From Sequences to Open Covers
Sequential compactness and open-cover compactness describe different ways that a space cannot spread out indefinitely. Sequential compactness concerns sequences: every sequence has a convergent subsequence whose limit lies in the space. Open-cover compactness concerns collections of open sets: every open cover has a finite subcover. In metric spaces, the first condition implies the second. The proof has two parts: sequential compactness provides a finite net, and it also gives a Lebesgue number for the cover. Together, these let us select finitely many cover members that cover the entire space.
A Lebesgue number for an open cover \(\mathcal{U}\) of a metric space is a number \(\delta>0\) such that, for every \(x\in X\), some \(U\in\mathcal{U}\) contains the whole ball \(B(x,\delta)\). The cover member may depend on \(x\). The important point is that the same positive radius \(\delta\) works at every point.
Two Consequences of Sequential Compactness
Proof. If \(X=\varnothing\), the empty set is a finite \(\varepsilon\)-net, so suppose \(X\ne\varnothing\). Fix \(\varepsilon>0\) and suppose, to obtain a contradiction, that no finite \(\varepsilon\)-net exists. Choose \(x_1\in X\). Once \(x_1,\ldots,x_n\) have been chosen, their \(\varepsilon\)-balls do not cover \(X\), since they would otherwise form a finite \(\varepsilon\)-net. Thus choose \(x_{n+1}\in X\setminus\bigcup_{i=1}^n B(x_i,\varepsilon)\). For every \(i<j\), this construction gives \(d(x_i,x_j)\ge\varepsilon\), because \(x_j\notin B(x_i,\varepsilon)\). Sequential compactness gives a convergent subsequence. Every convergent sequence in a metric space is Cauchy: if its limit is \(x\), then for any \(\eta>0\), sufficiently late terms have distance less than \(\eta/2\) from \(x\), so any two such terms have distance less than \(\eta\) from each other by the triangle inequality. But no subsequence of \((x_n)\) is Cauchy, since distinct terms are at distance at least \(\varepsilon\). This contradiction proves the proposition. \(\square\)
Proof. If \(X=\varnothing\), every \(\delta>0\) satisfies the definition vacuously. Suppose \(X\ne\varnothing\), and assume for contradiction that no positive Lebesgue number exists. For each \(n\in\mathbb{N}\), the number \(1/n\) then fails to be a Lebesgue number. Consequently there is \(x_n\in X\) such that no member of \(\mathcal{U}\) contains \(B(x_n,1/n)\). By sequential compactness, some subsequence \((x_{n_k})\) converges to a point \(x\in X\). Since \(\mathcal{U}\) covers \(X\), choose \(U\in\mathcal{U}\) with \(x\in U\). Openness gives \(r>0\) such that \(B(x,r)\subseteq U\). For sufficiently large \(k\), both \(d(x_{n_k},x)<r/2\) and \(1/n_k<r/2\). If \(y\in B(x_{n_k},1/n_k)\), then \(d(y,x)\le d(y,x_{n_k})+d(x_{n_k},x)<1/n_k+r/2<r\). Thus \(B(x_{n_k},1/n_k)\subseteq B(x,r)\subseteq U\), contradicting the choice of \(x_{n_k}\). Therefore a Lebesgue number exists. \(\square\)
The Finite-Subcover Argument
Proof. Let \((X,d)\) be sequentially compact and let \(\mathcal{U}\) be any open cover of \(X\). If \(X=\varnothing\), the empty subcollection is a finite subcover. Otherwise, Proposition 2 gives a Lebesgue number \(\delta>0\) for \(\mathcal{U}\). By Proposition 1, there is a finite \(\delta\)-net \(F=\{c_1,\ldots,c_m\}\subseteq X\). For each \(i\), the Lebesgue number property supplies a member \(U_i\in\mathcal{U}\) such that \(B(c_i,\delta)\subseteq U_i\). Because \(F\) is a \(\delta\)-net, every \(x\in X\) belongs to \(B(c_i,\delta)\) for at least one \(i\). It follows that \(x\in U_i\) for that \(i\). Hence \(X\subseteq\bigcup_{i=1}^m U_i\). Each \(U_i\) is a member of the original cover, so this finite collection covers \(X\) and is a finite subcover. \(\square\)
The two ingredients have separate jobs. The Lebesgue number says that a ball around each center fits inside a cover member. The finite net says that finitely many of those balls reach every point of the space. Neither ingredient alone completes the argument: a uniform ball size without finitely many centers does not produce a finite selection, and finitely many centers do not help unless their surrounding balls fit inside cover members.
Worked Example: A Three-Point Metric Space
Let \(X=\{p,q,r\}\) with the metric \(d(x,y)=1\) when \(x\ne y\), and \(d(x,x)=0\). Consider the open cover \(\mathcal{U}=\{U_1,U_2\}\), where \(U_1=\{p,q\}\) and \(U_2=\{q,r\}\). In this metric, every singleton is open, since \(B(x,1/2)=\{x\}\); consequently every subset of \(X\) is open.
Every sequence in \(X\) has a constant subsequence: among any three possible values, at least one value occurs infinitely many times. A constant subsequence converges to its repeated value. Thus \(X\) is sequentially compact. Set \(\delta=1/2\). For each \(x\in X\), the ball \(B(x,\delta)=\{x\}\) lies in a member of the cover: \(B(p,\delta)\subseteq U_1\), \(B(q,\delta)\subseteq U_1\), and \(B(r,\delta)\subseteq U_2\). The set of centers \(F=\{p,q,r\}\) is a finite \(\delta\)-net because the three balls are \(\{p\},\{q\},\{r\}\), whose union is \(X\). Selecting the indicated cover members gives \(U_1\) and \(U_2\), and \(U_1\cup U_2=\{p,q\}\cup\{q,r\}=\{p,q,r\}=X\).
Check: \(U_1\) alone misses \(r\), and \(U_2\) alone misses \(p\), so both cover members are needed for this cover. The finite-net and ball-containment steps produce exactly that finite subcover.
Worked Example: A Cover of the Unit Interval
Give \(X=[0,1]\) the metric inherited from \(\mathbb{R}\), and consider the subspace-open sets \(U_0=(-1/10,3/5)\cap[0,1]=[0,3/5)\) and \(U_1=(2/5,11/10)\cap[0,1]=(2/5,1]\). The interval \([0,1]\) is sequentially compact by the Bolzano–Weierstrass theorem for bounded real sequences: a sequence in \([0,1]\) has a convergent subsequence, and its limit lies in the closed interval \([0,1]\).
First check the cover, including the endpoint \(3/5\). If \(x<3/5\), then \(x\in U_0\). If \(x\ge 3/5\), then \(x>2/5\), and \(x\le1<11/10\), so \(x\in U_1\). Thus \(U_0\cup U_1=[0,1]\). In particular, \(3/5\notin U_0\) but \(3/5\in U_1\).
We can take \(\delta=1/10\) as a Lebesgue number. If \(x\le1/2\) and \(y\in B(x,1/10)\), then \(0\le y<x+1/10\le3/5\), so \(y\in U_0\). If \(x>1/2\) and \(y\in B(x,1/10)\), then \(y>x-1/10>2/5\) and \(y\le1\), so \(y\in U_1\). The strict inequalities in the open-ball condition ensure that the endpoints excluded by \(U_0\) and \(U_1\) cause no problem.
For a finite \(1/10\)-net, take \(F=\{j/10:j=0,1,\ldots,10\}\). If \(x<1\), choose \(j\) with \(j/10\le x<(j+1)/10\); then \(0\le x-j/10<1/10\), so \(x\in B(j/10,1/10)\). The remaining point \(x=1\) is the center \(10/10\), and hence also belongs to one of these balls. Every center \(j/10\le1/2\) has its \(1/10\)-ball contained in \(U_0\), while every center \(j/10>1/2\) has its ball contained in \(U_1\), by the bounds just proved. Selecting these members yields the finite subcover \(\{U_0,U_1\}\).
Check: the centers \(0\) and \(6/10\) are among the specified net points; \(0\in U_0\) and \(6/10>1/2\), so the selection procedure includes both \(U_0\) and \(U_1\). Their union covers \([0,1]\), including \(3/5\), which lies in \(U_1\).
Why the Ball-Containment Condition Matters
A frequent error is to take a finite net, choose for each center any cover member containing that center, and conclude that the chosen members cover the space. A center lying in a set does not imply that the whole net ball around it lies in that set. The proof needs the stronger Lebesgue-number guarantee.
Worked Example: A Net Center Does Not Control Its Whole Ball
Let \(X=\{0,1,2\}\) with the metric inherited from \(\mathbb{R}\), and let \(U=\{0,1\}\) and \(V=\{1,2\}\). These are open in \(X\), since for each point \(x\in X\), the ball \(B(x,1/2)=\{x\}\) is open. The collection \(\{U,V\}\) covers \(X\).
Take the single center \(1\) and radius \(11/10\). Its open ball is \(B(1,11/10)=\{0,1,2\}\), because the distances from \(1\) to \(0\) and to \(2\) are both \(1<11/10\). This one ball is a finite \(11/10\)-net. But choosing \(U\) merely because \(1\in U\) does not cover \(X\): \(U=\{0,1\}\) misses \(2\). The actual containment needed by the proof fails, since \(B(1,11/10)=X\) is not a subset of either \(U\) or \(V\).
Using radius \(1/2\) instead, the three centers \(0,1,2\) form a finite \(1/2\)-net, and each ball is a singleton contained in a cover member: \(B(0,1/2)=\{0\}\subseteq U\), \(B(1,1/2)=\{1\}\subseteq U\), and \(B(2,1/2)=\{2\}\subseteq V\). Selecting \(U\) and \(V\) then covers all of \(X\). The correction is to use the Lebesgue number to ensure ball containment, not just membership of the center.
The metric hypothesis is also essential to this particular implication. For contrast, let \(\mathbb{N}\) have the discrete metric, where distinct integers have distance \(1\). The cover by singleton sets \(\{\{n\}:n\in\mathbb{N}\}\) has no finite subcover: any finite selection contains only finitely many integers and misses another integer. The sequence \(1,2,3,\ldots\) has no convergent subsequence, since convergence in this metric forces a sequence to be eventually constant. Thus this example is not sequentially compact, so it does not contradict the theorem; it shows why the hypothesis cannot simply be omitted from the claim.
Check Your Understanding
Use the definitions of finite net, Lebesgue number, and open-cover compactness to answer these questions.
- In the proof of Proposition 1, why does the assumption that no finite \(\varepsilon\)-net exists allow the recursive choice of \(x_{n+1}\)?
- In the Lebesgue number proof, explain why \(d(x_{n_k},x)<r/2\) and \(1/n_k<r/2\) imply \(B(x_{n_k},1/n_k)\subseteq B(x,r)\).
- For the cover of \([0,1]\), which set contains \(3/5\), and why is it not in the other set?
- In the three-point example, find a finite subcover and state which point would be missed if either one of its two members were removed.
- For the discrete metric on \(\mathbb{N}\), verify directly that the sequence \(1,2,3,\ldots\) has no convergent subsequence.