Two Properties Working Together
Density and openness describe different features of a set. Density says that a set reaches arbitrarily close to every real number: equivalently, every nonempty open interval meets it. Openness says that each point in the set has some open interval around it that stays within the set. When both properties hold, the set meets every interval and also contains a whole smaller interval inside it.
The Density and Closure theorem from “Dense Sets” gives a useful test: \(U\) is dense if and only if \(\overline{U}=\mathbb{R}\), or, equivalently, every nonempty open interval intersects \(U\). The word “open” adds a local guarantee around each point of \(U\). Together, the properties give more than just a point of \(U\) in each interval: they give an entire open interval contained in \(U\).
The Complement Describes What Is Missing
An open dense set can be smaller than the whole real line. Its complement consists of the points that it misses. The complement cannot contain an open interval, because such an interval would fail to meet \(U\); and it is closed because \(U\) is open. This gives an exact characterization in terms of the earlier notion of nowhere denseness.
Proof. Suppose first that \(U\) is open and dense, and let \(F=\mathbb{R}\setminus U\). Since \(U\) is open, its complement \(F\) is closed, by the theorem “A Set Is Closed If and Only If Its Complement Is Open.” Because \(U\) is dense, every nonempty open interval intersects \(U\). Thus no nonempty open interval can be contained in \(F\), so \(\operatorname{int}(F)=\varnothing\). Since \(F\) is closed, \(\overline{F}=F\), and therefore $$ \operatorname{int}(\overline{F})=\operatorname{int}(F)=\varnothing. $$ This is exactly the definition of \(F\) being nowhere dense.
Conversely, suppose \(F\) is closed and nowhere dense, and put \(U=\mathbb{R}\setminus F\). Since \(F\) is closed, \(U\) is open. Now \(\overline{F}=F\), so nowhere denseness gives \(\operatorname{int}(F)=\varnothing\). Every nonempty open interval must therefore contain a point outside \(F\): if an interval were contained in \(F\), it would be a nonempty open subset of \(F\), contradicting its empty interior. Hence every nonempty open interval intersects \(U\), and the density characterization shows that \(U\) is dense. Thus \(U\) is open and dense. \(\square\)
One useful consequence is that the complement of an open dense set is not merely small in the sense of having empty interior: it is closed and nowhere dense. Conversely, removing a closed nowhere dense set always leaves an open dense set. This correspondence lets us use whichever description makes a problem easier to handle.
Every Interval Contains a Whole Patch of the Set
Density alone guarantees a point of \(U\) in each open interval, but not that the point comes with an interval around it still inside \(U\). Openness supplies that extra step. The result is a convenient way to use open dense sets locally.
Proof. Let \(I\) be a nonempty open interval. Since \(U\) is dense, choose \(x\in I\cap U\). Because \(I\) is open, there is \(r_1>0\) such that \(B_{r_1}(x)\subseteq I\). Because \(U\) is open and \(x\in U\), there is \(r_2>0\) such that \(B_{r_2}(x)\subseteq U\). Set \(r=\min\{r_1,r_2\}\), which is positive. Then $$ J=B_r(x)\subseteq I\cap U. $$ The ball \(J\) is a nonempty open interval, as required. \(\square\)
The order of the argument matters: density first provides a point in the given interval, and openness then provides a neighborhood of that point. Density by itself does not say that the set contains any interval. For example, the rationals are dense, but they are not open: every interval around a rational also contains irrational numbers.
Worked Example: Removing One Point
Let \(U=\mathbb{R}\setminus\{0\}\). The singleton \(\{0\}\) is closed, so \(U\) is open. To check density directly, let \(I\) be any nonempty open interval. The Infinitely Many Rationals in Every Open Interval theorem gives infinitely many rational points in \(I\). At most one of them is \(0\), so \(I\) contains a point other than \(0\). Hence \(I\cap U\ne\varnothing\), and \(U\) is dense.
The complement characterization gives the same conclusion: \(\{0\}\) is closed and has empty interior, so it is nowhere dense. The set \(U\) is therefore open and dense, even though it is not all of \(\mathbb{R}\). The interval-containment theorem further guarantees that every nonempty open interval contains a smaller open interval avoiding \(0\).
Worked Example: A Proper Dense Union of Rational-Centered Intervals
List the nonzero rationals as \(q_1,q_2,\ldots\), and define $$ r_n=\min\{2^{-n},|q_n|/2\},\qquad U=\bigcup_{n=1}^{\infty}(q_n-r_n,q_n+r_n). $$ Each interval in the union is open, so \(U\) is open by the Arbitrary Unions of Open Sets theorem.
To check density, take any nonempty open interval \(I\). It contains a nonzero rational \(q_n\), by the Density of the Rationals theorem (and, for example, by taking a smaller subinterval that avoids zero if necessary). Since \(q_n\) is the center of one of the intervals in the union, \(q_n\in U\cap I\). Thus every nonempty open interval meets \(U\), so \(U\) is dense.
This set is not the whole real line. For each \(n\), \(r_n\leq |q_n|/2\), so the interval around \(q_n\) stays strictly on the same side of zero as \(q_n\). In particular, \(0\notin(q_n-r_n,q_n+r_n)\) for every \(n\), and hence \(0\notin U\). The construction illustrates how intervals around a dense set of centers can form an open dense set while still omitting a point.
Closed Nowhere Dense Sets as Complements
The complement characterization can also produce open dense sets from familiar closed sets. For a countable example, include the limit point when forming a sequence set: a convergent sequence without its limit is not closed, so its complement need not be open.
Worked Example: Removing a Convergent Sequence and Its Limit
Let \(F=\{2\}\cup\{2+1/n:n\geq1\}\), and set \(U=\mathbb{R}\setminus F\). The sequence \(2+1/n\) converges to \(2\), and its range together with its limit is closed: this follows from the closure calculation for a convergent sequence in “Nowhere Dense Sets,” after translating each point by \(2\). Thus \(F\) is closed and \(U\) is open.
Every point of \(F\) is rational. The Density of the Irrationals theorem says that every nonempty open interval contains an irrational number, which cannot belong to \(F\). Consequently every such interval meets \(U\), and \(U\) is dense. Equivalently, \(F\) is closed with empty interior, so it is nowhere dense and the complement characterization applies.
If the limit point \(2\) were omitted from \(F\), the resulting sequence set would not be closed. Its complement would not be open: every ball around \(2\) contains terms of the sequence that have been removed. Including the limit is essential for this complement to be open.
What Openness and Density Do Not Say
An open dense set need not be connected, and it need not contain every real number. The examples above omit one point or a whole countable set, while still meeting every interval. Density is not a statement that every point belongs to the set; it is a statement about how the set meets neighborhoods throughout the line.
It is also important not to confuse a dense set with an open dense set. The rationals are dense but not open. Conversely, an open set need not be dense: the interval \((1,3)\) is open, but it misses, for instance, the open interval \((4,5)\). Both conditions are needed.
Earlier in this course, the Finite Intersection of Open Dense Sets theorem established that a finite intersection of open dense sets is again open and dense. This can be useful when several conditions must hold at once. The finite-intersection result does not, by itself, establish the corresponding claim for infinitely many open dense sets; an infinite intersection requires additional reasoning and hypotheses. For the present topic, the safe local tool is the proved interval-containment theorem: inside any interval, an open dense set contains a smaller interval.
The complement viewpoint and the local interval viewpoint are two ways to recognize the same structure. The complement viewpoint says that the points omitted by an open dense set form a closed nowhere dense set. The local viewpoint says that no matter how small an interval we start with, some whole open interval inside it lies in the set. Which description is more useful depends on whether a problem focuses on what is included or what is excluded.
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- Why must the complement of an open dense set be closed and have empty interior?
- In the interval-containment theorem, which step uses density, and which step uses openness?
- How does the rational-centered construction show that an open dense set can omit a point?
- Why is the complement of \(\{2+1/n:n\geq1\}\) not open, while the complement of \(\{2\}\cup\{2+1/n:n\geq1\}\) is open?
- Give an example of a dense set that is not open, and explain which property fails.