Why Closure Matters
A set can have empty interior and still be dense in the whole space. In that case its closure is the whole space, which certainly has nonempty interior. The definition of nowhere density therefore examines not just the set itself, but what remains after taking its closure. This distinction is useful whenever we want to describe sets that are small in a topological sense without claiming that they are empty or finite.
Throughout, \((X,\mathcal{T})\) is a topological space. A set is open if it belongs to \(\mathcal{T}\), and it is closed if its complement in \(X\) is open. The closure \(\overline{A}\) of \(A\subseteq X\) is the intersection of all closed subsets of \(X\) containing \(A\). The interior \(\operatorname{int}A\) is the union of all open subsets of \(X\) contained in \(A\). Thus \(\overline{A}\) is the smallest closed set containing \(A\), while \(\operatorname{int}A\) is the largest open set contained in \(A\).
The definition allows \(A\) itself to have interior only if that interior disappears on passing to the closure—which cannot happen, since \(A\subseteq\overline{A}\). In particular, a nowhere-dense set must have empty interior. The converse is not generally true: a set with empty interior may have a closure with nonempty interior.
A Useful Neighborhood Test
Proof. Suppose first that \(x\in\overline{A}\). If an open neighborhood \(U\) of \(x\) did not intersect \(A\), then \(A\subseteq X\setminus U\). The set \(X\setminus U\) is closed, so the definition of closure would give \(\overline{A}\subseteq X\setminus U\). This contradicts \(x\in U\cap\overline{A}\). Conversely, suppose every open neighborhood of \(x\) intersects \(A\). If \(x\notin\overline{A}\), then \(X\setminus\overline{A}\) is an open neighborhood of \(x\), because \(\overline{A}\) is closed. It misses \(A\), since \(A\subseteq\overline{A}\). This is a contradiction. \(\square\)
- \(A\) is nowhere dense in \(X\).
- Every nonempty open set \(U\subseteq X\) contains a nonempty open set \(V\) such that \(V\cap A=\varnothing\).
- \(X\setminus\overline{A}\) is dense in \(X\), meaning \(\overline{X\setminus\overline{A}}=X\).
Proof. Assume first that \(A\) is nowhere dense, so \(\operatorname{int}\overline{A}=\varnothing\). Given a nonempty open set \(U\), it cannot be contained in \(\overline{A}\), since that would put the nonempty open set \(U\) inside \(\operatorname{int}\overline{A}\). Choose \(x\in U\setminus\overline{A}\). Then \(V=U\cap(X\setminus\overline{A})\) is open and contains \(x\), so it is nonempty. Because \(A\subseteq\overline{A}\), this \(V\) is disjoint from \(A\). Thus (1) implies (2).
Now assume (2). If \(\operatorname{int}\overline{A}\) were nonempty, it would itself be a nonempty open set \(U\) contained in \(\overline{A}\). Condition (2) would give a nonempty open \(V\subseteq U\) disjoint from \(A\). Choose \(v\in V\). Since \(v\in\overline{A}\), the closure test says that every open neighborhood of \(v\), including \(V\), intersects \(A\). This contradicts \(V\cap A=\varnothing\). Hence \(\operatorname{int}\overline{A}=\varnothing\), proving (2) implies (1).
Finally, \(X\setminus\overline{A}\) is open because \(\overline{A}\) is closed. By the closure test, it is dense in \(X\) exactly when every nonempty open set in \(X\) intersects it. That happens exactly when no nonempty open set is contained in \(\overline{A}\), which is equivalent to \(\operatorname{int}\overline{A}=\varnothing\). Thus (1) and (3) are equivalent as well. These arguments include \(X=\varnothing\): in that case there are no nonempty open sets, and all three conditions hold. \(\square\)
The open-set criterion gives a practical interpretation: inside any nonempty open region, one can find a smaller nonempty open region that avoids \(A\). It does not say that the whole original region avoids \(A\), and it does not say that \(A\) is closed.
Closed Sets and Singletons in the Real Line
Worked Example: A Closed Set with Nonempty Interior
Work in \(\mathbb{R}\) with its standard topology, and let \(A=[0,1]\cup\{2\}\). Determine whether \(A\) is nowhere dense.
The complement is \((-\infty,0)\cup(1,2)\cup(2,\infty)\), which is open. Hence \(A\) is closed, so \(\overline{A}=A\). Every point of \((0,1)\) lies in an open interval contained in \(A\), so \((0,1)\subseteq\operatorname{int}A\). The points \(0\) and \(1\) are not interior points: every open interval about \(0\) contains negative numbers, and every open interval about \(1\) contains numbers greater than \(1\) and less than \(2\). The point \(2\) is not an interior point either, since every open interval about \(2\) contains points other than \(2\) that are not in \(A\).
Therefore \(\operatorname{int}\overline{A}=\operatorname{int}A=(0,1)\), which is nonempty. So \(A\) is not nowhere dense. Check: because \(A\) is closed, taking its closure changes nothing; the interval \((0,1)\) remains inside that closure.
For a closed set \(F\), the definition simplifies: \(\overline{F}=F\), so \(F\) is nowhere dense exactly when \(\operatorname{int}F=\varnothing\). This equivalence depends on closedness. For a set that is not closed, checking its own interior is not enough.
Worked Example: A Singleton and a Finite Set
In standard \(\mathbb{R}\), let \(A=\{3\}\). A singleton is closed: its complement \((-\infty,3)\cup(3,\infty)\) is open. Thus \(\overline{A}=A\). No nonempty open interval can be contained in \(\{3\}\), because every open interval containing \(3\) also contains points different from \(3\). Hence \(\operatorname{int}\overline{A}=\varnothing\), and \(\{3\}\) is nowhere dense.
The same reasoning applies to any finite set \(F=\{x_1,\ldots,x_n\}\subseteq\mathbb{R}\), including the empty set. For nonempty \(F\), its complement is open: for any \(y\notin F\), the finitely many distances \(|y-x_i|\) are all positive, so their minimum \(r=\min_{1\le i\le n}|y-x_i|\) is positive, and \((y-r/2,y+r/2)\) misses \(F\). Thus \(F\) is closed. No open interval is contained in a finite set, so \(\operatorname{int}F=\varnothing\), and \(F\) is nowhere dense. The empty set is also closed with empty interior, so it is nowhere dense.
Check: for any \(y\notin F\), the displayed radius is positive because it is the minimum of a finite list of positive numbers. The resulting interval contains \(y\) and avoids every \(x_i\), as required to show the complement is open.
Dense Subsets: Empty Interior Is Not Enough
A subset \(D\subseteq X\) is dense if \(\overline{D}=X\). If \(X\) is nonempty and \(D\) is dense, then \(\operatorname{int}\overline{D}=\operatorname{int}X=X\), which is nonempty. Therefore no dense subset of a nonempty space is nowhere dense. This conclusion follows from the closure in the definition, even when the dense subset itself has empty interior.
Worked Example: The Rationals and Irrationals
In standard \(\mathbb{R}\), both \(\mathbb{Q}\) and \(\mathbb{R}\setminus\mathbb{Q}\) are dense, and each has empty interior. We verify the density claims by checking that every nonempty open interval contains a rational and an irrational.
Let \((a,b)\) be any nonempty open interval. Choose \(n\in\mathbb{N}\) with \(n(b-a)>1\), and let \(m=\lfloor na\rfloor+1\), where \(\lfloor na\rfloor\) is the greatest integer at most \(na\). Then \(m>na\) and \(m\le na+1<nb\). Hence \(a<m/n<b\), and \(m/n\in\mathbb{Q}\). Thus every nonempty open interval intersects \(\mathbb{Q}\), so the closure test gives \(\overline{\mathbb{Q}}=\mathbb{R}\).
For an irrational in \((a,b)\), choose \(k\in\mathbb{N}\) large enough that \(\sqrt{2}/k<(b-a)/2\). The interval \((a,b-\sqrt{2}/k)\) is nonempty, since its length is \(b-a-\sqrt{2}/k>(b-a)/2>0\). By the rational argument just given, choose \(q\in\mathbb{Q}\) with \(a<q<b-\sqrt{2}/k\). Then \(q+\sqrt{2}/k\) lies strictly between \(a\) and \(b\). It is irrational, since otherwise subtracting the rational \(q\) would make \(\sqrt{2}/k\), and therefore \(\sqrt{2}\), rational. (The usual parity argument shows \(\sqrt{2}\) is irrational.) So every nonempty open interval intersects \(\mathbb{R}\setminus\mathbb{Q}\), and \(\overline{\mathbb{R}\setminus\mathbb{Q}}=\mathbb{R}\).
Neither set contains a nonempty open interval: every such interval contains both a rational and an irrational by the two arguments above. Thus both sets have empty interior. Nevertheless, their closures are \(\mathbb{R}\), whose interior is \(\mathbb{R}\), not the empty set. Neither set is nowhere dense. Check: for either set \(D\), \(\operatorname{int}\overline{D}=\operatorname{int}\mathbb{R}=\mathbb{R}\ne\varnothing\).
A Common Mistake
Common pitfall. The incorrect test “\(A\) is nowhere dense whenever \(\operatorname{int}A=\varnothing\)” would label \(\mathbb{Q}\) nowhere dense in \(\mathbb{R}\), because the rationals contain no open interval. The correction is to compute the closure before taking the interior: \(\overline{\mathbb{Q}}=\mathbb{R}\), so \(\operatorname{int}\overline{\mathbb{Q}}=\mathbb{R}\ne\varnothing\). The relevant condition is empty interior of the closure, not merely empty interior of the set.
For a closed set, the shortcut \(\operatorname{int}\overline{F}=\operatorname{int}F\) is valid because \(\overline{F}=F\). Without closedness, that shortcut can give the wrong answer. The open-set criterion offers a second check: to establish nowhere density, every nonempty open region must contain a smaller nonempty open region disjoint from the set.
Check Your Understanding
For each question, distinguish the interior of the set from the interior of its closure.
- In standard \(\mathbb{R}\), is \([2,4]\) nowhere dense? Explain using its closure and interior.
- Why is \(\{0\}\) nowhere dense in standard \(\mathbb{R}\), but not nowhere dense in \(\mathbb{R}\) with the discrete topology?
- Can a dense subset of a nonempty space be nowhere dense? Justify your answer from the definition.
- Give an example of a subset of \(\mathbb{R}\) with empty interior that is not nowhere dense, and compute the interior of its closure.
- State the open-set criterion for nowhere density and explain where the closure test is used in its proof.