Three Tests for Compactness
In a metric space, compactness can be recognized in three different ways. One test asks whether every open cover has a finite subcover. Another asks whether every sequence has a convergent subsequence. A third combines two metric conditions: completeness and total boundedness. The equivalence is useful because the right test depends on the problem: open covers are natural in topology, while sequences, Cauchy sequences, and finite nets are often more convenient in metric arguments.
Throughout, \((X,d)\) is a metric space with its metric topology. An open ball is \(B(x,r)=\{y\in X:d(x,y)<r\}\), where \(r>0\). A sequence \((x_n)\) is Cauchy if for every \(\varepsilon>0\) there is \(N\in\mathbb{N}\) such that \(d(x_m,x_n)<\varepsilon\) whenever \(m,n\geq N\).
In the definition of total boundedness, the centers \(x_i\) belong to \(X\), and the balls are open. The number of centers and their positions may depend on \(\varepsilon\). Completeness is also relative to \(X\): a Cauchy sequence must converge to a point of the space, not merely to a point in some larger space.
The Implications Behind the Equivalence
The component implications have been established in the earlier material on compactness in metric spaces. We record them precisely so that the direction of each step is visible:
- Open-cover compactness implies sequential compactness.
- Sequential compactness implies open-cover compactness.
- Open-cover compactness implies completeness and total boundedness.
- Completeness together with total boundedness implies sequential compactness.
Each implication has a distinct role. The first two give the equivalence between the two compactness definitions. The third shows that compactness supplies both metric conditions. The fourth uses both conditions to recover sequential compactness; combining it with sequential compactness implying open-cover compactness recovers compactness.
Proof. Suppose first that \(X\) is compact. By the established implication from open-cover compactness to sequential compactness, \(X\) is sequentially compact. By the established implication from compactness to the metric conditions, \(X\) is also complete and totally bounded.
Now suppose \(X\) is sequentially compact. The established implication from sequential compactness to open-cover compactness shows that \(X\) is compact. Applying the compactness-to-metric-conditions implication then shows that \(X\) is complete and totally bounded.
Finally, suppose \(X\) is complete and totally bounded. The established implication from these two conditions to sequential compactness gives sequential compactness. Sequential compactness implies open-cover compactness, so \(X\) is compact. These arguments prove that each of (1), (2), and (3) implies the other two. \(\square\)
This proof identifies the needed direction in each use. In particular, completeness and total boundedness are used together to obtain sequential compactness; they are not interchangeable conditions. Conversely, compactness yields both conditions, not just one of them.
Checking the Tests on Explicit Spaces
Worked Example: A Three-Point Metric Space
Let \(X=\{0,1,3\}\) have the metric inherited from \(\mathbb{R}\), so \(d(x,y)=|x-y|\). We check the conditions directly.
Compactness. Let \(\mathcal{U}\) be any open cover of \(X\). For each of the three points, choose one member of \(\mathcal{U}\) containing that point. The at most three chosen sets cover \(X\), so \(\mathcal{U}\) has a finite subcover.
Sequential compactness. Given any sequence in \(X\), at least one of \(0,1,3\) occurs infinitely many times. Otherwise each occurs only finitely many times, and the whole sequence would have only finitely many terms, which is impossible. The terms equal to the value occurring infinitely often form a constant subsequence, which converges to that value in \(X\).
Completeness and total boundedness. Distinct points in \(X\) are at distance at least \(1\). For a Cauchy sequence, apply the Cauchy condition with \(\varepsilon=1\). From some index onward, all its terms have pairwise distance less than \(1\). Since distinct points of \(X\) have distance at least \(1\), all those terms must be equal. Thus the sequence is eventually constant and converges in \(X\). For any \(\varepsilon>0\), the three balls \(B(0,\varepsilon)\), \(B(1,\varepsilon)\), and \(B(3,\varepsilon)\) cover \(X\), since each contains its center. Hence \(X\) is totally bounded.
Check. The finite-subcover argument proves compactness directly, and each sequence has a convergent subsequence. The metric conditions also hold, in agreement with the theorem.
Worked Example: The Closed Unit Interval
Let \(X=[0,1]\) with the metric \(d(x,y)=|x-y|\). We verify completeness and total boundedness, then use the theorem to conclude compactness and sequential compactness.
Completeness. A Cauchy sequence in \(X\) is a Cauchy sequence in \(\mathbb{R}\), so it converges in \(\mathbb{R}\) because the real line is complete. Its limit belongs to \([0,1]\): if the limit \(x\) were less than \(0\), then for sufficiently large indices the sequence terms would be less than \(0\); if \(x\) were greater than \(1\), sufficiently late terms would be greater than \(1\). Both contradict that every term lies in \([0,1]\). Thus \(X\) is complete.
Total boundedness at a specified scale. Take \(\varepsilon=0.3\), and choose \(N=4\), so \(1/N=0.25<0.3\). Use the centers \(0,0.25,0.5,0.75,1\), all of which lie in \(X\). For \(x\in[0,1)\), let \(k\) be the greatest integer not exceeding \(4x\). Then \(k\in\{0,1,2,3\}\) and \(0\leq x-k/4<1/4=0.25<0.3\). The point \(x=1\) is itself one of the centers. Therefore these five open balls of radius \(0.3\) cover the interval.
Total boundedness at every scale. Given any \(\varepsilon>0\), choose \(N\in\mathbb{N}\) with \(1/N<\varepsilon\). The centers \(k/N\), for \(k=0,\ldots,N\), cover \([0,1]\): for \(x<1\), choose \(k\) to be the greatest integer not exceeding \(Nx\), giving \(0\leq x-k/N<1/N<\varepsilon\); the point \(1\) is a center. This proves total boundedness.
Conclusion and check. Completeness and total boundedness imply sequential compactness, which implies open-cover compactness. Thus \([0,1]\) is compact in its metric topology. For the specified scale, each point was shown to lie within a distance strictly less than \(0.3\) of a listed center, as required for open balls.
Worked Example: Rational Points in the Unit Interval
Let \(X=\mathbb{Q}\cap[0,1]\) with the metric inherited from \(\mathbb{R}\). We show that it is totally bounded but not complete, and use the theorem to see why it is not compact.
Total boundedness. Given \(\varepsilon>0\), choose \(N\in\mathbb{N}\) with \(1/N<\varepsilon\). The rational centers \(k/N\), for \(k=0,\ldots,N\), belong to \(X\). If \(x\in X\) and \(x<1\), take \(k\) to be the greatest integer not exceeding \(Nx\). Then \(0\leq x-k/N<1/N<\varepsilon\). If \(x=1\), the center \(1\) covers it. Thus \(X\) is totally bounded.
A Cauchy sequence with no limit in \(X\). Set \(\alpha=\sqrt{2}/2\), which is irrational and lies in \([0,1]\). For each \(n\in\mathbb{N}\), let \(a_n=\lfloor 10^n\alpha\rfloor/10^n\), where \(\lfloor t\rfloor\) is the greatest integer not exceeding \(t\). The value \(a_n\) is rational and lies in \([0,1]\), so \(a_n\in X\). The definition of the floor gives \(0\leq\alpha-a_n<10^{-n}\). For \(m,n\geq N\), it follows that \[ |a_m-a_n|\leq |a_m-\alpha|+|\alpha-a_n|<10^{-m}+10^{-n}\leq 2\cdot 10^{-N}. \] Given any \(\varepsilon>0\), choose \(N\) large enough that \(2\cdot10^{-N}<\varepsilon\). Then \(|a_m-a_n|<\varepsilon\) for all \(m,n\geq N\), so \((a_n)\) is Cauchy.
As a real sequence, \((a_n)\) converges to \(\alpha\), since \(|a_n-\alpha|<10^{-n}\). If it converged in \(X\) to some rational \(q\), it would also converge in \(\mathbb{R}\) to \(q\). A real sequence cannot have two different limits, so \(q=\alpha\), contradicting that \(q\) is rational and \(\alpha\) is irrational. Hence \(X\) is not complete.
Conclusion and check. The theorem says that a compact metric space must be complete, so this \(X\) is not compact. The same sequence has no subsequence converging in \(X\): every subsequence still converges to \(\alpha\) in \(\mathbb{R}\), so it cannot converge to a point of \(X\). This agrees with the equivalent sequential compactness test.
How to Choose a Test
The equivalence lets us select a proof strategy without changing the meaning of compactness. If the problem gives an open cover, the finite-subcover definition may be the most direct route. If it gives a sequence and asks for a convergent subsequence, sequential compactness is the natural language. If it gives distance estimates, Cauchy sequences, or finite approximations at arbitrarily small scales, completeness and total boundedness often provide the cleanest route.
A finite \(\varepsilon\)-net is another name for a finite set of centers whose open \(\varepsilon\)-balls cover the space. Total boundedness says such a net exists at every positive scale. The scale matters: a cover by finitely many balls of one fixed radius does not by itself establish total boundedness. The condition requires the ability to make the radius as small as desired.
Common pitfall. A frequent error is to use total boundedness alone as if it implied compactness. In the rational-interval example, the rational grid gives finite covers at every positive scale, but the Cauchy sequence \((a_n)\) has no limit in the space. The missing condition is completeness. The correct implication is completeness and total boundedness \(\Rightarrow\) sequential compactness \(\Rightarrow\) compactness. In the other direction, compactness gives both completeness and total boundedness.
The empty metric space causes no exception. It is compact because the empty subfamily is a finite subcover of the empty open cover; it is sequentially compact because there are no sequences in it; and it is complete and totally bounded, with the Cauchy and finite-cover conditions holding vacuously.
Check Your Understanding
For each question, name the implication or definition that justifies your answer.
- Which implication turns sequential compactness into open-cover compactness?
- What two conditions must be combined to infer sequential compactness from metric estimates?
- For a given \(\varepsilon>0\), what must a finite \(\varepsilon\)-net cover, and where must its centers lie?
- Why does the sequence in \(\mathbb{Q}\cap[0,1]\) rule out convergence in that space, even though it converges in \(\mathbb{R}\)?
- State the three equivalent tests for a metric space to be compact.