Choose a Point, Then Choose a Neighborhood
Topological proofs on the real line often become manageable once the goal is translated into a statement about intervals. To prove that a set is dense, take an arbitrary nonempty open interval and find a point of the set in it. To prove that a set is open, start with one of its points and find an interval around that point that remains in the set. These choices are local, but they can establish a property that holds across the whole line.
The previous tutorial showed that an open dense set contains a smaller open interval inside every nonempty open interval. That result illustrates a useful order of operations: density first locates a point, and openness then supplies room around it. This tutorial develops related proof techniques, including how to use a dense set inside an open region and how to rule out the possibility that a dense set has only finitely many points in an interval.
When \(U=\mathbb{R}\), this is the usual definition of density. For an open set \(U\), density in \(U\) is a local condition: every point of \(U\) can be approximated by points of \(A\) that remain in \(U\). The distinction between being dense in \(U\) and being dense in all of \(\mathbb{R}\) matters. For example, \((1,4)\) is dense in itself, but it is not dense in \(\mathbb{R}\).
Restricting a Dense Set to an Open Region
A dense set does not lose its ability to approximate points merely because we restrict attention to an open region. The openness ensures that sufficiently small neighborhoods of a point in the region stay inside it; density then supplies points of the original dense set in those smaller neighborhoods.
Proof. If \(U=\varnothing\), the definition of density in \(U\) is vacuously satisfied. Suppose \(U\ne\varnothing\), and take any \(x\in U\) and any \(r>0\). Since \(U\) is open, there is a \(\delta>0\) such that \(B_\delta(x)\subseteq U\). Let \(\rho=\min\{r,\delta\}/2\), so \(0<\rho<r\) and \(B_\rho(x)\subseteq U\). Since \(D\) is dense in \(\mathbb{R}\), the nonempty open interval \(B_\rho(x)\) meets \(D\). Choose \(y\in D\cap B_\rho(x)\). Then \(y\in D\cap U\), and \(|y-x|<\rho<r\). Thus \(B_r(x)\cap(D\cap U)\ne\varnothing\). This holds for every \(x\in U\) and every \(r>0\), proving that \(D\cap U\) is dense in \(U\). \(\square\)
The proof uses openness at exactly one point: it provides a smaller neighborhood lying wholly within \(U\). Density is then applied to that smaller neighborhood, rather than to a larger interval that might extend outside \(U\). This is a common pattern whenever a proof must satisfy both a location requirement and an approximation requirement.
Worked Example: Rational Points Inside an Open Interval
Let \(D=\mathbb{Q}\) and \(U=(1,4)\). The Density of the Rationals theorem says that \(\mathbb{Q}\) is dense in \(\mathbb{R}\), and \(U\) is open. The theorem therefore gives that \(\mathbb{Q}\cap(1,4)\) is dense in \((1,4)\).
To see the neighborhood choices explicitly, take \(x\in(1,4)\) and \(r>0\). The number \(\delta=\min\{x-1,4-x\}/2\) is positive, and \(B_\delta(x)\subseteq(1,4)\). Set \(\rho=\min\{r,\delta\}/2>0\). By rational density, there is a rational \(q\) with \(|q-x|<\rho\). Then \(q\in(1,4)\), because \(q\in B_\rho(x)\subseteq B_\delta(x)\subseteq(1,4)\); also \(|q-x|<r\). Hence every neighborhood of \(x\) contains a rational point that remains in \((1,4)\).
Dense Sets Cannot Be Finite in an Open Interval
Density guarantees at least one point in every nonempty open interval. In fact, it guarantees infinitely many. The proof is a useful technique in its own right: if only finitely many points were available, choose a smaller open interval that avoids all of them. Density would then force another point to exist there.
Proof. Fix a nonempty open interval \(I\), and suppose, for contradiction, that \(K=D\cap I\) is finite. Since a nonempty open interval contains infinitely many real numbers, choose \(y\in I\setminus K\). Because \(I\) is open, there is an \(\eta>0\) such that \(B_\eta(y)\subseteq I\). If \(K\ne\varnothing\), each distance \(|y-k|\), for \(k\in K\), is positive because \(y\notin K\). The finite set of these distances therefore has a positive minimum. Choose \(s>0\) smaller than \(\eta\) and, when \(K\ne\varnothing\), smaller than every \(|y-k|\). Then \(B_s(y)\subseteq I\) and \(B_s(y)\cap K=\varnothing\). If \(K=\varnothing\), simply take any \(s\) with \(0<s<\eta\), and the same two conclusions hold.
The interval \(B_s(y)\) is nonempty and open, so density of \(D\) gives some \(z\in D\cap B_s(y)\). As \(B_s(y)\subseteq I\), we have \(z\in D\cap I=K\). But \(B_s(y)\cap K=\varnothing\), while \(z\in B_s(y)\). This contradiction proves that \(D\cap I\) is infinite. \(\square\)
The finite minimum in this argument is essential: it allows one neighborhood to avoid every point in \(K\) at once. The proof does not claim that there is a positive distance from \(y\) to an arbitrary infinite set. The finiteness assumption is what makes the choice of \(s\) possible.
Worked Example: A Translated Dense Set
Consider \(D=\{q+\sqrt{2}:q\in\mathbb{Q}\}\). This set is dense in \(\mathbb{R}\). Indeed, for any nonempty open interval \((a,b)\), the interval \((a-\sqrt{2},b-\sqrt{2})\) is nonempty and open, so it contains a rational \(q\). Adding \(\sqrt{2}\) to the inequalities gives \(a<q+\sqrt{2}<b\), and \(q+\sqrt{2}\in D\cap(a,b)\).
Now take \(I=(-2,0)\). By the Infinite Intersection with Every Open Interval theorem, \(D\cap(-2,0)\) is infinite. The proof does not require listing its points or finding a formula for them; it uses only the verified density of \(D\). This is useful when a set is specified by a construction but its points are inconvenient to enumerate.
Combining Density with an Open Dense Set
Two dense sets need not have a dense intersection. The useful extra condition is openness: an open dense set contains a whole interval inside every nonempty open interval. A second dense set must meet that contained interval. This yields a robust way to combine two different requirements.
Proof. Let \(I\) be any nonempty open interval. By the Interval Containment for Open Dense Sets theorem from “Open Dense Sets,” there is a nonempty open interval \(J\) such that \(J\subseteq U\cap I\). Since \(D\) is dense, \(J\cap D\ne\varnothing\). Any point in \(J\cap D\) also belongs to \(I\cap U\cap D\). Therefore \(I\cap(D\cap U)\ne\varnothing\). Since every nonempty open interval meets \(D\cap U\), the Density and Closure theorem shows that \(D\cap U\) is dense in \(\mathbb{R}\). \(\square\)
Worked Example: Rational Points Avoiding the Integers
Let \(D=\mathbb{Q}\) and \(U=\mathbb{R}\setminus\mathbb{Z}\). The set \(\mathbb{Z}\) is locally finite, so it is closed by the Locally Finite Sets Are Closed theorem. It has empty interior: no nonempty open interval can consist entirely of integers, since every such interval contains infinitely many real numbers whereas only finitely many integers lie in any bounded subinterval. Thus \(\mathbb{Z}\) is closed and nowhere dense, and the Complement Characterization of Open Dense Sets theorem shows that \(U\) is open and dense.
The theorem now applies to the dense set \(\mathbb{Q}\) and the open dense set \(U\). It follows that \(\mathbb{Q}\setminus\mathbb{Z}\) is dense in \(\mathbb{R}\). In practical terms, every nonempty open interval contains a rational number that is not an integer. The open dense condition is what ensures that the rational point can be chosen while avoiding all the integers at once.
A Pitfall: Density Alone Does Not Control Intersections
It is tempting to infer that the intersection of two dense sets must be dense. The rationals and irrationals provide an immediate counterexample: both are dense in \(\mathbb{R}\), but their intersection is empty. The preceding intersection theorem does not apply, because neither set is open. Its proof needs an open interval contained in one set, not just a point of that set in each interval.
Worked Example: Two Dense Sets with Empty Intersection
The Density of the Rationals and Density of the Irrationals theorems establish that \(\mathbb{Q}\) and \(\mathbb{R}\setminus\mathbb{Q}\) are both dense. By the definitions of rational and irrational numbers, no real number belongs to both sets. Therefore $$ \mathbb{Q}\cap(\mathbb{R}\setminus\mathbb{Q})=\varnothing. $$ The empty set is not dense, since, for example, the nonempty interval \((0,1)\) does not meet it. Thus two dense sets can have a very small intersection.
When an intersection must be shown dense, check the hypotheses carefully. One useful route is to identify an open dense set and invoke the intersection theorem. Another is to work directly with an arbitrary interval and construct a point satisfying both conditions. Merely knowing that each set separately meets every interval is not enough.
A Practical Proof Routine
For many real-line topology arguments, the following sequence of choices keeps the proof precise. It also helps reveal which hypothesis is doing the work.
For a density claim, take an arbitrary nonempty open interval. For a local claim at a point, fix the point and an arbitrary radius.
When a point must stay in an open set, choose a smaller ball around it contained in that set.
Apply density only after the interval has been chosen to satisfy the location constraints.
State explicitly why it belongs to the desired sets and why its distance from the chosen point is within the required radius.
A related contradiction strategy is to assume that a dense set misses an open interval, or has only finitely many points in one. The first assumption directly contradicts density; the second can be converted into the first by shrinking the interval to avoid the finitely many points. This illustrates a broader lesson: topological proofs often succeed by turning a global failure into a local open region where the defining property cannot hold.
Check Your Understanding
Use the definitions and proof techniques in this tutorial to answer the following questions.
- In the proof that a dense set remains dense in an open region, why must the smaller ball be chosen inside that region?
- How does the proof that a dense set meets every open interval infinitely often use the finiteness assumption?
- Why does an open dense set help prove that its intersection with another dense set is dense?
- Give an example of two dense subsets of \(\mathbb{R}\) whose intersection is empty, and identify the missing hypothesis in the intersection theorem.
- When proving that a set is dense, what arbitrary object should be chosen first, and what must the proof establish about it?