb{R}$. Each $A_n$ is open, so $A_n^\circ = A_n$. The intersection is: \begin{align*} \bigcap_{n=1}^\infty A_n = \{0\}. \end{align*} The singleton $\{0\}$ is not open in $\mathbb{R}$ (no open interval fits inside it), so: \begin{align*} \left(\bigcap_{n=1}^\infty A_n\right)^\circ = \{0\}^\circ = \varnothing \neq \bigcap_{n=1}^\infty A_n^\circ = \bigcap_{n=1}^\infty A_n = \{0\}. \end{align*} The interior of the intersection is strictly smaller than the intersection of the interiors. In general, we always have the **one-sided inclusion**: \begin{align*} \left(\bigcap_{\alpha \in I} A_\alpha\right)^\circ \subset \bigcap_{\alpha \in I} A_\alpha^\circ, \end{align*} which follows from monotonicity (property 5), since $\bigcap_\alpha A_\alpha \subset A_\beta$ for each $\beta$. The reverse inclusion fails whenever the infinite intersection "collapses" a family of open sets down to a set with no room left inside it. [/explanation] The situation for unions is asymmetric. While the interior of a finite intersection equals the intersection of the interiors, the interior of a union need not equal the union of the interiors. [explanation: Interior of Unions] For arbitrary sets $A, B \subset X$, we have: \begin{align*} A^\circ \cup B^\circ \subset (A \cup B)^\circ, \end{align*} but equality can fail. The inclusion follows from monotonicity: $A \subset A \cup B$ implies $A^\circ \subset (A \cup B)^\circ$, and similarly for $B$. For a counterexample to the reverse inclusion, take $A = \mathbb{Q}$ and $B = \mathbb{R} \setminus \mathbb{Q}$ in $\mathbb{R}$. Both sets have empty interior (by the density of irrationals in $\mathbb{R}$ and the density of rationals in $\mathbb{R}$, respectively), so $A^\circ \cup B^\circ = \varnothing$. But $A \cup B = \mathbb{R}$, which has interior $\mathbb{R}$. Thus: \begin{align*} A^\circ \cup B^\circ = \varnothing \subsetneq \mathbb{R} = (A \cup B)^\circ. \end{align*} The union of two sets with empty interior can have full interior, provided the two sets "fill in each other's gaps." [/explanation] ## Interior, Closure, and Boundary The interior $A^\circ$, the [closure](/page/Closed%20Set) $\overline{A}$, and the [boundary](/page/Boundary) $\partial A$ form a trio of operations that together describe the topological position of a set $A$ relative to the ambient space $X$. The three operations are not independent: any one can be defined in terms of the other two, and their interplay is governed by complement duality. The fundamental difficulty in understanding this relationship is that the interior and closure appear to measure different things. The closure adds all limit points to $A$ — it inflates $A$ to its "tightest closed envelope." The interior removes all boundary points from $A$ — it deflates $A$ to its "largest open core." These two operations are connected by a single principle: **the interior of $A$ is the complement of the closure of the complement**. [quotetheorem:1014] This duality is the topological manifestation of [De Morgan's laws](/theorems/622). The closure $\overline{X \setminus A}$ is the smallest closed set containing the complement of $A$. Its own complement $X \setminus \overline{X \setminus A}$ is therefore the largest open set contained in $A$ — which is precisely $A^\circ$. This relationship is the reason that the four properties of the interior operator (inclusion, idempotence, preservation of $X$, finite intersection) are exactly dual to the four Kuratowski closure axioms (extension, idempotence, preservation of $\varnothing$, finite union). Every statement about interiors translates, via complementation, into a statement about closures. The [boundary](/page/Boundary) occupies the region between the interior and the closure: [quotetheorem:1015] Part (3) decomposes the ambient space into three disjoint regions: the interior of $A$ (points "fully inside" $A$), the interior of the complement (points "fully outside" $A$), and the boundary (points on the edge, belonging to the closure of both $A$ and $X \setminus A$). This decomposition is exhaustive and exclusive — every point of $X$ falls into exactly one of the three regions. [example: The Decomposition for $\mathbb{Q} \subset \mathbb{R}$] Let $A = \mathbb{Q} \subset \mathbb{R}$. We computed above that $\mathbb{Q}^\circ = \varnothing$. By the same argument (the density of rationals among irrationals), $(\mathbb{R} \setminus \mathbb{Q})^\circ = \varnothing$. The closure $\overline{\mathbb{Q}} = \mathbb{R}$ (since every real number is the limit of a sequence of rationals). The decomposition $\mathbb{R} = \mathbb{Q}^\circ \cup \partial \mathbb{Q} \cup (\mathbb{R} \setminus \mathbb{Q})^\circ$ becomes: \begin{align*} \mathbb{R} = \varnothing \cup \partial \mathbb{Q} \cup \varnothing, \end{align*} so $\partial \mathbb{Q} = \mathbb{R}$. The boundary of $\mathbb{Q}$ is all of $\mathbb{R}$ — every real number, whether rational or irrational, sits on the boundary of the rationals. There is no region of $\mathbb{R}$ that is "purely rational" or "purely irrational." This reinforces the earlier observation that $\mathbb{Q}$ has empty interior: the rationals are so thoroughly interspersed with the irrationals that neither set contains any open set. [/example] The duality formula $A^\circ = X \setminus \overline{X \setminus A}$ is also a powerful computational tool. When computing the interior of a set is difficult directly, it is sometimes easier to compute the closure of the complement and then take the complement. [example: Interior via Complement–Closure–Complement] Let $A = \{(x, y) \in \mathbb{R}^2 : |x| + |y| \leq 1\}$ be the closed unit diamond (the closed unit ball in the $\ell^1$ norm on $\mathbb{R}^2$). We compute $A^\circ$ using the duality formula. The complement is $\mathbb{R}^2 \setminus A = \{(x, y) : |x| + |y| > 1\}$. To find $\overline{\mathbb{R}^2 \setminus A}$, we identify the limit points. Any point $(x, y)$ with $|x| + |y| > 1$ already belongs to $\mathbb{R}^2 \setminus A$. A point $(x_0, y_0)$ with $|x_0| + |y_0| = 1$ is a limit of points in $\mathbb{R}^2 \setminus A$: for instance, the sequence $(x_0, y_0) + \frac{1}{n}(x_0, y_0)/|(x_0, y_0)|$ lies in the complement for all large $n$ and converges to $(x_0, y_0)$. No point with $|x_0| + |y_0| < 1$ is a limit of points in the complement, because the ball $B((x_0, y_0), 1 - |x_0| - |y_0|)$ in the $\ell^1$ metric is contained in $A$. Therefore: \begin{align*} \overline{\mathbb{R}^2 \setminus A} = \{(x, y) : |x| + |y| \geq 1\}. \end{align*} Applying the duality formula: \begin{align*} A^\circ = \mathbb{R}^2 \setminus \overline{\mathbb{R}^2 \setminus A} = \mathbb{R}^2 \setminus \{(x, y) : |x| + |y| \geq 1\} = \{(x, y) : |x| + |y| < 1\}. \end{align*} The interior of the closed diamond is the open diamond — the boundary $\{|x| + |y| = 1\}$ is removed. This confirms the general pattern: the interior of the closed unit ball (in any norm on $\mathbb{R}^n$) is the open unit ball. [/example] ## Characterising Open Sets How does one verify that a given set is [open](/page/Open%20Set)? If the topology is defined by a metric, one must produce, for each point $x \in A$, an explicit ball $B(x, \varepsilon) \subset A$. If the topology is defined by a basis, one must exhibit $A$ as a union of basis elements. Both approaches require working with the specific generating structure of the topology. The interior provides a characterisation of openness that is independent of any particular basis or metric: a set is open if and only if it equals its own interior. [quotetheorem:1016] In one direction: if $A$ is open, then $A$ is an open set contained in itself, so $A \subset A^\circ$; combined with $A^\circ \subset A$ (property 1 of the interior operator), we get $A = A^\circ$. In the other direction: if $A = A^\circ$, then $A$ equals a union of open sets, hence is open. This characterisation converts the question "is $A$ open?" into the question "does $A$ equal its interior?" In practice, this is often the most efficient way to verify that a set is open in a topology defined by a basis: rather than exhibiting $A$ as a union of basis elements, one shows that every point of $A$ is an interior point. The dual statement is equally important: a set $F$ is [closed](/page/Closed%20Set) if and only if $F = \overline{F}$, which by duality is equivalent to $(X \setminus F)^\circ = X \setminus F$. [example: Verifying Openness via the Interior] Consider the set $A = \{(x, y) \in \mathbb{R}^2 : x^2 + y^2 < 1 \text{ and } y > 0\}$ — the upper half of the open unit disc. To verify that $A$ is open, we show every point of $A$ is an interior point. Let $(x_0, y_0) \in A$, so $x_0^2 + y_0^2 < 1$ and $y_0 > 0$. Set: \begin{align*} \varepsilon := \min\left(\frac{1 - (x_0^2 + y_0^2)}{2(|x_0| + |y_0| + 1)},\; \frac{y_0}{2}\right) > 0. \end{align*} We need to verify $B((x_0, y_0), \varepsilon) \subset A$. For any $(x, y) \in B((x_0, y_0), \varepsilon)$: First, $|y - y_0| < \varepsilon \leq y_0/2$, so $y > y_0 - y_0/2 = y_0/2 > 0$. Second, writing $x = x_0 + h$ and $y = y_0 + k$ with $h^2 + k^2 < \varepsilon^2$: \begin{align*} x^2 + y^2 &= x_0^2 + 2x_0 h + h^2 + y_0^2 + 2y_0 k + k^2 \\ &\leq x_0^2 + y_0^2 + 2(|x_0| + |y_0|)\varepsilon + 2\varepsilon^2 \\ &\leq x_0^2 + y_0^2 + 2(|x_0| + |y_0| + 1)\varepsilon \\ &\leq x_0^2 + y_0^2 + (1 - x_0^2 - y_0^2) = 1. \end{align*} (In the third line, we used $\varepsilon < 1$ so $2\varepsilon^2 < 2\varepsilon$.) Hence $(x, y) \in A$, confirming that $A = A^\circ$ and $A$ is open. [/example] ## Interior and Continuity The interior provides a natural language for expressing [continuity](/page/Continuity) in terms of preimages. In a first course, continuity is often defined via epsilon-delta conditions (in metric spaces) or via the preimage condition (in topological spaces). The interior offers a third perspective that is sometimes more natural, particularly when dealing with continuity at a single point. The standard topological definition requires that the preimage of every [open set](/page/Open%20Set) is open: $f^{-1}(V)$ is open whenever $V$ is open. Using the interior, we can reformulate this without explicitly mentioning open sets on the domain side. [quotetheorem:1017] Condition (2) says that the preimage operation "respects interiors": if $y = f(x)$ is interior to a set $A$ in the codomain, then $x$ is interior to $f^{-1}(A)$ in the domain. The equivalence with (1) is a consequence of the fact that as $A$ ranges over all subsets of $Y$, the sets $A^\circ$ range over all open subsets of $Y$. Taking $A = V$ for an open set $V$ gives $V^\circ = V$, so condition (2) requires $f^{-1}(V) \subset (f^{-1}(V))^\circ$; since the reverse inclusion always holds, this forces $f^{-1}(V) = (f^{-1}(V))^\circ$, i.e., $f^{-1}(V)$ is open. Note that the corresponding "forward" statement — $f(A^\circ) \subset (f(A))^\circ$ for all $A \subset X$ — characterises **open maps**, not continuous maps. An open map sends open sets to open sets, which is a different (and independent) condition from continuity. The interior also gives a clean formulation of continuity at a single point. In metric spaces, $f: X \to Y$ is continuous at $x_0$ if for every $\varepsilon > 0$, there exists $\delta > 0$ with $f(B(x_0, \delta)) \subset B(f(x_0), \varepsilon)$. In terms of the interior: [definition: Continuity at a Point via Interior] Let $(X, \tau_X)$ and $(Y, \tau_Y)$ be [topological spaces](/page/Topology) and let \begin{align*} f: X &\to Y \end{align*} be a map. Let $x_0 \in X$. Then $f$ is **continuous at $x_0$** if and only if for every subset $V \subset Y$ with $f(x_0) \in V^\circ$, we have $x_0 \in (f^{-1}(V))^\circ$. [/definition] This reformulation says: if $f(x_0)$ is interior to $V$ in the codomain, then $x_0$ is interior to $f^{-1}(V)$ in the domain. In metric spaces, this recovers the epsilon-delta condition: taking $V = B(f(x_0), \varepsilon)$ (which is open, hence equal to its interior), the condition requires $x_0 \in (f^{-1}(B(f(x_0), \varepsilon)))^\circ$, meaning there exists a ball $B(x_0, \delta)$ mapped into $B(f(x_0), \varepsilon)$. ## Nowhere Dense Sets and the Baire Category The concept of a set with **empty interior** — a set so "thin" that it contains no open set — plays a central role in the descriptive theory of topological spaces. When the closure of a set also has empty interior, the set is called **nowhere dense**, and collections of nowhere dense sets provide the basis for [Baire category](/theorems/630) theory. The motivating question is: how "large" can a set be while still being topologically negligible? A finite set in $\mathbb{R}$ is sparse — it has empty interior and its closure (itself) also has empty interior. But can a dense set be topologically negligible? Can a set be "everywhere present" yet still contain no open set? [definition: Nowhere Dense Set] Let $(X, \tau)$ be a [topological space](/page/Topology) and let $A \subset X$. The set $A$ is **nowhere dense** if the interior of its closure is empty: \begin{align*} (\overline{A})^\circ = \varnothing. \end{align*} Equivalently, $A$ is nowhere dense if and only if for every nonempty open set $U \subset X$, there exists a nonempty open set $V \subset U$ with $V \cap A = \varnothing$. [/definition] The equivalence of the two conditions captures the meaning of "nowhere dense" intuitively: no matter where you look in $X$ (no matter which open set $U$ you choose), you can find a region (a nonempty open $V \subset U$) that completely avoids $A$. The set $A$ is dense nowhere — it fails to approximate every point in any open region. The distinction between "empty interior" and "nowhere dense" is subtle but important. A set has empty interior if it contains no open set. A set is nowhere dense if even its closure contains no open set. The set $\mathbb{Q} \subset \mathbb{R}$ has empty interior ($\mathbb{Q}^\circ = \varnothing$), but it is **not** nowhere dense because $\overline{\mathbb{Q}} = \mathbb{R}$, and $\mathbb{R}^\circ = \mathbb{R} \neq \varnothing$. The rationals are too densely distributed to be nowhere dense — their closure fills the entire line. [example: Nowhere Dense Sets in $\mathbb{R}$] **The integers.** The set $\mathbb{Z} \subset \mathbb{R}$ is nowhere dense. Its closure is $\overline{\mathbb{Z}} = \mathbb{Z}$ (it is already closed, since the complement $\mathbb{R} \setminus \mathbb{Z} = \bigcup_{n \in \mathbb{Z}} (n, n+1)$ is a union of open intervals). The interior $\mathbb{Z}^\circ = \varnothing$ because no interval $(a, b)$ with $a < b$ is contained in $\mathbb{Z}$. Therefore $(\overline{\mathbb{Z}})^\circ = \varnothing$, confirming that $\mathbb{Z}$ is nowhere dense. **The Cantor set.** The standard Cantor set $C \subset [0, 1]$ is closed (it is an intersection of closed sets) and has empty interior. To see that $C^\circ = \varnothing$, observe that every interval of length greater than $3^{-n}$ is split during the $n$-th step of the construction, so $C$ contains no interval of positive length. Since $C$ is closed, $\overline{C} = C$, so $(\overline{C})^\circ = C^\circ = \varnothing$. The Cantor set is nowhere dense. Despite being nowhere dense, the Cantor set is uncountable (it bijects with $\{0, 2\}^{\mathbb{N}}$ via the ternary expansion). This shows that nowhere dense sets can be "large" in a cardinality sense while being "thin" topologically. [/example] The connection to the [Baire Category Theorem](/theorems/630) is the following: in a complete [metric space](/page/Metric%20Space), the space cannot be expressed as a countable union of nowhere dense sets. Equivalently, the complement of a countable union of nowhere dense sets is dense. Sets that can be written as countable unions of nowhere dense sets are called **meagre** (or **of the first category**), and the Baire Category Theorem asserts that a complete metric space is **not** meagre. [quotetheorem:630] This is the contrapositive of the Baire Category Theorem in its "category form." It provides a powerful tool for existence proofs: to show that a set $A$ is nonempty (or even "large"), express its complement as a countable union of closed sets and show each has empty interior. The theorem then forces $A$ to be dense. The classical application is the proof that the set of continuous nowhere-differentiable functions is dense in $C([0,1])$ with the supremum norm. The set of functions that are differentiable at even one point can be expressed as a countable union of closed sets with empty interior — and so, by the Baire Category Theorem, "most" continuous functions are nowhere differentiable. ## Standard Techniques for Computing Interiors Working with interiors in practice requires a toolkit of standard methods. The following techniques appear repeatedly in topology and analysis. ### Direct Verification via Open Balls In a [metric space](/page/Metric%20Space), the most elementary approach to computing $A^\circ$ is to characterise the set of all $x \in A$ for which an explicit $\varepsilon > 0$ exists with $B(x, \varepsilon) \subset A$. This requires two steps: 1. **Show inclusion:** For each $x$ in the candidate set, construct an explicit $\varepsilon = \varepsilon(x) > 0$ with $B(x, \varepsilon) \subset A$. 2. **Show exclusion:** For each $x \in A$ not in the candidate set, show that for every $\varepsilon > 0$, the ball $B(x, \varepsilon)$ contains a point outside $A$. The $\varepsilon$ in step 1 typically depends on the distance from $x$ to the boundary of $A$. In the examples above, we used $\varepsilon = \min(x, 1 - x)$ for the interval $[0, 1]$ and $\varepsilon = r - |x - x_0|$ for the closed ball $\overline{B}(x_0, r)$. ### Complement–Closure–Complement When the set $A$ is defined implicitly (as a preimage, a level set, or a Boolean combination of other sets), it is often easier to compute the interior via the duality formula: \begin{align*} A^\circ = X \setminus \overline{X \setminus A}. \end{align*} This converts the problem of finding the largest open subset of $A$ into the problem of finding the smallest closed superset of the complement — which may be more tractable if the closure is easier to compute. ### Preimage of Open Sets If $A = f^{-1}(V)$ for a continuous map $f: X \to Y$ and an open set $V \subset Y$, then $A$ is open, so $A^\circ = A$. More generally, if $A = f^{-1}(B)$ for an arbitrary set $B \subset Y$, then: \begin{align*} f^{-1}(B^\circ) \subset (f^{-1}(B))^\circ = A^\circ, \end{align*} by continuity of $f$. This provides a lower bound on the interior: the preimage of the interior of $B$ is contained in the interior of the preimage. [example: Interior of a Sublevel Set] Let $f: \mathbb{R}^n \to \mathbb{R}$ be a [continuous function](/page/Continuity) and let $A = \{x \in \mathbb{R}^n : f(x) \leq 0\} = f^{-1}((-\infty, 0])$. What is $A^\circ$? Since $f$ is continuous, the preimage of the open set $(-\infty, 0) = ((-\infty, 0])^\circ$ is open: \begin{align*} f^{-1}((-\infty, 0)) = \{x \in \mathbb{R}^n : f(x) < 0\} \end{align*} is an open set contained in $A$, so $\{x : f(x) < 0\} \subset A^\circ$. The remaining points of $A$ satisfy $f(x) = 0$. Whether such a point is interior to $A$ depends on the local behaviour of $f$. If $f(x_0) = 0$ and $\nabla f(x_0) \neq 0$, then $f$ changes sign near $x_0$ (by the Implicit Function Theorem), so every ball around $x_0$ contains points where $f > 0$ — hence $x_0 \notin A^\circ$. On the other hand, if $f(x_0) = 0$ and $f$ attains a local maximum at $x_0$ (so $f(x) \leq 0$ near $x_0$), then $x_0 \in A^\circ$. In the generic case where $0$ is a regular value of $f$ (i.e., $\nabla f(x) \neq 0$ for all $x$ with $f(x) = 0$), we get: \begin{align*} A^\circ = \{x \in \mathbb{R}^n : f(x) < 0\}. \end{align*} The interior of the sublevel set $\{f \leq 0\}$ is the strict sublevel set $\{f < 0\}$. [/example] ### Intersection and Union Rules When $A$ is a finite intersection $A = A_1 \cap \cdots \cap A_k$, the interior distributes: \begin{align*} (A_1 \cap \cdots \cap A_k)^\circ = A_1^\circ \cap \cdots \cap A_k^\circ. \end{align*} This reduces the problem of computing the interior of an intersection to computing the interiors of the individual sets. For unions, the inclusion $A_1^\circ \cup \cdots \cup A_k^\circ \subset (A_1 \cup \cdots \cup A_k)^\circ$ holds but can be strict. However, if each $A_i$ is closed and the $A_i$ cover a region without overlap (except at boundaries), the interior of the union can sometimes be computed by removing the boundaries of the individual pieces. ## References - Munkres, J., *Topology* (2nd ed., 2000). Chapters 2 and 3. - Willard, S., *General Topology* (1970). Chapter 2. - Kelley, J., *General Topology* (1955). Chapter 1. - Rudin, W., *Principles of Mathematical Analysis* (3rd ed., 1976). Chapter 2. - Evans, L. C., *Partial Differential Equations* (2nd ed., 2010). Appendix A and C (for applications of interior in PDE domains).