What “Nowhere Dense” Means
In the previous tutorial, the Baire Category Theorem was used to turn certain countable closed covers into conclusions on open sets. The term nowhere dense describes sets that are too small, topologically, to contain any open region, even after their limit points are added. It is the closure of a set—not necessarily the set itself—that matters.
All closures and interiors below are taken in the stated ambient metric space \(X\). This qualification matters: a set can be nowhere dense in a larger space and not nowhere dense in a smaller subspace containing it. For example, a line is nowhere dense in \(\mathbb{R}^2\), but is not nowhere dense in the smaller subspace consisting of that line, with its relative topology.
A set can fail to be closed and still be nowhere dense. Taking the closure in the definition accounts for points that can be approached by points of the set, even if they are not in the set itself. The definition also rules out a common mistaken shortcut: having empty interior by itself is not enough. The rational numbers have empty interior in \(\mathbb{R}\), but their closure is all of \(\mathbb{R}\), so they are not nowhere dense.
Proof. The set \(\overline{A}^{\,X}\) is closed, so its complement \(X\setminus\overline{A}^{\,X}\) is open. It remains to check density. A subset \(G\) of \(X\) is dense precisely when every nonempty open subset of \(X\) intersects \(G\). Thus \(X\setminus\overline{A}^{\,X}\) is dense precisely when there is no nonempty open set \(U\) contained in \(\overline{A}^{\,X}\). That is precisely the condition \(\operatorname{int}_X(\overline{A}^{\,X})=\varnothing\), which is the definition of nowhere denseness. \(\square\)
This characterization produces an open dense set by taking the complement of the closure. The complement \(X\setminus A\) is dense whenever \(A\) is nowhere dense, but it need not be open. The distinction is important when applying results, such as the Baire Category Theorem, whose hypotheses require open dense sets.
Local Avoidance
Another useful way to recognize nowhere dense sets is to ask whether every open region contains a smaller open region that avoids the set. The smaller region can be chosen to avoid not only \(A\), but its closure as well.
Proof. Suppose first that \(A\) is nowhere dense, and let \(U\) be a nonempty open subset of \(X\). Since \(\overline{A}^{\,X}\) has empty interior, \(U\) cannot be contained in \(\overline{A}^{\,X}\). Choose \(x\in U\setminus\overline{A}^{\,X}\). The set \(X\setminus\overline{A}^{\,X}\) is open, and \(U\) is open, so their intersection \(U\cap(X\setminus\overline{A}^{\,X})\) is a nonempty open set. Taking this intersection as \(V\) gives \(V\subseteq U\) and \(V\cap\overline{A}^{\,X}=\varnothing\).
Conversely, suppose the stated local avoidance property holds. If \(\overline{A}^{\,X}\) had nonempty interior, its interior would be a nonempty open set \(U\) contained in \(\overline{A}^{\,X}\). The property would give a nonempty open \(V\subseteq U\) disjoint from \(\overline{A}^{\,X}\). But \(V\subseteq U\subseteq\overline{A}^{\,X}\), so \(V\) would be both contained in and disjoint from \(\overline{A}^{\,X}\), an impossibility because \(V\) is nonempty. Therefore \(\overline{A}^{\,X}\) has empty interior, and \(A\) is nowhere dense. \(\square\)
In a metric space, this result can be read as a practical construction method: inside any nonempty open set, one can find a nonempty open set lying outside the closure of \(A\). This is stronger than merely finding a point outside \(A\). A point outside \(A\) might still be a limit point of \(A\); the open set \(V\) avoids all such limit points.
Worked Example: A Sequence with a Limit Point
Consider \(A=\{1/n:n\geq1\}\) in \(\mathbb{R}\). Its closure is \(\overline{A}=A\cup\{0\}\): the sequence \(1/n\) converges to \(0\), and no other points can be limits of distinct elements of \(A\). To see the latter, fix \(x>0\) with \(x\notin A\). Choose an integer \(N\) so large that \(1/N<x/2\). All terms with \(n\geq N\) lie below \(x/2\); the remaining terms form a finite set not containing \(x\), so a sufficiently small neighborhood of \(x\) avoids them and the tail. Points \(x<0\) have neighborhoods disjoint from \(A\), and the only possible limit point not already in \(A\) is therefore \(0\).
The set \(A\cup\{0\}\) has empty interior. Indeed, if an open interval \(J\) were contained in it, then \(J\) would contain a smaller open interval \((a,b)\) with \(0<a<b\). There are only finitely many terms \(1/n\) in \([a,b]\), since \(1/n\geq a\) implies \(n\leq1/a\). A nonempty open interval contains infinitely many real numbers, so \((a,b)\) cannot be contained in that finite set. Hence \(\operatorname{int}(\overline{A})=\varnothing\), and \(A\) is nowhere dense.
This example also shows why \(A\) itself need not be closed. Its complement is not open: \(0\notin A\), but every neighborhood of \(0\) contains some \(1/n\in A\). In contrast, \(\mathbb{R}\setminus\overline{A}\) is open and dense, as the Open Dense Complement Characterization guarantees.
Worked Example: The Cantor Set
Begin with \(C_0=[0,1]\). At each stage, remove the open middle third of every interval remaining from the preceding stage, and call the union of the closed intervals left at stage \(n\) the set \(C_n\). Then \(C=\bigcap_{n=0}^{\infty}C_n\) is the Cantor set. Each \(C_n\) is a finite union of closed intervals, so it is closed; consequently \(C\), an intersection of closed sets, is closed. At stage \(n\), the set \(C_n\) consists of \(2^n\) closed intervals, each of length \(3^{-n}\).
We show that \(C\) has empty interior. Suppose instead that \(C\) contained a nonempty open interval \(J\), and choose a closed interval \([a,b]\subseteq J\) with \(a<b\). Choose \(n\) large enough that \(3^{-n}<b-a\). Since \(C\subseteq C_n\), the interval \([a,b]\) would be contained in \(C_n\), which is a union of pairwise separated closed intervals of length \(3^{-n}\). A connected interval contained in that union must lie in one of its component intervals: if it met two distinct component intervals, it would also contain points in the gap between them, which is not in \(C_n\). But no component interval has length as large as \(b-a\), contradicting \([a,b]\subseteq C_n\). Thus \(C\) contains no nonempty open interval. Since \(C\) is closed, \(\overline{C}=C\), and so \(C\) is nowhere dense in \(\mathbb{R}\).
The Cantor set is not empty: for example, \(0\) and \(1\) remain at every stage. Nowhere dense therefore does not mean empty. It means that even after all limit points are included, the set still contains no open interval.
Worked Example: A Line in the Plane
Let \(L=\{(x,y)\in\mathbb{R}^2:y=2x\}\), with the usual Euclidean metric. The function \(F(x,y)=y-2x\) is continuous and \(L=F^{-1}(\{0\})\), so \(L\) is closed. It has empty interior: given any point \((x,2x)\in L\) and any radius \(r>0\), choose \(0<\delta<r\). The point \((x,2x+\delta)\) is within distance \(\delta<r\) of \((x,2x)\), but it is not in \(L\), since \(2x+\delta\neq2x\). Thus no open ball centered at a point of \(L\) is contained in \(L\); an open set contained in \(L\) would have to contain such a ball centered at one of its points. Therefore \(\operatorname{int}(L)=\varnothing\), and, as \(L\) is closed, \(L\) is nowhere dense.
Finite Unions and a Countable-Union Pitfall
Nowhere dense sets remain nowhere dense under finite unions. The local avoidance characterization explains why: starting with any nonempty open region, avoid the first closure by passing to a smaller open region, then avoid the next closure, and continue through the finite list.
Proof. Let \(U\subseteq X\) be any nonempty open set. Since \(A_1\) is nowhere dense, the Local Avoidance Characterization gives a nonempty open set \(U_1\subseteq U\) disjoint from \(\overline{A_1}^{\,X}\). Apply the same characterization to \(U_1\) and \(A_2\) to obtain a nonempty open set \(U_2\subseteq U_1\) disjoint from \(\overline{A_2}^{\,X}\). Continue this finite process. At step \(j\), obtain a nonempty open set \(U_j\subseteq U_{j-1}\) disjoint from \(\overline{A_j}^{\,X}\), with \(U_0=U\). The final set \(U_n\) is nonempty and open, is contained in \(U\), and is disjoint from every \(A_j\), because it is disjoint from each \(\overline{A_j}^{\,X}\). Thus every nonempty open \(U\) contains a nonempty open subset disjoint from \(\bigcup_{j=1}^{n}A_j\). Applying the Local Avoidance Characterization shows that this finite union is nowhere dense. \(\square\)
The finiteness condition cannot simply be replaced by “countably many.” In \(\mathbb{R}\), each singleton \(\{q\}\) is nowhere dense: it is closed and contains no nonempty open interval. The rational numbers are the countable union of such singletons, but \(\mathbb{Q}\) is dense in \(\mathbb{R}\), so it is not nowhere dense. The next tutorial develops the term meager for countable unions of nowhere dense sets and studies the consequences of that broader notion.
Why the Closure and the Ambient Space Matter
There are two frequent errors to avoid. First, empty interior alone does not establish nowhere denseness. For example, \(\mathbb{Q}\) has empty interior but is dense, so \(\overline{\mathbb{Q}}=\mathbb{R}\) and its closure has nonempty interior. The required test is whether the closure has empty interior.
Second, the complement of a nowhere dense set need not be open. The set \(A=\{1/n:n\geq1\}\) is nowhere dense in \(\mathbb{R}\), but \(0\) belongs to \(\mathbb{R}\setminus A\) and every neighborhood of \(0\) meets \(A\); therefore \(\mathbb{R}\setminus A\) is not open. What is always open and dense is \(\mathbb{R}\setminus\overline{A}\). The complements of the sets themselves are dense, but they need not be open. Keeping these statements separate prevents an invalid use of open-dense hypotheses.
Finally, always specify the ambient space. In \(\mathbb{R}\), the set \([0,1]\) is not nowhere dense, because its closure has interior \((0,1)\). In the metric space \(X=[0,1]\), however, the same set is all of \(X\), so it is also not nowhere dense there. By contrast, the set \(\{0\}\) is nowhere dense in \([0,1]\): it is closed in the relative topology and contains no nonempty relatively open subset. Relative openness and closure are taken within \(X\), not automatically within \(\mathbb{R}\).
Check Your Understanding
Use the closure-based definition and the proved characterizations to answer the following questions.
- Why does a nowhere dense set have an open dense complement after its closure is removed?
- For \(A=\{1/n:n\geq1\}\), what point must be added to \(A\) to obtain its closure, and why is the complement of \(A\) not open?
- Which feature of the stage intervals proves that the Cantor set contains no nonempty open interval?
- How does local avoidance prove that a finite union of nowhere dense sets is nowhere dense?
- Why does the fact that every singleton in \(\mathbb{R}\) is nowhere dense not imply that \(\mathbb{Q}\) is nowhere dense?
- Why must closure and interior be interpreted relative to the stated ambient space?