Compactness Produces a Single Bound
A subset of the real line is bounded when all its points fit inside one interval of finite length. Compactness, by contrast, is defined through open covers: every open cover must have a finite subcover. The connection between these ideas is especially direct in \(\mathbb{R}\), with its standard topology. The intervals \((-n,n)\), for positive integers \(n\), cover the whole real line. If a subset is compact, a finite number of these intervals already cover it; the largest index among them gives one interval that contains the entire subset.
Throughout, \(\mathbb{R}\) has its standard topology. A family of open sets is an open cover of \(K\subseteq\mathbb{R}\) if its union contains \(K\). A finite subcover is a finite subfamily that still covers \(K\). The subset \(K\) is compact if every open cover of \(K\) has a finite subcover. Equivalently, \(K\) is compact as a subset of \(\mathbb{R}\) when it is compact with the subspace topology: an ambient open set \(U\) covers \(K\) through its intersection \(K\cap U\).
The intervals in the argument are open, and their endpoints matter: \(x\in(-n,n)\) means \(-n<x<n\), or equivalently \(|x|<n\). We use the Archimedean property of the real numbers: for every real number \(r\), there is a positive integer \(n\) with \(n>r\). In particular, every real number belongs to some interval \((-n,n)\).
Proof. Suppose first that \(A\) is bounded. Then there is \(M>0\) such that \(|x|\le M\) for every \(x\in A\). By the Archimedean property, choose a positive integer \(N>M\). For each \(x\in A\), \(|x|\le M<N\), so \(-N<x<N\). Thus \(A\subseteq(-N,N)\). Conversely, if \(A\subseteq(-N,N)\) for a positive integer \(N\), then every \(x\in A\) satisfies \(|x|<N\), and hence \(|x|\le N\). Taking \(M=N>0\) shows that \(A\) is bounded. If \(A=\varnothing\), both conditions hold as well. \(\square\)
The Compactness Argument
Proof. If \(K=\varnothing\), it is bounded by definition. Suppose \(K\ne\varnothing\). For each \(n\in\mathbb{N}\), let \(U_n=(-n,n)\). Each \(U_n\) is open in the standard topology on \(\mathbb{R}\). The family \(\{U_n:n\in\mathbb{N}\}\) covers \(K\): for every \(x\in K\), the Archimedean property gives a positive integer \(n\) with \(n>|x|\), so \(-n<x<n\) and \(x\in U_n\).
Since \(K\) is compact, this open cover has a finite subcover. Write its members as \(U_{n_1},\ldots,U_{n_r}\), where \(r\ge1\) and each \(n_i\) is a positive integer. Let \(N=\max\{n_1,\ldots,n_r\}\). For each \(i\), \(n_i\le N\), so \((-n_i,n_i)\subseteq(-N,N)\). The finite subcover covers \(K\), and every set in it is contained in \((-N,N)\); therefore \(K\subseteq(-N,N)\). The lemma now implies that \(K\) is bounded. \(\square\)
The key step is not merely that each point lies in some interval from the cover. The finite-subcover property gives a finite list of indices, and a finite list of positive integers has a largest member. That single largest index supplies a bound for every point of \(K\) at once. Without compactness, the index needed to contain a point may grow without any one index working for the entire set.
Worked Example: A Closed Interval
Take \(K=[-2,3]\) in \(\mathbb{R}\). The earlier open-cover proof for closed intervals establishes that \(K\) is compact. The intervals \(U_n=(-n,n)\) form an open cover of \(K\), and \(U_4=(-4,4)\) alone covers it: for every \(x\in[-2,3]\), \(-4<-2\le x\le3<4\). Thus \(\{U_4\}\) is a finite subcover, and the resulting bound is \(|x|<4\) for all \(x\in K\).
Check: The radius \(4\) works, and radius \(3\) does not work for an interval of the form \((-n,n)\), because \(3\notin(-3,3)\). This verifies both that the displayed finite subcover covers the endpoint \(3\) and that \(4\) is the smallest positive-integer radius that works.
Worked Example: A Finite Compact Set
Let \(F=\{-7,2,9\}\subseteq\mathbb{R}\). This set is compact: given any open cover of \(F\), choose one member containing \(-7\), one containing \(2\), and one containing \(9\). These at most three open sets form a finite subcover.
For the particular cover \(U_n=(-n,n)\), the interval \(U_{10}=(-10,10)\) contains all three points because \(-10<-7<10\), \(-10<2<10\), and \(-10<9<10\). The smaller integer-radius interval \(U_9=(-9,9)\) misses \(9\), since the endpoint \(9\) is not in the open interval. Therefore the smallest positive-integer radius giving one interval that covers \(F\) is \(10\).
Check: The absolute values of the three elements are \(7\), \(2\), and \(9\). Their maximum is \(9\), so the strict inequality \(|x|<N\) requires \(N>9\); the smallest positive integer satisfying that requirement is \(N=10\).
Worked Example: A Compact Image with Negative Endpoints
Let \(I=[0,1]\), and define \(f:I\to\mathbb{R}\) by \(f(t)=5t-8\). The interval \(I\) is compact, and \(f\) is continuous, so the continuous-image theorem for compact spaces implies that \(K=f(I)\) is compact. As \(t\) ranges from \(0\) to \(1\), \(5t-8\) ranges from \(-8\) to \(-3\), so \(K=[-8,-3]\).
For this compact set, \(U_9=(-9,9)\) is a one-member subcover from the family \(U_n=(-n,n)\): its endpoints satisfy \(-9<-8\) and \(-3<9\), so \([-8,-3]\subseteq(-9,9)\). The interval \(U_8=(-8,8)\) does not contain \(-8\), because \(-8\) is an endpoint and the interval is open. Thus the smallest positive-integer radius that works is \(9\).
Check: The endpoints come directly from the formula: \(f(0)=5(0)-8=-8\) and \(f(1)=5(1)-8=5-8=-3\). Every point of \(K\) therefore has absolute value at most \(8\), and \(8<9\), as required for containment in \((-9,9)\).
What the Conclusion Does—and Does Not—Say
The theorem gives a necessary condition for compactness in \(\mathbb{R}\): a compact set must be bounded. The cover used in the proof also provides a direct test for some noncompact sets. If a set requires arbitrarily large indices in order to be covered by the intervals \((-n,n)\), then this particular cover has no finite subcover.
Worked Example: The Integers Are Not Compact
Give \(\mathbb{Z}\) its subspace topology from \(\mathbb{R}\). The family \(V_n=\mathbb{Z}\cap(-n,n)\), for \(n\in\mathbb{N}\), consists of open subsets of \(\mathbb{Z}\) and covers \(\mathbb{Z}\): each integer \(k\) satisfies \(|k|<n\) for some positive integer \(n\).
No finite subfamily covers \(\mathbb{Z}\). Indeed, if a finite nonempty selection has largest index \(N\), then every selected set is contained in \(V_N\). But the integer \(N+1\) is not in \(V_N\), since \(N+1\notin(-N,N)\). Thus the finite selection misses \(N+1\). An empty selection also cannot cover the nonempty set \(\mathbb{Z}\). This open cover has no finite subcover, so \(\mathbb{Z}\) is not compact.
Check: The strict endpoint condition is essential here: \(N\notin(-N,N)\) as well, and \(N+1\) is certainly outside it. The chosen integer \(N+1\) is a specific point missed by every member of the proposed finite subcover.
Boundedness is not, by itself, enough to conclude compactness. For example, consider \(A=(0,1)\) with the subspace topology from \(\mathbb{R}\). It is bounded because \(0<x<1\) implies \(|x|<1\). For each integer \(n\ge2\), let \(W_n=(1/n,1)\), viewed as an open subset of \(A\). These sets cover \(A\): for any \(x\in(0,1)\), choose an integer \(n>\max\{2,1/x\}\); then \(1/n<x<1\), so \(x\in W_n\). A finite selection has a largest index \(m\), and its union is contained in \(W_m=(1/m,1)\). It misses \(1/(2m)\), which belongs to \(A\) and satisfies \(1/(2m)\le1/m\). Thus this cover has no finite subcover, and \(A\) is not compact.
Common pitfall. One may see that every point of a set lies in some interval \((-n,n)\) and incorrectly conclude that one of those intervals covers the whole set. The quantifiers are different: the cover property says that for each point there is a suitable index, while the compactness argument produces finitely many indices and then takes their maximum. For the integers, each point has a suitable index, but any finite selection has a largest index \(N\) and misses \(N+1\). It is compactness—not merely being covered by the expanding intervals—that forces one common bound.
Check Your Understanding
For each question, attend to whether the intervals are open and whether their endpoints are included.
- Why does \(\{(-n,n):n\in\mathbb{N}\}\) cover every subset of \(\mathbb{R}\)? State the real-number property used.
- If a nonempty compact set has a finite subcover with indices \(2,5,8\), which single interval \((-N,N)\) contains all three selected intervals?
- What changes in the main theorem’s proof when \(K=\varnothing\)?
- For \(K=\{-4,6\}\), what is the smallest positive integer \(N\) for which \(K\subseteq(-N,N)\)? Explain why the preceding integer radius fails.
- Why does the cover \(W_n=(1/n,1)\), \(n\ge2\), of \((0,1)\) have no finite subcover?