Open-Cover Compactness

From Metric Estimates to Finite Subcovers

In the previous tutorial, we saw how total boundedness and completeness combine to control sequences: total boundedness provides a Cauchy subsequence, and completeness ensures that it converges in the space. Open-cover compactness expresses a related kind of finite control, but it begins with an arbitrary family of open sets rather than a sequence. The goal here is to connect these viewpoints directly.

Definition: A collection \(\mathcal{U}\) of subsets of a metric space \(X\) is an open cover of \(X\) if every member of \(\mathcal{U}\) is open in \(X\) and \(\bigcup_{U\in\mathcal{U}}U=X\). The space \(X\) is compact if every open cover of \(X\) has a finite subcollection that still covers \(X\). Such a subcollection is called a finite subcover.

The word “open” is relative to the space under consideration. For a subset \(X\) of a larger metric space, an open set in \(X\) need not be open in that larger space. Also, compactness does not say that an open cover is itself finite; it says that some finite part of it suffices.

Earlier, the tutorial Compactness in Metric Spaces established that an open-cover compact metric space is complete and totally bounded. We now prove the converse. The proof turns a hypothetical open cover with no finite subcover into a nested sequence of smaller sets. Their diameters shrink, so completeness forces points chosen from them to converge. Openness then contradicts the assumption that no finite subcover exists.

Completeness and Total Boundedness Give Compactness

Theorem (Complete and Totally Bounded Metric Spaces Are Compact): A metric space is compact in the open-cover sense if and only if it is complete and totally bounded.

Proof. The implication from compactness to completeness and total boundedness was established in Compactness in Metric Spaces. For the converse, let \(X\) be complete and totally bounded. If \(X\) is empty, every open cover has the empty finite subcover, so \(X\) is compact. Suppose from now on that \(X\) is nonempty, and let \(\mathcal{U}\) be an open cover of \(X\).

Assume, seeking a contradiction, that \(\mathcal{U}\) has no finite subcover. We construct nested nonempty sets \(A_n\subseteq X\), none of which can be covered by finitely many members of \(\mathcal{U}\). Set \(A_0=X\), which has this property by assumption.

For each integer \(n\geq1\), total boundedness gives finitely many centers \(c_{n,1},\ldots,c_{n,m_n}\) such that the open balls of radius \(2^{-n}\) around those centers cover \(X\). The corresponding closed balls also cover \(X\):

X\subseteq\bigcup_{j=1}^{m_n}\{x\in X:d(x,c_{n,j})\leq 2^{-n}\}.

Suppose \(A_{n-1}\) has been chosen and cannot be covered by finitely many members of \(\mathcal{U}\). At least one intersection of \(A_{n-1}\) with these finitely many closed balls also cannot be covered by finitely many members of \(\mathcal{U}\). Otherwise, each intersection would have a finite cover from \(\mathcal{U}\), and combining those finitely many covers would give a finite cover of \(A_{n-1}\). Choose such an intersection and call it \(A_n\). Thus \(A_n\) is nonempty, \(A_n\subseteq A_{n-1}\), and \(A_n\) lies in one closed ball of radius \(2^{-n}\).

Choose \(x_n\in A_n\) for each \(n\geq1\). If \(m,k\geq n\), nesting gives \(x_m,x_k\in A_n\). Since \(A_n\) lies in a closed ball of radius \(2^{-n}\), the triangle inequality yields

d(x_m,x_k)\leq 2^{1-n}.

Given \(\varepsilon>0\), choose \(N\) so large that \(2^{1-N}<\varepsilon\). Then \(d(x_m,x_k)<\varepsilon\) whenever \(m,k\geq N\). Hence \((x_n)\) is Cauchy. By completeness, it converges to some \(x\in X\).

Because \(\mathcal{U}\) covers \(X\), some \(U\in\mathcal{U}\) contains \(x\). Since \(U\) is open, there is \(r>0\) such that \(B_r(x)\subseteq U\). Choose \(n\) large enough that \(d(x_n,x)<r/2\) and \(2^{1-n}<r/2\). For any \(y\in A_n\), both \(y\) and \(x_n\) lie in \(A_n\), so \(d(y,x_n)\leq 2^{1-n}\). Consequently,

d(y,x)\leq d(y,x_n)+d(x_n,x)<r.

This holds for every \(y\in A_n\), so \(A_n\subseteq B_r(x)\subseteq U\). The single member \(U\) of the cover covers \(A_n\), contradicting how \(A_n\) was chosen. Therefore \(\mathcal{U}\) has a finite subcover, and \(X\) is compact. \(\square\)

The shrinking sets are the key to the argument. Total boundedness lets us repeatedly localize the part of the space that still resists a finite subcover. Their shrinking diameters make the chosen points Cauchy; completeness supplies a limit, and openness gives a neighborhood of that limit that eventually contains one entire set in the construction.

Worked Examples: Reading Open Covers

Worked Example: A Finite Metric Space Is Compact

Let \(X=\{a,b,c\}\) with any metric, and let \(\mathcal{U}\) be an open cover of \(X\). Since \(a\in X\), some \(U_a\in\mathcal{U}\) contains \(a\). Choose \(U_b\in\mathcal{U}\) containing \(b\), and \(U_c\in\mathcal{U}\) containing \(c\). Then \(U_a,U_b,U_c\) cover \(X\), so they form a finite subcover. If two of these sets coincide, the subcover is even smaller. This argument uses only that \(X\) has finitely many points; no special feature of the metric is needed.

Worked Example: A Convergent Sequence Together with Its Limit

Consider \(K=\{0\}\cup\{1/n:n\in\mathbb{N},\,n\geq1\}\) with the usual metric inherited from \(\mathbb{R}\). Let \(\mathcal{U}\) be an open cover of \(K\), and choose \(U_0\in\mathcal{U}\) with \(0\in U_0\). Openness in \(K\) means that for some \(\varepsilon>0\),

K\cap(-\varepsilon,\varepsilon)\subseteq U_0.

Choose a positive integer \(N\) with \(1/N<\varepsilon\). For every \(n>N\), we have \(0<1/n<\varepsilon\), so \(1/n\in U_0\). Only the finitely many points \(1,1/2,\ldots,1/N\) remain. For each of these points choose one member of \(\mathcal{U}\) containing it. Together with \(U_0\), these finitely many cover members cover all of \(K\). Thus \(K\) is compact.

Worked Example: An Unbounded Space Has a Cover with No Finite Subcover

Let \(X=(0,\infty)\) with the usual metric. For each positive integer \(n\), set \(U_n=(0,n)\). Each \(U_n\) is open in \(X\), and their union is \(X\): given \(x>0\), an integer \(n>x\) satisfies \(x\in U_n\).

Any finite selection \(U_{n_1},\ldots,U_{n_k}\) has a largest index \(N\). Since the sets increase with their indices, their union is \(U_N=(0,N)\). It does not cover \(X\), because \(N+1\in X\) but \(N+1\notin U_N\). The open cover \(\{U_n:n\geq1\}\) therefore has no finite subcover, so \(X\) is not compact.

The Lebesgue Number Lemma

Compactness does more than guarantee a finite subcover. It also ensures that an open cover has a uniform scale: sufficiently small subsets of the space fit inside a single member of the cover. The number that controls this scale is called a Lebesgue number.

Definition: A positive number \(\delta\) is a Lebesgue number for an open cover \(\mathcal{U}\) of a metric space \(X\) if every nonempty subset \(E\subseteq X\) with diameter less than \(\delta\) is contained in some member of \(\mathcal{U}\). The diameter is \(\operatorname{diam}(E)=\sup\{d(u,v):u,v\in E\}\).

Theorem (Lebesgue Number Lemma): Every open cover of a compact metric space has a Lebesgue number.

Proof. Let \(X\) be a nonempty compact metric space and \(\mathcal{U}\) an open cover. For each \(x\in X\), choose a set \(U_x\in\mathcal{U}\) containing \(x\). Since \(U_x\) is open, choose \(s_x>0\) such that \(B_{s_x}(x)\subseteq U_x\). Put \(r_x=s_x/2\). The balls \(B_{r_x}(x)\), for \(x\in X\), form an open cover of \(X\). Compactness gives a finite subcover, say \(B_{r_{x_1}}(x_1),\ldots,B_{r_{x_m}}(x_m)\).

Set \(\delta=\min\{r_{x_1},\ldots,r_{x_m}\}\), which is positive. Let \(E\subseteq X\) be nonempty with \(\operatorname{diam}(E)<\delta\), and choose \(z\in E\). Since the finite balls cover \(X\), there is an index \(i\) for which \(z\in B_{r_{x_i}}(x_i)\). For any \(y\in E\), the diameter bound and \(\delta\leq r_{x_i}\) give

d(y,x_i)\leq d(y,z)+d(z,x_i)<\delta+r_{x_i}\leq 2r_{x_i}=s_{x_i}.

Thus every \(y\in E\) lies in \(B_{s_{x_i}}(x_i)\subseteq U_{x_i}\), so \(E\subseteq U_{x_i}\). The number \(\delta\) is a Lebesgue number for \(\mathcal{U}\). \(\square\)

The empty space causes no exception: every positive \(\delta\) satisfies the defining condition, since there are no nonempty subsets to test. The lemma is useful when a cover must be handled uniformly at many points. Instead of choosing a different neighborhood size at each point, one positive number works across the entire compact space.

Why Uniform Scale Matters

A common mistake is to infer a Lebesgue number merely from the fact that every point belongs to some open set in the cover. Openness provides a suitable radius around each point, but those radii may vary and may become arbitrarily small across a noncompact space. Compactness is what allows the pointwise neighborhoods to be reduced to finitely many, after which their positive radii have a positive minimum.

The lemma concerns the diameter of an entire subset, not just the distance of one selected point from the boundary of a cover member. If a subset has diameter less than \(\delta\), the proof first places one of its points in a smaller ball and then uses the diameter bound to place every other point of that subset in a larger ball contained in one cover member.

Choose local neighborhoods.
At each point, use openness to find a ball contained in one member of the cover.

Use compactness.
Select finitely many smaller balls that still cover the space.

Take a positive minimum.
The smallest of the finitely many radii gives a uniform scale for all small-diameter subsets.

Together, the open-cover criterion and the Lebesgue number lemma express two complementary forms of finite control. Completeness and total boundedness guarantee that every open cover has a finite subcover; compactness then supplies a uniform size below which subsets cannot spread across multiple cover members.

Check Your Understanding

Use the definitions and proofs to answer the following questions.

What does it mean for a collection of open sets to have a finite subcover?
In the proof that completeness and total boundedness imply compactness, why can the chosen points be shown to form a Cauchy sequence?
Where does openness enter the contradiction in that proof?
Why does the cover \(U_n=(0,n)\) of \((0,\infty)\) have no finite subcover?
What property of the finite subcover is used to construct a Lebesgue number?
Does a Lebesgue number require each point to have the same distance from the boundary of its cover member? Explain the role of the finite selection instead.

What You'll Learn