Examples and Nonexamples of Nowhere Dense Sets

Closure, Interior, and the Test for Nowhere Denseness

A set can be small in one sense and large in another. The integers are spread through the real line but leave open intervals between their points; the rationals meet every open interval, even though they do not contain any open interval. To classify such sets, we must inspect not just a set’s interior, but the interior of its closure.

Throughout, \(\mathbb{R}\) has its usual topology, whose open neighborhoods contain open intervals around each of their points. For \(A\subseteq\mathbb{R}\), \(\overline{A}\) denotes its closure and \(\operatorname{int}A\) its interior. Recall that \(x\in\overline{A}\) exactly when every open neighborhood of \(x\) meets \(A\), and that \(\operatorname{int}A\) is the union of all open sets contained in \(A\).

This definition immediately gives a useful distinction. If \(A\) is closed, then \(\overline{A}=A\), so \(A\) is nowhere dense exactly when \(\operatorname{int}A=\varnothing\). If \(A\) is not closed, its closure may have a nonempty interior even when \(A\) itself has none.

Singletons and Finite Sets

Proof. Let \(F\subseteq\mathbb{R}\) be finite. If \(F=\varnothing\), it is closed and has empty interior, so it is nowhere dense. Now suppose \(F\ne\varnothing\), and write \(m\) for its number of elements. To show \(F\) is closed, take any \(y\in\mathbb{R}\setminus F\). For each \(f\in F\), the distance \(|y-f|\) is positive. The minimum \(d=\min_{f\in F}|y-f|\) exists because \(F\) is finite, and \(d>0\). The interval \((y-d/2,y+d/2)\) contains \(y\) and misses \(F\), since every \(f\in F\) is at distance at least \(d\) from \(y\). Thus every point of \(\mathbb{R}\setminus F\) has an open neighborhood in \(\mathbb{R}\setminus F\), so the complement is open and \(F\) is closed. Hence \(\overline{F}=F\).

To show that \(F\) has empty interior, suppose an open interval \((x-r,x+r)\), with \(r>0\), were contained in \(F\). The \(m+1\) numbers \(x+r/(j+2)\), for \(j=1,\ldots,m+1\), are distinct and all lie in \((x,x+r)\subseteq(x-r,x+r)\). A set with \(m\) elements cannot contain \(m+1\) distinct numbers, a contradiction. No nonempty open interval is contained in \(F\), so \(\operatorname{int}F=\varnothing\). Because \(F\) is closed, \(\operatorname{int}\overline{F}=\operatorname{int}F=\varnothing\). Therefore \(F\) is nowhere dense. \(\square\)

The Integers: Closed but with Empty Interior

Proof. First we show that \(\mathbb{Z}\) is closed. If \(y\notin\mathbb{Z}\), there is an integer \(k\) with \(k<y<k+1\). Both \(y-k\) and \(k+1-y\) are positive, so let \(r=\frac12\min(y-k,k+1-y)>0\). Then \((y-r,y+r)\subset(k,k+1)\), which contains no integers. Thus every point outside \(\mathbb{Z}\) has a neighborhood disjoint from \(\mathbb{Z}\), and \(\mathbb{R}\setminus\mathbb{Z}\) is open. Hence \(\mathbb{Z}\) is closed and \(\overline{\mathbb{Z}}=\mathbb{Z}\).

Next, no nonempty open interval is contained in \(\mathbb{Z}\). Indeed, any open interval contains a bounded open subinterval, which contains infinitely many distinct real numbers. Only finitely many integers can lie in a bounded interval: if the interval is contained in \((a,b)\), its integers must lie between two fixed integers that bound \(a\) and \(b\). Therefore the subinterval contains a point not in \(\mathbb{Z}\). This proves \(\operatorname{int}\mathbb{Z}=\varnothing\). Since \(\mathbb{Z}\) is closed, \(\operatorname{int}\overline{\mathbb{Z}}=\varnothing\), as required. \(\square\)

The Rationals: Dense but Not Nowhere Dense

The rationals provide the contrasting case. We will verify both relevant facts: every nonempty open interval contains a rational, so the closure is all of \(\mathbb{R}\); but every nonempty open interval also contains an irrational, so the rationals themselves have empty interior.

Proof. Let \((a,b)\) be any nonempty open interval, so \(a<b\). Choose a positive integer \(n\) with \(n(b-a)>2\), using the Archimedean property of \(\mathbb{R}\). Let \(m=\lfloor na\rfloor+1\), where \(\lfloor na\rfloor\) is the greatest integer not exceeding \(na\). Then \(na<m\le na+1<nb\), because \(nb-na>2\). Dividing by \(n>0\) gives \(a<m/n<b\). Thus every nonempty open interval contains a rational number. Every neighborhood of every real number contains such an interval, so every neighborhood meets \(\mathbb{Q}\). By the neighborhood test for closure, \(\overline{\mathbb{Q}}=\mathbb{R}\).

To show \(\operatorname{int}\mathbb{Q}=\varnothing\), take any nonempty open interval \((a,b)\). By the result just proved, choose \(q\in\mathbb{Q}\cap(a,b)\). Choose a positive integer \(n\) large enough that \(\sqrt{2}/n<b-q\). Then \(q+\sqrt{2}/n\) is still in \((a,b)\), since it is greater than \(q>a\) and less than \(b\). It is irrational: if it were rational, subtracting the rational number \(q\) and multiplying by the integer \(n\) would make \(\sqrt{2}\) rational. To recall why this is impossible, if \(\sqrt{2}=p/s\) in lowest terms with positive integers \(p,s\), then \(p^2=2s^2\), so \(p\) is even; writing \(p=2t\) gives \(s^2=2t^2\), so \(s\) is even too, contradicting that \(p/s\) was in lowest terms. Thus every nonempty open interval contains a point outside \(\mathbb{Q}\), and \(\operatorname{int}\mathbb{Q}=\varnothing\).

Finally, \(\operatorname{int}\overline{\mathbb{Q}}=\operatorname{int}\mathbb{R}=\mathbb{R}\), which is not empty. By definition, \(\mathbb{Q}\) is therefore not nowhere dense. \(\square\)

A Common Pitfall: Checking the Wrong Set

A frequent error is to reason, “\(\operatorname{int}\mathbb{Q}=\varnothing\), so \(\mathbb{Q}\) is nowhere dense.” The first statement is correct, but it checks only whether \(\mathbb{Q}\) contains an open interval. The definition asks whether \(\overline{\mathbb{Q}}\) contains an open interval. Since \(\overline{\mathbb{Q}}=\mathbb{R}\), its interior is \(\mathbb{R}\), not the empty set. Therefore the rationals are not nowhere dense.

The comparison can be summarized as follows. The closure of a finite set is itself and has empty interior; the closure of \(\mathbb{Z}\) is itself and has empty interior; but the closure of \(\mathbb{Q}\) is the whole real line. The closure step is essential, especially when the set under consideration is not closed.