How Examples Reveal Closedness
The definition of closedness gives a useful test, but applying it from scratch to each set can be cumbersome. The Sequential Characterization of Closed Sets offers another route: a set is closed if and only if it contains the limit of every convergent sequence of its points. In this tutorial, we use that characterization to study several different kinds of subsets of the real line.
Some examples have only finitely many points; others extend indefinitely. A helpful new criterion will cover sets that have only finitely many points in every sufficiently small neighborhood. This local condition lets us prove closedness without having to describe the complement explicitly or identify every possible convergent sequence individually.
The neighborhood in this definition is allowed to depend on \(x\). Local finiteness does not require one fixed radius to work everywhere, nor does it require the whole set to be finite. It says that near each individual real number, only finitely many points of \(E\) occur.
A Local Finiteness Criterion
A locally finite set cannot have a convergent sequence of its points whose limit lies outside the set. Indeed, terms sufficiently far along such a sequence must lie in a neighborhood that contains only finitely many points of the set. Among finitely many possible values, at least one must occur infinitely often. The resulting constant subsequence forces the limit to be that value.
Proof. Let \(E\subseteq\mathbb{R}\) be locally finite. Take any convergent sequence \((x_n)\) with \(x_n\in E\) for every \(n\), and suppose \(x_n\to x\). By local finiteness at \(x\), there is an \(r>0\) such that \(B_r(x)\cap E\) is finite. Since \(x_n\to x\), there is an \(N\in\mathbb{N}_0\) such that \(x_n\in B_r(x)\) for every \(n\geq N\). Thus all terms in this tail belong to the finite set \(B_r(x)\cap E\).
At least one value \(c\) in this finite set must occur as \(x_n\) for infinitely many indices \(n\geq N\). Otherwise, each of the finitely many possible values would occur only finitely many times, so the whole tail would have only finitely many terms, which is impossible. Select the indices where \(x_n=c\). They give a constant subsequence with limit \(c\). Every subsequence of a convergent sequence has the same limit as the original sequence, so uniqueness of limits gives \(x=c\). Since \(c\in E\), we have \(x\in E\). The Sequential Characterization of Closed Sets now implies that \(E\) is closed. \(\square\)
Worked Example: A Finite Set
Let \(F=\{-4,0,7\}\). This is a finite subset of \(\mathbb{R}\), so the Complement of a Finite Set Is Open theorem from “Examples of Open Sets” says that \(\mathbb{R}\setminus F\) is open. By the definition of a closed set, \(F\) is closed.
The local finiteness criterion gives another way to view the example. For any \(x\in\mathbb{R}\), take \(r=1\). Then \(B_1(x)\cap F\) is a subset of the three-element set \(F\), and hence is finite. Thus \(F\) is locally finite and is closed by the theorem. The same reasoning works for any finite subset, including the empty set.
Uniform Separation Gives Closed Sets
Some infinite sets have a fixed positive gap between any two distinct points. Such sets cannot crowd infinitely many of their points into a bounded neighborhood. The following result formalizes that observation.
Proof. Let \(E\) be uniformly separated, and choose \(\delta>0\) such that distinct points of \(E\) are at least \(\delta\) apart. Fix any \(x\in\mathbb{R}\), and consider \(B_{\delta/3}(x)\). Its diameter is \(2\delta/3\): if \(u,v\in B_{\delta/3}(x)\), then the triangle inequality gives \[ |u-v|\leq|u-x|+|x-v|<\frac{2\delta}{3}<\delta. \] Consequently, this ball cannot contain two distinct points of \(E\), since any such pair would have to be at least \(\delta\) apart. Therefore \(B_{\delta/3}(x)\cap E\) has at most one point and is finite. This holds for every \(x\), so \(E\) is locally finite. The Locally Finite Sets Are Closed theorem now implies that \(E\) is closed. \(\square\)
Worked Example: An Arithmetic Progression
Consider \(E=\{3k+1:k\in\mathbb{Z}\}\), where \(\mathbb{Z}\) denotes the integers. If \(u=3k+1\) and \(v=3\ell+1\) are distinct points of \(E\), then \(k\neq\ell\), and
Thus \(E\) is uniformly separated with \(\delta=3\), so it is closed by the Uniformly Separated Sets Are Closed theorem. For a direct local check, every ball of radius \(1\) has diameter \(2\), which is less than the gap \(3\); it can therefore contain at most one point of \(E\). This example is unbounded, but closedness does not require boundedness.
Unbounded Sets Can Also Be Locally Finite
Uniform separation is a convenient sufficient condition, but local finiteness can apply even when it is easier to count points in bounded regions than to find a minimum gap. For example, an unbounded sequence of distinct points may have only finitely many terms in any bounded interval. The definition then gives local finiteness, because every ball in \(\mathbb{R}\) is bounded.
Worked Example: A Closed Set of Increasing Values
Let \(E=\{n+1/n:n\text{ is a positive integer}\}\). Fix \(x\in\mathbb{R}\), and choose any \(r>0\). If a point \(n+1/n\) belongs to \(B_r(x)\), then
There are only finitely many positive integers \(n\) satisfying \(n<x+r\); if \(x+r\leq1\), there are none. Hence \(B_r(x)\cap E\) is finite. This holds at every \(x\), so \(E\) is locally finite and therefore closed.
The argument does not depend on calculating a distance from an arbitrary \(x\) to the whole set. It uses only a bound on the indices of points that could occur in a given neighborhood. In particular, it verifies local finiteness even though \(E\) is unbounded.
A Set Can Be Bounded and Still Fail to Be Closed
Local finiteness is sufficient for closedness, but a set that is not locally finite need not automatically be nonclosed. More generally, no single feature such as boundedness, having many points, or having isolated points decides the question. The sequential characterization directly detects a missing limit.
Worked Example: An Open Interval Is Not Closed
Consider \(E=(2,5)\). For each positive integer \(n\), let \(x_n=2+1/n\). Since \(0<1/n\leq1\), we have \(2<x_n\leq3<5\), so \(x_n\in E\). Also,
and therefore \(x_n\to2\). But \(2\notin(2,5)\). This sequence of points in \(E\) has a limit outside \(E\), so the Sequential Characterization of Closed Sets shows that \(E\) is not closed.
This example also illustrates why finite-point reasoning cannot be applied indiscriminately. Every individual point of \((2,5)\) has other points of the interval arbitrarily near it, and a sequence can approach the omitted endpoint. The interval is bounded, but boundedness does not prevent it from omitting a limit point.
What the Examples Have in Common
The examples suggest a practical way to choose a closedness test. For a finite set, the complement result from “Examples of Open Sets” is immediate. For an infinite set with a fixed gap, check uniform separation. For a set whose points escape every bounded region except for finitely many, check local finiteness. If a likely limit is easier to identify than the local structure, try the sequential characterization instead.
A common pitfall is to confuse “each point of the set is isolated” with local finiteness. Isolation only says that each point of \(E\) has a neighborhood containing no other points of \(E\). Local finiteness makes a demand at every real number, including points outside \(E\), and asks that some neighborhood contain only finitely many points of \(E\). Another pitfall is to treat closedness and boundedness as opposites or equivalents: the arithmetic progression above is unbounded and closed, while the bounded interval \((2,5)\) is not closed.
Check Your Understanding
Use the local finiteness criterion, uniform separation, and the sequential characterization to answer the following questions.
- In the proof that locally finite sets are closed, why must some value in the finite tail range occur infinitely often?
- What separation constant works for the set \(\{5k-2:k\in\mathbb{Z}\}\), and which theorem then proves it is closed?
- Why does a ball of radius \(\delta/3\) contain at most one point of a set whose distinct points are at least \(\delta\) apart?
- For \(E=\{n+1/n:n\text{ is a positive integer}\}\), what inequality bounds the possible indices of points in \(B_r(x)\)?
- Which sequence of points in \((2,5)\) converges to a point outside the interval?