Compactness: A Finite Cover Hidden in an Arbitrary Cover
Openness describes a local property: each point of an open set has room around it. Compactness expresses a different kind of control over a whole set. It asks whether every open cover, no matter how many sets it uses, can be reduced to finitely many sets that still cover the set. This change from arbitrary covers to finite subcovers is the central idea.
We work in the real line with its usual metric and open sets. The definition is stated for a subset \(K\subseteq\mathbb{R}\), and its open covers will consist of open subsets of \(\mathbb{R}\). This is equivalent to using sets that are open relative to \(K\): a relatively open set has the form \(K\cap O\) for some open \(O\subseteq\mathbb{R}\), and such sets cover \(K\) exactly when the corresponding ambient open sets do.
The indexing set \(A\) may be finite or infinite. Compactness does not require the original cover to be finite; it requires that one can select finitely many of its members while preserving the cover. The empty set is compact: every family covers it, and the empty subfamily is a finite subcover. When proving a nonempty set is covered by finitely many chosen sets, the chosen subfamily must, of course, be nonempty.
Finite Sets and a Compact Infinite Set
Proof. Let \(K=\{x_1,\ldots,x_m\}\) be a nonempty finite set, and let \(\{U_\alpha:\alpha\in A\}\) be an open cover of \(K\). For each \(j\in\{1,\ldots,m\}\), choose an index \(\alpha_j\) such that \(x_j\in U_{\alpha_j}\); such an index exists because the family covers \(K\). Then the finite subfamily \(U_{\alpha_1},\ldots,U_{\alpha_m}\) covers every point of \(K\). If some of these sets coincide, removing repetitions still leaves a finite cover. Thus \(K\) is compact. The empty set is compact by the definition, so the result also holds for every finite set. \(\square\)
The proof uses only that there are finitely many points: choose one cover member for each point. An infinite set needs a different argument, because choosing one member per point might require infinitely many sets.
Worked Example: A Sequence Together with Its Limit
Consider \(K=\{0\}\cup\{1/n:n\geq 1\}\). We show directly that this infinite set is compact. Let \(\{U_\alpha:\alpha\in A\}\) be any open cover of \(K\). Some member \(U_{\alpha_0}\) contains \(0\). Since \(U_{\alpha_0}\) is open, there is an \(\varepsilon>0\) such that \((-\varepsilon,\varepsilon)\subseteq U_{\alpha_0}\).
Choose a positive integer \(N\) large enough that \(1/(N+1)<\varepsilon\). For every \(n>N\), positivity and \(n\geq N+1\) give $$ 0<\frac{1}{n}\leq\frac{1}{N+1}<\varepsilon. $$ Thus \(1/n\in U_{\alpha_0}\) for every \(n>N\), and \(0\in U_{\alpha_0}\) as well. The remaining points \(1,1/2,\ldots,1/N\) are finite in number. For each of them, choose one member of the cover that contains it. Together with \(U_{\alpha_0}\), these finitely many cover members cover all of \(K\). Since the original cover was arbitrary, \(K\) is compact.
The key is that a single neighborhood of \(0\) captures all but finitely many points of the sequence. The argument does not claim that every open cover has one member covering \(K\); it constructs a finite subcover that may use several members.
Open Covers That Cannot Be Reduced to Finitely Many Sets
To prove that a set is not compact, it is enough to find one open cover for which every finite subfamily fails to cover. Two common constructions use intervals that expand to cover an unbounded set, or intervals that approach a missing endpoint.
Worked Example: The Real Line Is Not Compact
For each positive integer \(n\), let \(U_n=(-n,n)\). These are open intervals, and they cover \(\mathbb{R}\): for any real \(x\), the Archimedean property gives a positive integer \(n>|x|\), so \(x\in(-n,n)\).
Now consider a finite selection from this cover. If the selection is empty, its union is empty and does not cover \(\mathbb{R}\). If it is nonempty, let \(N\) be the largest selected index. Since the intervals are nested, every selected interval is contained in \((-N,N)\), and the selected interval \(U_N\) itself is \((-N,N)\). Hence the union of the selected family is \((-N,N)\), which does not contain \(N+1\). No finite subfamily covers \(\mathbb{R}\), so \(\mathbb{R}\) is not compact.
Worked Example: The Open Interval \((0,1)\) Is Not Compact
For each integer \(n\geq 2\), take \(V_n=(1/n,1)\). These are open subsets of \(\mathbb{R}\). They cover \((0,1)\): if \(x\in(0,1)\), choose an integer \(n>1/x\). Then \(1/n<x<1\), so \(x\in V_n\).
If a finite selection is empty, it does not cover \((0,1)\). If the selection is nonempty, let \(N\) be its largest index. Since \(V_n\subseteq V_N\) whenever \(n\leq N\), the union of the selected intervals is \(V_N=(1/N,1)\). The point \(1/(2N)\) belongs to \((0,1)\), because \(N\geq2\), but it does not belong to \(V_N\), because \(1/(2N)<1/N\). Thus no finite selection covers \((0,1)\), and the interval is not compact.
These examples illustrate why checking only that a particular set is covered is not enough: the definition quantifies over every open cover. To disprove compactness, the cover must be chosen so that each finite selection leaves at least one point out.
Compact Sets Are Bounded
Proof. The empty set is bounded, so suppose \(K\) is nonempty and compact. Consider the family of open intervals \(\{(-n,n):n\text{ is a positive integer}\}\). The Archimedean property shows that this family covers \(\mathbb{R}\), and therefore covers \(K\). By compactness, a finite subfamily covers \(K\). Since \(K\ne\varnothing\), this subfamily cannot be empty. Let \(N\) be its largest index. The intervals are nested, so the union of the finite subfamily is \((-N,N)\). Consequently \(K\subseteq(-N,N)\), which proves that \(K\) is bounded. \(\square\)
This proof is a useful model for applying the definition: first choose an open cover adapted to the property you want to prove, then use a finite subcover to obtain a single bound. The crucial fact about this particular cover is that every finite selection lies inside one interval \((-N,N)\).
Compact Sets Are Closed
Proof. The empty set is closed, so let \(K\) be a nonempty compact set. We show that each \(x\in\mathbb{R}\setminus K\) has an open neighborhood disjoint from \(K\). Fix such an \(x\). For each \(y\in K\), let \(d_y=|x-y|>0\), and consider the two open intervals $$ B_{d_y/3}(y)\quad\text{and}\quad B_{d_y/3}(x). $$ They are disjoint. Indeed, if a point \(z\) belonged to both, the triangle inequality would give $$ |x-y|\leq |x-z|+|z-y|<\frac{d_y}{3}+\frac{d_y}{3}=\frac{2d_y}{3}, $$ contradicting \(d_y=|x-y|\).
The family \(\{B_{d_y/3}(y):y\in K\}\) is an open cover of \(K\), since \(y\in B_{d_y/3}(y)\) for every \(y\in K\). Compactness gives a finite subcover, say \(B_{d_{y_1}/3}(y_1),\ldots,B_{d_{y_m}/3}(y_m)\). Here \(m\geq1\), since \(K\) is nonempty. Set $$ \delta=\min_{1\leq i\leq m}\frac{d_{y_i}}{3}. $$ Each \(d_{y_i}\) is positive, and there are only finitely many of them, so \(\delta>0\).
We claim that \(B_\delta(x)\cap K=\varnothing\). If \(z\in B_\delta(x)\cap K\), the finite subcover puts \(z\) in some \(B_{d_{y_i}/3}(y_i)\). But then $$ |x-y_i|\leq |x-z|+|z-y_i|<\delta+\frac{d_{y_i}}{3}\leq\frac{2d_{y_i}}{3}, $$ again contradicting \(d_{y_i}=|x-y_i|\). Therefore \(B_\delta(x)\) is disjoint from \(K\). Every point outside \(K\) has such a neighborhood, so \(\mathbb{R}\setminus K\) is open. By the characterization of closed sets as complements of open sets, \(K\) is closed. \(\square\)
The proof separates the point \(x\notin K\) from each individual point \(y\in K\), then uses compactness to reduce the resulting cover of \(K\) to finitely many neighborhoods. A positive minimum radius can then be chosen. Without compactness, the pointwise separation radii need not have a positive lower bound that works for the entire set.
What the Definition Does—and Does Not—Say
Compactness is a global property expressed through open sets. It does not say that a set has only finitely many points: the set \(\{0\}\cup\{1/n:n\geq1\}\) is an infinite compact example. Nor does compactness mean that one open set from every cover must cover the whole set. It guarantees a finite subfamily, which may have more than one member.
The two theorems above give useful necessary conditions in the real line: compact sets are bounded and closed. They are not themselves the definition of compactness. Later results can investigate when these conditions are also sufficient; for now, the open-cover definition is the criterion to apply, and the examples show how a carefully chosen cover can settle the question directly.
Start with an arbitrary open cover, not just a convenient one, and explain how finitely many of its members cover the set.
Give one open cover and show that every finite selection misses a point; handle the empty selection separately.
Design a cover whose finite subcover yields a bounded interval or finitely many neighborhoods, then use compactness.
Check Your Understanding
Use the definition and proofs in this tutorial to answer the following questions.
- What distinguishes an open cover from a finite subcover?
- Why does the proof that finite sets are compact require only finitely many choices?
- In the cover of \((0,1)\) by \(V_n=(1/n,1)\), why does a nonempty finite selection have a largest index, and which point does its union miss?
- How does a finite subcover of \(\{(-n,n):n\geq1\}\) give one bound for a compact set?
- In the proof that compact sets are closed, why is it important that the finite subcover is nonempty?