Compactness in Euclidean Space
In Euclidean space, compactness can often be decided by two geometric checks: whether a set is closed and whether it is bounded. These checks are separate. A set can be bounded but not closed, or closed but unbounded; either failure rules out compactness. The Heine-Borel theorem combines the two checks into an exact criterion for subsets of \(\mathbb{R}^n\).
Throughout, \(\mathbb{R}^n\) has its standard topology from the Euclidean metric \(d(x,y)=\|x-y\|\). A set \(A\subseteq\mathbb{R}^n\) is bounded if there is some \(M>0\) such that \(\|x\|\le M\) for every \(x\in A\). It is closed if its complement is open in \(\mathbb{R}^n\). A set is compact if every cover of it by open sets has a finite subcover.
For a subset of \(\mathbb{R}^n\), the criterion has two directions to keep in view. If a set is closed and bounded, Heine-Borel guarantees compactness. If either closedness or boundedness fails, the set is not compact. The theorem applies to subsets of finite-dimensional Euclidean space with its standard topology; it is not a general criterion for arbitrary metric spaces.
What Compact Subsets Must Satisfy
The following facts help explain why the two tests are natural. We give direct proofs from the open-cover definition. They also provide ways to rule out compactness without trying to construct a finite subcover for an arbitrary cover.
Proof. If \(K=\varnothing\), it is closed and bounded, so suppose \(K\ne\varnothing\). For boundedness, consider the open cover \(\{B(0,m):m\in\mathbb{N}\}\) of \(\mathbb{R}^n\), where \(B(0,m)=\{x:\|x\|<m\}\). It covers \(K\), since every vector has finite norm. Compactness gives a finite subcover of \(K\). If \(N\) is the largest index among its finitely many balls, then every point of \(K\) belongs to one of those balls and hence has norm less than \(N\). Thus \(K\) is bounded.
To prove that \(K\) is closed, fix \(p\in\mathbb{R}^n\setminus K\). For each \(x\in K\), let \(r_x=\|x-p\|/3\), which is positive because \(x\ne p\). The balls \(B(x,r_x)\), as \(x\) ranges over \(K\), form an open cover of \(K\). Choose a finite subcover \(B(x_1,r_{x_1}),\ldots,B(x_j,r_{x_j})\). The set \(V=\bigcap_{i=1}^j B(p,r_{x_i})\) is an open neighborhood of \(p\). For each \(i\), the two balls \(B(x_i,r_{x_i})\) and \(B(p,r_{x_i})\) are disjoint: if \(z\) belonged to both, the triangle inequality would give \(\|x_i-p\|<2r_{x_i}=2\|x_i-p\|/3\), a contradiction. So \(V\) misses every ball in the finite cover and therefore misses \(K\). Every point outside \(K\) has an open neighborhood missing \(K\), which means \(\mathbb{R}^n\setminus K\) is open. Hence \(K\) is closed. \(\square\)
Proof. Let \(\{U_\alpha\}\) be any cover of \(C\) by sets open in \(C\). By the definition of the subspace topology, for each \(\alpha\) there is an open set \(V_\alpha\) in \(K\) such that \(U_\alpha=C\cap V_\alpha\). The sets \(V_\alpha\), together with \(K\setminus C\), form an open cover of \(K\), because \(C\) is closed in \(K\). Compactness of \(K\) gives a finite subcover. Removing \(K\setminus C\) from that finite list leaves finitely many \(V_\alpha\) that cover \(C\); their corresponding \(U_\alpha\) cover \(C\) as well. Thus every open cover of \(C\) has a finite subcover, so \(C\) is compact. This argument also covers \(C=\varnothing\), which has the empty finite subcover. \(\square\)
Open Balls and Spheres
Worked Example: An Open Ball Is Not Compact
Consider the open ball \(B(0,2)=\{x\in\mathbb{R}^2:\|x\|<2\}\). It is bounded: every \(x\) in it satisfies \(\|x\|<2\). But it is not closed. To check this directly, define \(x_k=(2-1/k,0)\) for \(k\in\mathbb{N}\). Since \(1/k>0\), we have \(\|x_k\|=2-1/k<2\), so \(x_k\in B(0,2)\) for every \(k\). The sequence converges to \(p=(2,0)\), because \(\|x_k-p\|=1/k\to0\). Yet \(\|p\|=2\), so \(p\notin B(0,2)\). Thus the ball does not contain this limit and is not closed. By Heine-Borel, \(B(0,2)\) is not compact.
Check: the corresponding closed ball \(\{x:\|x\|\le2\}\) is bounded by its defining inequality and is closed. For closedness, if \(\|x\|>2\), then every \(y\) with \(\|y-x\|<(\|x\|-2)/2\) satisfies \(\|y\|\ge\|x\|-\|y-x\|>(\|x\|+2)/2>2\). The complement is therefore open. Heine-Borel says the closed ball is compact. The strict inequality in the open ball is exactly what excludes its boundary.
Worked Example: A Sphere Is Compact
Let \(S=\{x\in\mathbb{R}^3:\|x\|=3\}\), the sphere of radius \(3\) centered at the origin. It is bounded because every point in it has norm exactly \(3\). To check that it is closed, take any \(x\notin S\), so \(\|x\|\ne3\), and set \(\delta=|\|x\|-3|/2>0\). If \(\|y-x\|<\delta\), the triangle inequality in both directions gives \(|\|y\|-\|x\||\le\|y-x\|<\delta\). Consequently \(\|y\|\) cannot equal \(3\): if it did, then \(|\|y\|-\|x\||=|3-\|x\||=2\delta\), contradicting the strict bound by \(\delta\). Thus the ball \(B(x,\delta)\) misses \(S\), so the complement of \(S\) is open and \(S\) is closed. Heine-Borel now gives that \(S\) is compact.
Check: the radius condition has not been weakened: every point of \(S\) has norm \(3\), while points in the open ball of radius \(3\) have norm strictly less than \(3\). The sphere is compact as a set in its own right, even though it does not include the points inside it.
Closed Boxes and Unbounded Closed Sets
Worked Example: A Closed Box in \(\mathbb{R}^3\)
Consider \(Q=[-1,2]\times[0,3]\times[-2,1]\). Its points have coordinates \(x=(x_1,x_2,x_3)\) satisfying \(-1\le x_1\le2\), \(0\le x_2\le3\), and \(-2\le x_3\le1\). These bounds give \(|x_1|\le2\), \(|x_2|\le3\), and \(|x_3|\le2\), and hence \(\|x\|^2=x_1^2+x_2^2+x_3^2\le4+9+4=17\). Therefore \(Q\) is bounded.
It is also closed. If \(x\notin Q\), at least one coordinate lies outside its specified closed interval. For example, if \(x_1<-1\), put \(\eta=(-1-x_1)/2>0\). Whenever \(\|y-x\|<\eta\), we have \(|y_1-x_1|\le\|y-x\|<\eta\), so \(y_1<x_1+\eta=(-1+x_1)/2<-1\). This ball misses \(Q\). The same argument applies to a coordinate below its interval’s lower endpoint or above its upper endpoint, using half the positive distance to that endpoint. Thus the complement of \(Q\) is open. Heine-Borel gives that \(Q\) is compact.
Check: the computed bound is \(\|x\|^2\le17\), so \(\|x\|\le\sqrt{17}\) for every \(x\in Q\). The argument checks both required conditions rather than inferring compactness from the word “box.”
Worked Example: A Closed Set That Is Not Compact
In \(\mathbb{R}^2\), let \(L=\{(t,0):t\in\mathbb{R}\}\), the horizontal axis. It is closed: if \((a,b)\notin L\), then \(b\ne0\), and every point \((u,v)\) with \(\|(u,v)-(a,b)\|<|b|/2\) satisfies \(|v-b|<|b|/2\), so \(v\ne0\). That open ball misses \(L\). Hence its complement is open.
The set is unbounded. For each \(m\in\mathbb{N}\), the point \((m,0)\) belongs to \(L\) and has norm \(\sqrt{m^2+0^2}=m\). No finite \(M\) can bound all these norms. Since \(L\) is closed but not bounded, Heine-Borel implies that \(L\) is not compact.
A Common Mistake: Bounded Does Not Mean Compact
A common error is to check only that a set lies inside a large ball, then conclude it is compact. The open ball \(B(0,2)\) is bounded, since all its points have norm less than \(2\), but it is not compact because it is not closed. The sequence \(x_k=(2-1/k,0)\) gives a concrete witness: \(\|x_k\|=2-1/k<2\) for each \(k\), while \(x_k\to(2,0)\) and \(\|(2,0)\|=2\), so the limit is excluded.
The correction is to check both conditions independently. Boundedness controls how far points can spread; closedness ensures that boundary points approached by the set are included. Conversely, closedness alone is not enough: the horizontal axis is closed and unbounded, so it is not compact. In \(\mathbb{R}^n\), it is the combination of closedness and boundedness that gives compactness.
Check Your Understanding
For each set, decide whether it is compact in the stated Euclidean space, and justify the decision using closedness and boundedness.
- Is the open interval \((0,1)\) compact as a subset of \(\mathbb{R}\)? Identify which Heine-Borel condition fails.
- Is the set \(\{x\in\mathbb{R}^2:\|x\|=1\}\) compact? Check both conditions.
- Is the box \([0,1]\times[-2,2]\) compact in \(\mathbb{R}^2\)? Give a bound for the norm of each point.
- Is the closed ray \([0,\infty)\) compact in \(\mathbb{R}\)? Which condition fails?
- Explain why every closed subset of a compact subset \(K\subseteq\mathbb{R}^n\) is compact, specifying which topology is used for closedness.