A Local Definition of Nowhere Dense
To say that a set is nowhere dense is to say that it occupies no open region, even after we fill in all the points that its neighborhoods approach. The closure matters: a set can have empty interior and still be dense, in which case its closure is the whole space. The local definition below turns “occupies no open region” into a test using open sets.
Throughout, \((X,\mathcal{T})\) is a topological space, \(A\subseteq X\), and open neighborhoods are open sets containing the point in question. A set is closed if its complement in \(X\) is open. The closure \(\overline{A}\) is the intersection of all closed sets containing \(A\); the interior \(\operatorname{int}A\) is the union of all open sets contained in \(A\). In particular, \(\operatorname{int}A\) is the largest open subset of \(A\).
The open set \(V\) may depend on \(U\). The definition does not require \(U\) itself to miss \(A\); it requires that some open part of \(U\) does. If \(X=\varnothing\), there are no nonempty open sets \(U\), so the condition holds vacuously for the only subset \(A=\varnothing\).
The Closure Test
We first record the neighborhood test for closure, which will connect the local definition to \(\overline{A}\).
Proof. Suppose \(x\in\overline{A}\), and let \(U\) be an open neighborhood of \(x\). If \(U\cap A=\varnothing\), then \(A\subseteq X\setminus U\). The set \(X\setminus U\) is closed and contains \(A\), so the definition of closure gives \(\overline{A}\subseteq X\setminus U\). This contradicts \(x\in U\cap\overline{A}\). Thus \(U\cap A\ne\varnothing\).
Conversely, suppose every open neighborhood of \(x\) intersects \(A\). If \(x\notin\overline{A}\), then \(X\setminus\overline{A}\) is an open neighborhood of \(x\), since \(\overline{A}\) is closed. It misses \(A\), because \(A\subseteq\overline{A}\). This contradicts the assumption. Hence \(x\in\overline{A}\). \(\square\)
Proof. First suppose \(\operatorname{int}\overline{A}=\varnothing\). Let \(U\) be any nonempty open set. It cannot be contained in \(\overline{A}\), since that would make \(U\) a nonempty open subset of \(\overline{A}\), contrary to the empty interior. Therefore \(V=U\setminus\overline{A}\) is nonempty. It is open because both \(U\) and \(X\setminus\overline{A}\) are open. Since \(A\subseteq\overline{A}\), we have \(V\cap A=\varnothing\). This gives the required open set \(V\) inside every such \(U\), so \(A\) is nowhere dense.
Conversely, suppose \(A\) is nowhere dense. If \(\operatorname{int}\overline{A}\ne\varnothing\), then there is a nonempty open set \(W\subseteq\overline{A}\). By the definition of nowhere dense, there must be a nonempty open set \(V\subseteq W\) with \(V\cap A=\varnothing\). Choose \(x\in V\). Since \(V\subseteq\overline{A}\), the neighborhood test says that every open neighborhood of \(x\), including \(V\), intersects \(A\). This contradicts \(V\cap A=\varnothing\). Thus \(\operatorname{int}\overline{A}=\varnothing\). \(\square\)
The proof uses no separation assumption: the criterion holds in every topological space. Its two directions serve different purposes. Empty interior of the closure lets us find an open region avoiding the set inside any given open region. Conversely, an open region inside the closure cannot contain an open subset disjoint from the set, because every point of the closure must have all its open neighborhoods meet the set.
Working Through the Criterion
Worked Example: A Finite Subset of the Real Line
Let \(\mathbb{R}\) have its standard topology, and take \(A=\{0,2\}\). We check that \(A\) is nowhere dense by computing its closure and interior.
First, \(A\) is closed. For any \(x\in\mathbb{R}\setminus A\), both \(|x|\) and \(|x-2|\) are positive. Set \(r=\min(|x|,|x-2|)/2\), which is positive. The interval \((x-r,x+r)\) contains neither \(0\) nor \(2\), so it lies in \(\mathbb{R}\setminus A\). Thus the complement is open and \(A\) is closed. Consequently \(\overline{A}=A=\{0,2\}\).
The interior of \(A\) is empty: every open neighborhood of either point contains an interval with infinitely many real points, so it cannot be a subset of \(\{0,2\}\). Therefore \(\operatorname{int}\overline{A}=\varnothing\), and the criterion shows that \(A\) is nowhere dense.
Check: for an open interval \(U\) containing \(0\), there is a smaller nonempty open interval inside \(U\) that avoids both \(0\) and \(2\); the same holds if \(U\) contains \(2\). If \(U\) contains neither point, \(U\) itself already avoids \(A\). This agrees with the local definition.
Worked Example: An Open Interval
In \(\mathbb{R}\) with its standard topology, take \(A=(0,1)\). Its closure is \([0,1]\). Indeed, every open neighborhood of a point in \((0,1)\) meets \((0,1)\), and every open neighborhood of \(0\) or \(1\) contains points strictly between \(0\) and \(1\). If \(x<0\), the open interval \((x-1,x/2)\) is a neighborhood of \(x\) disjoint from \(A\); if \(x>1\), the interval \(((x+1)/2,x+1)\) is such a neighborhood. The closure test therefore gives exactly \(\overline{A}=[0,1]\).
The interior of \([0,1]\) in \(\mathbb{R}\) is \((0,1)\): every point strictly between the endpoints has an open interval inside \([0,1]\), but no neighborhood of either endpoint is contained in \([0,1]\). Hence \(\operatorname{int}\overline{A}=(0,1)\ne\varnothing\), so \(A\) is not nowhere dense.
Check: the open set \(U=(0,1)\) cannot contain a nonempty open subset disjoint from \(A\), because every point of \(U\) belongs to \(A\). This directly violates the local definition.
Worked Example: The Rationals and the Empty-Interior Pitfall
Let \(A=\mathbb{Q}\subseteq\mathbb{R}\), with the standard topology on \(\mathbb{R}\). We show that \(\operatorname{int}\mathbb{Q}=\varnothing\) but \(\overline{\mathbb{Q}}=\mathbb{R}\). This is a useful test of why the criterion concerns the interior of the closure, not just the interior of the original set.
Every nonempty open interval contains a rational number. To see this, let \((a,b)\) be such an interval and choose \(n\in\mathbb{N}\) with \(n(b-a)>1\). There is an integer \(m\) with \(na<m\le na+1\); hence \(na<m<nb\), so \(m/n\in(a,b)\). Applying this to every neighborhood shows that every open neighborhood of every real number meets \(\mathbb{Q}\). By the closure test, \(\overline{\mathbb{Q}}=\mathbb{R}\).
Every nonempty open set in \(\mathbb{R}\) contains a nonempty open interval \((a,b)\). By the argument above, choose a rational \(q\in(a,b)\). Choose \(N\in\mathbb{N}\) large enough that \(0<\sqrt{2}/N<b-q\). Then \(q+\sqrt{2}/N\in(a,b)\), and this number is irrational: if \(\sqrt{2}\) were rational, writing it as a reduced fraction would imply that both numerator and denominator are even. Thus \((a,b)\) is not a subset of \(\mathbb{Q}\), proving \(\operatorname{int}\mathbb{Q}=\varnothing\).
Nevertheless, \(\operatorname{int}\overline{\mathbb{Q}}=\operatorname{int}\mathbb{R}=\mathbb{R}\), which is nonempty. Therefore \(\mathbb{Q}\) is not nowhere dense. The error in the tempting test “empty interior means nowhere dense” is that it ignores the points added when taking closure.
Two Consequences
For a closed set, taking the closure changes nothing. This gives a particularly simple version of the criterion.
Proof. Since \(C\) is closed, its closure is \(C\). Substituting \(\overline{C}=C\) into the nowhere-dense criterion gives \(C\) is nowhere dense if and only if \(\operatorname{int}C=\varnothing\). \(\square\)
Another consequence is that finite unions preserve nowhere density. This conclusion concerns finite unions; the argument below relies on being able to avoid the closures one at a time.
Proof. The closure of a finite union satisfies \(\overline{A\cup B}=\overline{A}\cup\overline{B}\). Indeed, the right-hand side is a closed set containing \(A\cup B\), so it contains \(\overline{A\cup B}\); and since each of \(A\) and \(B\) is contained in \(A\cup B\), each of their closures is contained in \(\overline{A\cup B}\). Now suppose, seeking a contradiction, that there is a nonempty open set \(W\subseteq\overline{A}\cup\overline{B}\). Since \(\operatorname{int}\overline{A}=\varnothing\), the open set \(V=W\setminus\overline{A}\) is nonempty. Since \(\operatorname{int}\overline{B}=\varnothing\), the open set \(V\setminus\overline{B}\) is also nonempty. But it is contained in \(W\) and misses both \(\overline{A}\) and \(\overline{B}\), contradicting \(W\subseteq\overline{A}\cup\overline{B}\). Thus \(\overline{A\cup B}\) has empty interior, and the criterion shows that \(A\cup B\) is nowhere dense. \(\square\)
The key distinction is between a set and its closure. Empty interior says that the set itself contains no open region. Nowhere density says that even its closure contains no open region. When testing a claim, first identify which of these sets is being examined; confusing them would incorrectly classify the rationals as nowhere dense.
Check Your Understanding
Use the local definition and the closure criterion to justify your answers.
- Show that the empty set is nowhere dense in every topological space, including the empty space.
- Can a dense subset of a nonempty space be nowhere dense? Explain using the criterion.
- Why does the closed-set corollary not imply that every set with empty interior is nowhere dense?
- If \(A\) and \(B\) are nowhere dense, what facts about their closures are used to prove that \(A\cup B\) is nowhere dense?
- In \(\mathbb{R}\), explain why \([0,1]\) is not nowhere dense, while \(\{0,2\}\) is.