A Point Outside Every Set in the List
A countable collection of sets can cover many points, but in a complete metric space it cannot cover any nonempty open region if each set is nowhere dense. The useful conclusion is pointwise: given a nonempty open set, we can construct a point in it that avoids every set on the list. Nested closed balls provide the construction, while completeness guarantees that the successively smaller choices have a common point.
Let \((X,d)\) be a metric space. A set \(A\subseteq X\) is nowhere dense if \(\operatorname{int}\overline{A}=\varnothing\). It need not be closed. The closure is important: even if \(x\notin A\), a point in \(\overline{A}\) has not been safely separated from \(A\) by an open region.
The theorem makes no claim when there is no nonempty open set to start from. Otherwise it applies even if some \(A_n\) are not closed, or if some of the sets are empty. Its stronger conclusion, avoiding each closure, is what makes the construction work.
Why the Nested-Ball Construction Works
For each \(n\), set \(G_n=X\setminus\overline{A_n}\). Since \(\overline{A_n}\) is closed, \(G_n\) is open. It is also dense: if a nonempty open set \(V\) missed \(G_n\), then \(V\subseteq\overline{A_n}\), contradicting \(\operatorname{int}\overline{A_n}=\varnothing\). Thus each \(G_n\) is an open region in which the \(n\)th set can be avoided.
For clarity, write \(C(x,r)=\{y\in X:d(x,y)\le r\}\) for a closed ball of radius \(r>0\). A closed ball is closed: if \(y\notin C(x,r)\), then \(d(x,y)>r\); any point \(z\) with \(d(y,z)<d(x,y)-r\) satisfies \(d(x,z)\ge d(x,y)-d(y,z)>r\), so a neighborhood of \(y\) misses the ball.
Proof. Choose \(x_0\in U\). Since \(U\) is open, there is \(\delta_0>0\) such that the open ball \(B(x_0,\delta_0)\) is contained in \(U\). Choose \(r_0\) with \(0<r_0<\delta_0\), and put \(C_0=C(x_0,r_0)\). Then \(C_0\subseteq U\).
We construct closed balls \(C_n=C(x_n,r_n)\) for \(n\ge1\), so that \(C_n\subseteq B(x_{n-1},r_{n-1})\cap G_n\) and \(r_n<2^{-n}\). Suppose \(C_{n-1}\) has been chosen. Its open ball \(B(x_{n-1},r_{n-1})\) is nonempty and open. Because \(G_n\) is dense, the intersection \(W_n=B(x_{n-1},r_{n-1})\cap G_n\) is nonempty; it is open because both sets in the intersection are open. Choose \(x_n\in W_n\). Openness gives some \(\delta_n>0\) for which \(B(x_n,\delta_n)\subseteq W_n\). Choose \(0<r_n<\min(\delta_n,2^{-n})\). Then the closed ball \(C_n\) is contained in \(B(x_n,\delta_n)\), hence in \(W_n\), as required.
It remains to show that these choices yield a point. For \(m\ge n\), nesting gives \(x_m\in C_n\), so \(d(x_m,x_n)\le r_n\). If \(m,k\ge n\), the triangle inequality gives \(d(x_m,x_k)\le 2r_n<2^{1-n}\). Since this bound tends to zero as \(n\) increases, \((x_n)\) is Cauchy. Completeness supplies a limit \(x\in X\).
Fix \(n\). Every \(x_m\) with \(m\ge n\) belongs to \(C_n\). As \(C_n\) is closed and \(x_m\to x\), we have \(x\in C_n\). Also \(x\in C_0\), by the same argument, so \(x\in U\). Finally, \(C_n\subseteq G_n=X\setminus\overline{A_n}\), and therefore \(x\notin\overline{A_n}\), hence \(x\notin A_n\). This holds for every \(n\in\mathbb{N}\). \(\square\)
The radii need not be chosen by a formula. At each stage, openness gives some positive amount of room, and we choose a smaller radius that also satisfies \(r_n<2^{-n}\). The shrinking bound ensures the centers are Cauchy; it does not require a uniform distance from all the sets at once.
Worked Constructions and Explicit Avoidance
Worked Example: Avoiding a Sequence of Points in the Real Line
In \(\mathbb{R}\) with its standard topology, let \(A_n=\{1/(n+1)\}\) for \(n\in\mathbb{N}\), and start with \(U=(0,1)\). Each singleton is closed. It has empty interior because every open interval containing its one point also contains other points. Thus each \(A_n\) is nowhere dense, and \(\mathbb{R}\) is complete.
Here are the first choices of closed intervals. Start with \(I_0=[1/8,7/8]\subset U\). To avoid \(A_1=\{1/2\}\), choose \(I_1=[1/4,3/8]\). This lies inside the interior of \(I_0\), and every point of \(I_1\) is at most \(3/8<1/2\). To avoid \(A_2=\{1/3\}\), choose \(I_2=[11/32,23/64]\). Its lower endpoint is greater than \(1/3\), since \(11/32>1/3\) is equivalent to \(33>32\). Its upper endpoint is \(23/64<24/64=3/8\), so \(I_2\) lies inside the interior of \(I_1\).
For \(A_3=\{1/4\}\), choose \(I_3=[45/128,91/256]\). The endpoints of \(I_2\), in denominator \(128\), are \(44/128\) and \(46/128\); the endpoints of \(I_3\) are \(45/128\) and \(45.5/128\). Thus \(I_3\) lies strictly inside \(I_2\). Every point of \(I_3\) is greater than \(1/4\), since \(45/128>32/128=1/4\). At each later stage, choose a closed interval inside the interior of the preceding interval that misses the next point, with length less than \(2^{-n}\). This is possible because removing one point from a nonempty open interval leaves a nonempty open subinterval.
The theorem says the nested intervals have a common point \(x\in(0,1)\) that is not any of \(1/2,1/3,1/4,\ldots\). The interval calculations check the first three avoidances directly; the construction at every later stage ensures the corresponding avoidance, and completeness supplies the common point.
Worked Example: Avoiding Countably Many Lines in the Plane
Give \(X=\mathbb{R}^2\) its Euclidean metric, and let \(A_n=\{(s,t)\in\mathbb{R}^2:s=n\}\) for \(n\in\mathbb{N}\). Each \(A_n\) is closed: if \((s,t)\notin A_n\), then \(s\ne n\), and every point \((u,v)\) with Euclidean distance less than \(|s-n|/2\) has \(|u-s|<|s-n|/2\), hence \(u\ne n\). Thus the complement is open.
Each line has empty interior. Given any open ball with center \((s,t)\) and radius \(r>0\), if \(s\ne n\) its center is already off \(A_n\). If \(s=n\), the point \((n+r/2,t)\) lies in that ball because its distance from the center is \(r/2<r\), and it is not in \(A_n\). Therefore no open ball is contained in \(A_n\), proving that \(A_n\) is nowhere dense.
Take \(U=B((1,0),2)\) and \(p=(1/2,0)\). The distance from \(p\) to \((1,0)\) is \(\sqrt{(1/2-1)^2+(0-0)^2}=\sqrt{1/4}=1/2<2\), so \(p\in U\). For every \(n\in\mathbb{N}\), the first coordinate of \(p\) is \(1/2\), which is not the positive integer \(n\); hence \(p\notin A_n\). This explicit point verifies the conclusion for this family and region. The theorem also applies to any other nonempty open region, where the same point need not work.
Worked Example: Avoiding Finite Grids in the Unit Interval
Let \(X=[0,1]\) with the metric inherited from \(\mathbb{R}\), and define \(A_n=\{k/2^n:k=0,1,\ldots,2^n\}\). The space \(X\) is complete: a Cauchy sequence in \(X\) converges in \(\mathbb{R}\), and its limit remains between \(0\) and \(1\). Each \(A_n\) is finite, hence closed in \(X\). It has empty interior in \(X\), because every nonempty relatively open subset of \([0,1]\) contains infinitely many points: around any one of its points it contains a sufficiently small interval intersected with \([0,1]\). Thus every \(A_n\) is nowhere dense in \(X\).
For example, \(A_1=\{0,1/2,1\}\) and \(A_2=\{0,1/4,1/2,3/4,1\}\). Let \(x=\sqrt{2}-1\). Since \(1<\sqrt{2}<2\), we have \(0<x<1\). Also \(x\) is irrational: if \(x\) were rational, then \(\sqrt{2}=x+1\) would be rational. To check the latter is impossible, write \(\sqrt{2}=p/q\) in lowest terms with positive integers \(p,q\). Then \(p^2=2q^2\), so \(p\) is even; writing \(p=2j\) gives \(q^2=2j^2\), so \(q\) is even as well, contradicting lowest terms.
Every member \(k/2^n\) of \(A_n\) is rational, whereas \(x\) is irrational, so \(x\notin A_n\) for every \(n\). This checks the conclusion directly for the open set \(X\) itself, which is open in its own subspace topology. The theorem further guarantees an avoiding point in every nonempty relatively open subset of \(X\).
A Common Pitfall: Avoid the Closure, Not Just the Set
A nowhere-dense set need not be closed, so \(X\setminus A_n\) need not be open. The construction needs open sets in which to place closed balls; that is why it uses \(X\setminus\overline{A_n}\), not merely \(X\setminus A_n\).
For a concrete example, in \(\mathbb{R}\) let \(A=\{1/n:n\in\mathbb{N}\}\). Its closure is \(A\cup\{0\}\): the points \(1/n\) approach \(0\), and every nonzero point not in \(A\) has a small interval missing \(A\). Thus \(A\) is not closed, and \(\mathbb{R}\setminus A\) is not open at \(0\), since every interval around \(0\) contains some \(1/n\). The set \(A\) is nowhere dense because \(A\cup\{0\}\) contains no open interval. Using \(\mathbb{R}\setminus\overline{A}\) gives an open set that avoids not only \(A\) but also its limit point \(0\).
A second error is to choose each new ball merely to meet the complement of the next set. Meeting the complement does not keep the eventual limit there. The construction instead places the entire closed ball \(C_n\) inside \(X\setminus\overline{A_n}\). Since the limit belongs to every \(C_n\), it inherits every avoidance condition.
Check Your Understanding
For each question, keep track of whether the sets are closed, nowhere dense, or both, and state where completeness enters.
- In \(\mathbb{R}\), why is each singleton nowhere dense, and why does that fact alone not imply that a countable union of singletons is closed?
- In the interval construction, explain why the upper endpoint \(3/8\) ensures that \(I_1=[1/4,3/8]\) avoids \(1/2\).
- Show that \(A=\{1/n:n\in\mathbb{N}\}\) is nowhere dense in \(\mathbb{R}\), even though it is not closed.
- In the nested-ball proof, where is density used, and where is completeness used?
- Why does placing \(C_n\) inside \(X\setminus\overline{A_n}\) give a stronger conclusion than merely placing it inside \(X\setminus A_n\)?