Compact Sets Are Totally Bounded

Compactness Supplies Finite Approximation

Total boundedness asks for a finite approximation at every positive scale: for each \(\varepsilon>0\), finitely many balls of radius \(\varepsilon\) must cover the set. Compactness is defined differently. It says that every open cover has a finite subcover. The connection between these conditions comes from considering a particular open cover: all the balls of one fixed radius, centered at points of the set.

Recall that a finite \(\varepsilon\)-net for \(K\) is a finite set of points in \(K\) such that every point of \(K\) is at distance less than \(\varepsilon\) from at least one of them. The centers are required to belong to \(K\). This requirement matters: a finite cover by balls centered somewhere in a larger ambient space is not, by itself, a net for \(K\) in this sense.

Theorem: Every compact subset of a metric space is totally bounded.

Proof. Let \(K\) be compact in a metric space \((X,d)\). If \(K\) is empty, it is totally bounded by the convention that the empty set has the empty finite net. Suppose, then, that \(K\) is nonempty, and fix \(\varepsilon>0\).

For each \(x\in K\), the ball \(B_\varepsilon(x)\) is open in \(X\), and hence \(B_\varepsilon(x)\cap K\) is open in the subspace \(K\). These sets cover \(K\), because each \(x\in K\) belongs to its own ball: \(d(x,x)=0<\varepsilon\). Compactness gives a finite subcover, say

K\subseteq B_\varepsilon(x_1)\cup\cdots\cup B_\varepsilon(x_m), \qquad x_1,\ldots,x_m\in K.

The centers \(x_1,\ldots,x_m\) belong to \(K\), and every point of \(K\) lies in at least one of the displayed balls. Thus \(\{x_1,\ldots,x_m\}\) is a finite \(\varepsilon\)-net for \(K\). Since the argument works for every \(\varepsilon>0\), \(K\) is totally bounded. \(\square\)

The proof uses compactness separately at each chosen scale. It does not claim that one finite set of centers works for every scale. As \(\varepsilon\) becomes smaller, the finite subcover—and therefore the net—may need more centers.

Worked Examples of Finite Nets

Worked Example: A Convergent Sequence Together with Its Limit

Consider

K=\{0\}\cup\{1/n:n\in\mathbb{N},\ n\geq1\}

in the real line with its usual metric. First, \(K\) is compact. Let \(\mathcal{U}\) be any open cover of \(K\), and choose a member \(U_0\in\mathcal{U}\) containing \(0\). Since \(U_0\) is open in the subspace \(K\), there is a \(\delta>0\) such that \(K\cap(-\delta,\delta)\subseteq U_0\). Choose a positive integer \(N\) large enough that \(1/(N+1)<\delta\). Whenever \(n>N\), we have \(0<1/n\leq1/(N+1)<\delta\), so \(1/n\in U_0\). Only the finitely many points \(1,1/2,\ldots,1/N\) remain. For each of them choose one member of \(\mathcal{U}\) containing it. Together with \(U_0\), these finitely many sets cover \(K\). Thus every open cover has a finite subcover, proving compactness.

Now fix \(\varepsilon>0\). Choose a positive integer \(N\) such that \(1/(N+1)<\varepsilon\), and take

F=\{0,1,1/2,\ldots,1/N\}\subseteq K.

Each of \(0,1,\ldots,1/N\) is itself a center in \(F\), so its distance to \(F\) is zero. If \(n>N\), then

d(1/n,0)=1/n\leq1/(N+1)<\varepsilon.

Therefore every point of \(K\) lies within \(\varepsilon\) of a point of \(F\). This gives an explicit finite net; the centers near the limit handle the entire tail of the sequence.

Worked Example: Equally Spaced Points on the Unit Circle

Let \(C=\{(\cos t,\sin t):0\leq t\leq2\pi\}\), with the Euclidean metric. The map \(t\mapsto(\cos t,\sin t)\) is continuous, and \([0,2\pi]\) is compact. By the theorem on continuous images of compact sets, \(C\) is compact. We can also describe finite nets directly.

Fix \(\varepsilon>0\), and choose a positive integer \(N\) so large that \(2\pi/N<\varepsilon\). Use the \(N\) points

p_j=\left(\cos\frac{2\pi j}{N},\sin\frac{2\pi j}{N}\right), \qquad j=0,\ldots,N-1.

For a point on the circle, choose its angle \(t\in[0,2\pi]\). If \(t<2\pi\), there is an index \(j\in\{0,\ldots,N-1\}\) such that \(2\pi j/N\leq t<2\pi(j+1)/N\). If \(t=2\pi\), the corresponding point is \(p_0\). In the first case, put \(s=2\pi j/N\); then \(0\leq t-s<2\pi/N\). The distance between the two circle points satisfies

\begin{aligned} d\big((\cos t,\sin t),(\cos s,\sin s)\big) &=\sqrt{(\cos t-\cos s)^2+(\sin t-\sin s)^2}\\ &=2\left|\sin\frac{t-s}{2}\right|\\ &\leq |t-s|\\ &<\frac{2\pi}{N}<\varepsilon. \end{aligned}

Here the identity follows by expanding the squares and using \(\cos(t-s)=\cos t\cos s+\sin t\sin s\); the inequality uses \(|\sin u|\leq|u|\). The endpoint \(t=2\pi\) represents the same point as \(s=0\), so its distance to \(p_0\) is zero. Consequently the displayed \(N\) points form a finite \(\varepsilon\)-net for \(C\).

Separated Points Cannot Accumulate Without Bound

A finite net does more than approximate every point. It also limits how many points can remain a fixed positive distance apart. Such a restriction is useful when trying to show that a proposed compact set cannot contain infinitely many well-separated points.

Definition: A subset \(S\) of a metric space is \(\varepsilon\)-separated, for \(\varepsilon>0\), if \(d(x,y)\geq\varepsilon\) whenever \(x,y\in S\) are distinct.

Theorem: Every \(\varepsilon\)-separated subset of a compact metric space is finite. More precisely, if \(F\) is a finite \(\varepsilon/2\)-net for \(K\), then every \(\varepsilon\)-separated subset of \(K\) has at most \(|F|\) points.

Proof. By the theorem that compact metric sets are totally bounded, \(K\) has a finite \(\varepsilon/2\)-net \(F\subseteq K\). For each \(x\) in an \(\varepsilon\)-separated subset \(S\), choose a center \(a(x)\in F\) such that \(d(x,a(x))<\varepsilon/2\). No two distinct points \(x,y\in S\) can have the same chosen center. If they did, the triangle inequality would give

d(x,y)\leq d(x,a(x))+d(a(y),y) =d(x,a(x))+d(a(x),y) <\frac{\varepsilon}{2}+\frac{\varepsilon}{2} =\varepsilon,

contradicting \(d(x,y)\geq\varepsilon\). Thus the assignment \(x\mapsto a(x)\) sends distinct points of \(S\) to distinct members of the finite set \(F\). It follows that \(S\) is finite and \(|S|\leq|F|\). \(\square\)

Worked Example: Compact Subsets of a Discrete Metric Space

Let \(X\) be an infinite set with the discrete metric \(\delta\), where \(\delta(x,y)=1\) for distinct points and \(\delta(x,x)=0\). Suppose \(K\subseteq X\) is compact. Every singleton \(\{x\}\) is open in \(X\), since it is the ball \(B_{1/2}(x)\). The family of singletons \(\{\{x\}:x\in K\}\) is therefore an open cover of \(K\). If \(K\) were infinite, no finite selection of these singletons could cover it. Compactness rules this out, so \(K\) must be finite.

For a concrete scale, take \(\varepsilon=1/2\). Distinct points of \(K\) have distance \(1\), so \(K\) is \(1/2\)-separated. The separated-set theorem shows that \(K\) is finite. Conversely, if \(K\) is finite, its points themselves form a finite \(\varepsilon\)-net for every \(\varepsilon>0\), since each point has distance zero from itself. Also, every finite set is compact: from any open cover, choose for each point one member containing it; these finitely many members cover the set. Thus in a discrete metric space the compact subsets are exactly the finite subsets.

What the Implication Does—and Does Not—Say

Compactness guarantees finite nets because balls of a fixed positive radius form an open cover. The finite-subcover property then turns that possibly infinite family into a finite one. The proof works at every scale, which is precisely what total boundedness requires. The centers are automatically in the set because the cover was formed using a ball centered at each point of the set.

Do not confuse this conclusion with ordinary boundedness. Boundedness supplies one ball of some finite radius containing the set; it does not guarantee finite covers by balls of arbitrarily small radius. The infinite discrete metric space is bounded, but its balls of radius \(1/2\) are singletons, so no finite collection of them covers the space. Total boundedness is the stronger finite-approximation condition.

The separated-set theorem gives another way to recognize the force of compactness: at any fixed positive separation, only finitely many points can fit in a compact metric set. Its proof depends on using net balls of radius \(\varepsilon/2\), not \(\varepsilon\). Two points in the same open ball of radius \(\varepsilon/2\) are at distance strictly less than \(\varepsilon\), which contradicts \(\varepsilon\)-separation. The strict inequality is essential to this argument.

Total boundedness is one of the two metric properties that, together with completeness, characterize compactness. Here the direction runs from compactness to total boundedness. This connection lets open-cover compactness produce a concrete approximation tool, and that tool can then be used in sequence arguments and in estimates that depend on controlling all points at a chosen scale.

Check Your Understanding

Use the compactness argument and the finite-net properties developed here to answer the following questions.

Why do the balls centered at every point of a compact set form an open cover for each fixed positive radius?
How does a finite subcover of those balls give a net whose centers belong to the compact set?
In the convergent-sequence example, why can one center at the limit handle all sufficiently late terms?
Why does an \(\varepsilon/2\)-net have at most one point from an \(\varepsilon\)-separated set in each of its balls?
Why does compactness force a subset of a discrete metric space to be finite?