From Sequence Control to Compactness
Completeness and total boundedness control two different steps in an argument about sequences. Total boundedness ensures that every sequence has a Cauchy subsequence; completeness ensures that this subsequence converges to a point of the space. Compactness, defined by finite subcovers of open covers, may at first seem to be a different kind of property. In metric spaces, these ideas fit together: the combination of completeness and total boundedness guarantees compactness.
The implication itself was established earlier in the course as the Complete and Totally Bounded Metric Spaces Are Compact theorem. Its proof route is worth keeping in view. The Sequential Characterization of Total Boundedness gives a Cauchy subsequence from any sequence. Completeness makes that subsequence convergent, so the space is sequentially compact. The Theorem that Sequentially Compact Metric Spaces Are Compact then gives compactness in the open-cover sense. Each step has a distinct job; neither completeness nor total boundedness alone supplies the whole argument.
A finite \(\varepsilon\)-net is a finite set of centers in the space such that every point is at distance strictly less than \(\varepsilon\) from at least one center. Total boundedness means that such a net exists for every \(\varepsilon>0\). The finite-net condition is much stronger than boundedness: it gives control at arbitrarily small scales, which is exactly what the subsequence argument needs.
A Criterion for Subsets of Complete Spaces
Suppose a metric space \(X\) is already known to be complete. For a subset \(A\subseteq X\), the criterion below reduces compactness to two properties that can often be checked directly: closedness in \(X\) and total boundedness in the restricted metric. It also explains why closedness matters: a Cauchy sequence in \(A\) converges in \(X\), and closedness ensures that its limit remains in \(A\).
Proof. First suppose \(A\) is compact. A metric space is Hausdorff, so the theorem that Compact Sets Are Closed in Hausdorff Spaces shows that \(A\) is closed in \(X\). The theorem that Compact Sets Are Totally Bounded shows that \(A\) is totally bounded.
Conversely, suppose \(A\) is closed in \(X\) and totally bounded. If \(A\) is empty, it is compact: the empty family is a finite subcover of the empty set. Otherwise, the Closed Subset of a Complete Space theorem shows that \(A\), with the restricted metric, is complete. It is also totally bounded by hypothesis. The Complete and Totally Bounded Metric Spaces Are Compact theorem therefore shows that \(A\) is compact. This proves both directions. \(\square\)
The necessity direction uses results about compact subsets directly. The sufficiency direction is where completeness of the ambient space is used: it passes to a closed subset, after which the complete-and-totally-bounded theorem applies. Without that ambient completeness assumption, closedness and total boundedness need not suffice.
Worked Examples: Applying the Criterion
Worked Example: A Closed Interval
Consider \(A=[2,5]\) inside the complete metric space \(\mathbb{R}\) with its usual metric. The interval is closed in \(\mathbb{R}\). To verify total boundedness, let \(\varepsilon>0\), and choose a positive integer \(N\) such that \(3/N<\varepsilon\). Define
Every center in \(F\) belongs to \([2,5]\). For any \(x\in[2,5]\), choose \(j\in\{0,\ldots,N-1\}\) with \(2+3j/N\leq x\leq 2+3(j+1)/N\); when \(x=5\), take \(j=N-1\). Then
Thus \(F\) is a finite \(\varepsilon\)-net. The subset criterion now gives that \([2,5]\) is compact.
Worked Example: A Closed Sequence Set
Let \(E=\{0\}\cup\{1/n:n\in\mathbb{N},\,n\geq1\}\subseteq\mathbb{R}\). We first check that \(E\) is closed. Suppose a sequence \((x_k)\) in \(E\) converges to \(x\in\mathbb{R}\). If some value \(a\in E\) occurs infinitely often, the corresponding constant subsequence converges to \(a\); since subsequences of convergent sequences have the same limit and metric limits are unique, \(x=a\in E\).
If no value occurs infinitely often, then, for each fixed positive integer \(M\), only finitely many terms of \((x_k)\) can belong to the finite set \(\{0,1,1/2,\ldots,1/M\}\). Consequently, we can choose a subsequence whose terms are \(1/n_j\) with \(n_j>M\) for its \(j\)-th selected term. In particular, choose the indices so that \(n_j\geq j\). Then \(0\leq1/n_j\leq1/j\), so this subsequence converges to \(0\). Uniqueness of limits gives \(x=0\in E\). The Sequential Characterization of Closed Sets therefore shows that \(E\) is closed.
To check total boundedness, fix \(\varepsilon>0\) and choose \(N\geq1\) with \(1/(N+1)<\varepsilon\). Use the finite set
The points \(0,1,\ldots,1/N\) are themselves centers. For any remaining point \(1/n\), where \(n>N\), its distance to the center \(0\) satisfies \(d(1/n,0)=1/n\leq1/(N+1)<\varepsilon\). Thus \(F\) is a finite \(\varepsilon\)-net. Since \(\mathbb{R}\) is complete, the criterion shows that \(E\) is compact.
Worked Example: An Incomplete Totally Bounded Space
Consider \(A=(0,1)\) with the usual metric. This space is totally bounded. Given \(\varepsilon>0\), choose an integer \(N\geq2\) so large that \(1/(N+1)<\varepsilon\), and use the centers \(j/(N+1)\) for \(j=1,\ldots,N\). They belong to \(A\), and the gaps between consecutive centers, as well as the gaps from the endpoints \(0\) and \(1\) to the nearest center, are at most \(1/(N+1)\). Since points of \(A\) do not equal the endpoints, every point is at distance strictly less than \(\varepsilon\) from one of these centers.
Nevertheless, \(A\) is not compact. For each integer \(n\geq2\), let \(U_n=(1/n,1)\), an open subset of \(A\) in its restricted metric. The family \(\{U_n:n\geq2\}\) covers \(A\): for any \(x\in(0,1)\), an integer \(n\) can be chosen with \(1/n<x\), and then \(x\in U_n\). But a finite collection of these sets has a largest index \(N\), and its union is \(U_N\), which misses, for example, \(1/(2N)\in A\). There is no finite subcover.
The sequence \(1/n\) is Cauchy in \(A\) but has no limit there, so \(A\) is not complete. This example shows why total boundedness alone is insufficient for compactness and why the closedness condition in the subset criterion cannot be dropped.
Finite Products of Compact Metric Spaces
The compactness theorem also gives a useful conclusion for products. Equip \(X\times Y\) with the maximum metric \(d((x_1,y_1),(x_2,y_2))=\max\{d_X(x_1,x_2),d_Y(y_1,y_2)\}\). The previous tutorial established that the product of two complete totally bounded spaces is complete and totally bounded in this metric. We can therefore transfer compactness from each factor to the product.
Proof. If either factor is empty, then \(X\times Y\) is empty and hence compact. Suppose both are nonempty. A compact metric space is complete: by Compact Metric Spaces Are Sequentially Compact, every sequence has a convergent subsequence; if the original sequence is Cauchy, the Cauchy Sequence with a Convergent Subsequence Converges theorem implies that the whole sequence converges. Thus \(X\) and \(Y\) are complete.
Both factors are totally bounded by the theorem that Compact Sets Are Totally Bounded. The product theorem from the previous tutorial now shows that \(X\times Y\), with the maximum metric, is complete and totally bounded. Applying the Complete and Totally Bounded Metric Spaces Are Compact theorem proves that \(X\times Y\) is compact. \(\square\)
Worked Example: The Unit Square
The interval \([0,1]\) is compact by the closed-interval argument, so the product theorem shows that \([0,1]\times[0,1]\) is compact in the maximum metric. For instance, the distance between \((1/5,3/4)\) and \((2/5,1/2)\) is
The maximum metric records the larger coordinate change. A finite net for each interval coordinate gives a finite net for the square, while completeness follows from convergence in each coordinate. The product theorem combines these two facts, and the complete-and-totally-bounded compactness theorem converts them into open-cover compactness.
What the Hypotheses Contribute
The results here give a practical order of attack. To prove a subset of a complete metric space compact, first establish closedness, then construct finite nets at an arbitrary positive scale. Closedness protects limits from leaving the set; total boundedness provides finite approximation at every scale. Once both have been checked, the compactness theorem supplies finite subcovers without requiring a separate argument for each open cover.
A common pitfall is to replace total boundedness with boundedness. Boundedness places the set inside one ball, but it does not guarantee finite covers by balls of every small radius. Another is to use total boundedness without checking completeness or closedness. The open-cover argument for \((0,1)\) demonstrates the consequence: a space may admit finite nets at every scale and still fail to have a finite subcover for a particular open cover.
Check Your Understanding
Use the compactness criterion and the examples above to answer the following questions.
- Which two conditions characterize compactness for a subset of a complete metric space?
- Where does ambient completeness enter the proof of the subset criterion?
- Why is it convenient to include the center \(0\) in the finite \(\varepsilon\)-net constructed for \(E=\{0\}\cup\{1/n:n\geq1\}\)?
- Why does the cover \(U_n=(1/n,1)\) of \((0,1)\) have no finite subcover?
- How do completeness and total boundedness of the factors lead to compactness of their product?