Dense Sets

What It Means for a Set to Be Dense

A set can be spread throughout the real line even when it does not contain every real number. The idea of density makes this precise: no matter where we look, and no matter how small an interval we choose, that interval contains a point of the set. This gives a local test for a global property, and it connects naturally to closure and limit points.

Definition: A set \(D\subseteq\mathbb{R}\) is dense in \(\mathbb{R}\) if every nonempty open interval in \(\mathbb{R}\) intersects \(D\). Equivalently, for every \(a<b\), there is some \(d\in D\) such that \(a<d<b\).

The interval in this definition can be as short as we please. Density does not mean that an interval lies inside \(D\); it means only that the interval contains at least one point of \(D\). It is also useful to keep the ambient space in view: here, density is always being tested against open intervals in the whole real line.

Worked Example: Removing Finitely Many Points

Let \(F\subseteq\mathbb{R}\) be finite and set \(D=\mathbb{R}\setminus F\). We show that \(D\) is dense. Take any nonempty open interval \((a,b)\), and suppose \(F\) has \(m\) points. Inside this interval, consider the \(m+1\) points

x_j=a+\frac{j(b-a)}{m+2},\qquad j=1,\ldots,m+1.

For each such \(j\), we have \(0<j/(m+2)<1\), so \(a<x_j<b\). The points are distinct because \(b-a>0\) and different values of \(j\) give different values of \(x_j\). At most \(m\) of these \(m+1\) points can belong to \(F\). Thus at least one belongs to \((a,b)\setminus F=(a,b)\cap D\). Since this works for every nonempty open interval, \(D\) is dense.

Density and Closure

The definition tests whether every interval meets \(D\). The Neighborhood Characterization of Closure from “Closure of a Set” gives the corresponding pointwise description: \(x\in\overline{D}\) exactly when every open ball centered at \(x\) intersects \(D\). Since open balls in \(\mathbb{R}\) are open intervals, density says that every real number belongs to the closure.

Theorem (Density and Closure): For \(D\subseteq\mathbb{R}\), the following are equivalent:

\(D\) is dense in \(\mathbb{R}\).
\(\overline{D}=\mathbb{R}\).
Every open ball in \(\mathbb{R}\) intersects \(D\).

Proof. Suppose first that \(D\) is dense. Let \(x\in\mathbb{R}\) and \(r>0\). The ball \(B_r(x)=(x-r,x+r)\) is a nonempty open interval, so density implies \(B_r(x)\cap D\neq\varnothing\). This holds for every \(r>0\). By the Neighborhood Characterization of Closure, \(x\in\overline{D}\). Since \(x\) was arbitrary, \(\mathbb{R}\subseteq\overline{D}\). The reverse inclusion holds because \(\overline{D}\subseteq\mathbb{R}\); hence \(\overline{D}=\mathbb{R}\).

Now suppose that \(\overline{D}=\mathbb{R}\). For any \(x\in\mathbb{R}\) and any \(r>0\), we have \(x\in\overline{D}\). The Neighborhood Characterization of Closure therefore gives \(B_r(x)\cap D\neq\varnothing\). In particular, every nonempty open interval \((a,b)\) intersects \(D\): choose its midpoint \(x=(a+b)/2\) and radius \(r=(b-a)/2>0\), so that \(B_r(x)=(a,b)\). Thus \(D\) is dense. The equivalence with the open-ball condition follows from these same arguments. \(\square\)

This theorem offers two equivalent ways to work with density. The interval definition is often easiest for constructing a point of \(D\), while the closure formulation can be more convenient when closure has already been computed. In particular, density is a statement about all points of \(\mathbb{R}\) being approximable by \(D\), not a statement that all those points belong to \(D\).

Worked Example: A Proper Dense Set

Consider \(D=\mathbb{R}\setminus\{0\}\). The finite-removal argument above shows that \(D\) is dense. In particular, \(D\neq\mathbb{R}\), because \(0\notin D\), but its closure is all of \(\mathbb{R}\). Directly, every ball around \(0\) contains a nonzero point: for \(r>0\), the point \(r/2\) satisfies \(0<r/2<r\), so \(r/2\in B_r(0)\cap D\). For a center \(x\neq 0\), the center itself is in \(D\) and in every ball around \(x\). Thus every real number belongs to \(\overline{D}\), including the point omitted from \(D\).

The example highlights an important distinction. A dense set need not contain each point of the real line; it must instead come arbitrarily close to each point. The omitted point \(0\) is a limit point of \(D\), even though it is not an element of \(D\).

Approximating Points by Sequences

Density also has a sequential description. For a point \(x\), one can ask whether there are points of \(D\) that approach \(x\) as the sequence index increases. The radii \(1/n\) provide a direct construction: choose one point of \(D\) in each ball of radius \(1/n\) around \(x\). The resulting points converge to \(x\).

Theorem (Sequential Characterization of Density): A set \(D\subseteq\mathbb{R}\) is dense in \(\mathbb{R}\) if and only if for every \(x\in\mathbb{R}\), there is a sequence \((d_n)\) with \(d_n\in D\) for every \(n\) and \(d_n\to x\).

Proof. Suppose \(D\) is dense, and fix \(x\in\mathbb{R}\). For each positive integer \(n\), the interval \((x-1/n,x+1/n)\) is nonempty and open. By density, it contains some \(d_n\in D\). The choices satisfy

|d_n-x|<\frac{1}{n}.

To verify convergence, let \(\varepsilon>0\). Choose a positive integer \(N\) such that \(1/N<\varepsilon\). For every \(n\geq N\), we have \(1/n\leq 1/N<\varepsilon\), and therefore \(|d_n-x|<\varepsilon\). Hence \(d_n\to x\).

Conversely, suppose that for every real \(x\) there is a sequence \((d_n)\) in \(D\) converging to \(x\). Fix \(x\in\mathbb{R}\) and \(r>0\). By convergence, there is an \(N\) such that \(n\geq N\) implies \(|d_n-x|<r\). In particular, \(d_N\in B_r(x)\cap D\), so every open ball centered at \(x\) intersects \(D\). This holds for every \(x\), and the Density and Closure Theorem implies that \(D\) is dense. \(\square\)

Worked Example: Approximating an Omitted Point

For \(D=\mathbb{R}\setminus\{0\}\), define \(d_n=1/n\) for each positive integer \(n\). Each \(d_n\) is nonzero, so \(d_n\in D\). Also,

|d_n-0|=\frac{1}{n}\longrightarrow 0.

Thus this sequence in \(D\) converges to \(0\), even though \(0\notin D\). For a general target \(x\), the sequence \(d_n=x+1/n\) also lies in \(D\) when \(x=0\), but it may equal \(0\) for some \(n\) when \(x\neq 0\). A construction that works uniformly for this particular set is \(d_n=x+1/(n+|x|+1)\): its added term is positive, so it is never zero when \(x=0\); when \(x\neq 0\), it can equal \(-x\) for at most one value of \(n\). If that happens, replace that single term by any other nonzero point within distance \(1/n\) of \(x\). Alternatively, the density theorem already guarantees a sequence for every \(x\); the simple sequence \(1/n\) verifies the key case of the omitted point.

Finite Intersections of Open Dense Sets

Density is not generally preserved by taking intersections of arbitrary dense sets. But openness supplies a useful additional condition: a nonempty open set contains room to keep intersecting the next dense set. Repeating this argument a finite number of times proves that the intersection of finitely many open dense sets remains dense.

Theorem (Finite Intersection of Open Dense Sets): Let \(U_1,\ldots,U_m\) be open and dense subsets of \(\mathbb{R}\), where \(m\) is a positive integer. Then \(\bigcap_{j=1}^{m}U_j\) is open and dense in \(\mathbb{R}\).

Proof. The intersection is open by the Finite Intersections of Open Sets Theorem. To prove it is dense, take any nonempty open interval \(I\). Set \(V_0=I\), which is nonempty and open. Since \(U_1\) is dense, \(V_0\cap U_1\neq\varnothing\). The set \(V_1=V_0\cap U_1\) is open, because it is an intersection of open sets, and it is nonempty by the preceding observation.

Now suppose \(1\leq k<m\) and \(V_k=I\cap\bigcap_{j=1}^{k}U_j\) is nonempty and open. Density of \(U_{k+1}\) implies that \(V_k\cap U_{k+1}\neq\varnothing\). Define \(V_{k+1}=V_k\cap U_{k+1}\). It is open as a finite intersection of open sets, and it is nonempty. By induction, \(V_m=I\cap\bigcap_{j=1}^{m}U_j\) is nonempty. Therefore \(I\) intersects \(\bigcap_{j=1}^{m}U_j\). Since this is true for every nonempty open interval \(I\), the intersection is dense. \(\square\)

Worked Example: Intersecting Two Open Dense Sets

Let \(U_1=\mathbb{R}\setminus\{0\}\) and \(U_2=\mathbb{R}\setminus\{2\}\). Each is open, since the complement of a singleton is open, and each is dense by the finite-removal example. Their intersection is

U_1\cap U_2=\mathbb{R}\setminus\{0,2\}.

This intersection is open and dense by the Finite Intersection of Open Dense Sets Theorem. The finite-removal argument also verifies density directly: no nonempty open interval can be contained in the two-point set \(\{0,2\}\), so each such interval contains a point other than \(0\) and \(2\). The theorem is useful because it applies even when the intersection is not as simple to describe as this one.

What Density Does and Does Not Guarantee

Density should not be confused with openness. An open set contains a small ball around each of its points; a dense set is required to meet every open interval, whether or not it contains a ball around any of its own points. Nor does density imply that the set is closed. The set \(\mathbb{R}\setminus\{0\}\), for example, is open and dense but not closed, because its closure contains \(0\).

There is one immediate consequence when closedness is present: a set that is both closed and dense must be the whole real line. Indeed, density gives \(\overline{D}=\mathbb{R}\), while closedness gives \(\overline{D}=D\), so \(D=\mathbb{R}\). Thus a proper dense set cannot be closed. This is a useful check when considering examples or proving that a set has a point in its closure that it does not contain.

A common mistake is to read “dense” as “contains many points” without checking the interval condition, or to infer that a dense set contains every point. The decisive test is local: every nonempty open interval must intersect the set. To disprove density, it is enough to find just one nonempty open interval that misses the set entirely.

Worked Example: An Interval That Shows a Set Is Not Dense

Let \(E=\{1/n:n\text{ is a positive integer}\}\). The interval \((3/5,4/5)\) contains no point of \(E\). Indeed, \(1/1=1>4/5\), \(1/2=1/2<3/5\), and for \(n\geq 2\), \(1/n\leq 1/2<3/5\). Hence \((3/5,4/5)\cap E=\varnothing\), so \(E\) is not dense.

Notice that this argument does not need a full description of \(\overline{E}\). The interval criterion gives a short, conclusive test: one missed nonempty open interval is enough to rule out density.

In practice, use the formulation that best fits the problem. Use open intervals to establish density by finding a point in each one; use closure when closure has already been computed; and use sequences when approximating a specified point is the natural task. For intersections, remember the extra hypothesis in the finite-intersection theorem: the sets must be open as well as dense.

Check Your Understanding

Use the definitions and results in this tutorial to answer the following questions.

State the open-interval definition of a set dense in \(\mathbb{R}\).
How is density expressed using the closure of a set?
How can density be used to construct a sequence in \(D\) converging to a specified real number \(x\)?
Why does a single nonempty open interval disjoint from \(E\) prove that \(E\) is not dense?
What two hypotheses on each set ensure that a finite intersection of sets is dense?
Why must a set that is both closed and dense equal \(\mathbb{R}\)?

What You'll Learn