Compactness of Spheres From Compact Cubes

From a Cube to a Sphere

A sphere in Euclidean space is described by an equation, while a cube is described by bounds on each coordinate. These descriptions fit together: every point on the unit sphere has each coordinate between \(-1\) and \(1\), and the sphere is a closed subset of the resulting cube. Compactness of the cube then gives compactness of the sphere. The argument uses open-cover compactness and works in every finite dimension, including dimension zero.

For \(n\geq 0\), give \(\mathbb{R}^{n+1}\) its standard Euclidean topology and write \(\|x\|=(x_0^2+\cdots+x_n^2)^{1/2}\) for the Euclidean norm. Define the unit sphere and the cube by

The cube \(C_n\) has the product topology from its \(n+1\) factors \([-1,1]\). For a finite product of Euclidean spaces, the product topology agrees with the standard Euclidean topology. Indeed, for \(u\in\mathbb{R}^{n+1}\), the maximum coordinate size \(\|u\|_\infty=\max_{0\leq i\leq n}|u_i|\) satisfies \(\|u\|_\infty\leq\|u\|\leq\sqrt{n+1}\|u\|_\infty\). Thus balls in either metric contain balls in the other, with appropriately rescaled radii.

Why the Cube Is Compact

First, the interval \([-1,1]\) is compact in its standard topology. Here is the open-cover argument. Let \(\mathcal{U}\) be an open cover of \([-1,1]\), where openness is relative to \([-1,1]\). Let \(E\) consist of those \(t\in[-1,1]\) for which \([-1,t]\) is covered by finitely many members of \(\mathcal{U}\). The set \(E\) is nonempty: a member of the cover contains \(-1\), and therefore covers \([-1,t]\) for some \(t>-1\) in \([-1,1]\). Let \(c=\sup E\), which exists by completeness of \(\mathbb{R}\).

Choose \(U\in\mathcal{U}\) containing \(c\). Since \(U\) is relatively open, there is \(\delta>0\) such that \((c-\delta,c+\delta)\cap[-1,1]\subseteq U\). If \(c<1\), choose \(t\in E\) with \(t>c-\delta/2\); this is possible from the definition of supremum. Also \(t\leq c\), so the same finite cover of \([-1,t]\), together with \(U\), covers \([-1,c+\eta]\), where \(\eta=\min(\delta/2,(1-c)/2)>0\). This contradicts \(c=\sup E\). Hence \(c=1\). In that case choose \(t\in E\) with \(t>1-\delta\). A finite cover of \([-1,t]\), together with \(U\), covers all of \([-1,1]\). Thus \(\mathcal{U}\) has a finite subcover, as required.

The finite-product theorem for compact spaces says that a product of finitely many compact spaces, with the product topology, is compact. This theorem was established in the earlier material on finite products. Applying it to \(n+1\) copies of the compact interval gives the needed fact about the cube.

Proof. Each factor \([-1,1]\) is compact by the open-cover argument above. There are \(n+1\) factors, a finite number, and \(C_n\) has their product topology. The finite-product theorem therefore implies that \(C_n\) is compact. \(\square\)

The Sphere Is a Closed Subset of the Cube

Proof. If \(x\in S^n\), then \(\sum_{i=0}^n x_i^2=1\). Every term \(x_i^2\) is nonnegative, so \(x_i^2\leq1\) for each \(i\), and hence \(-1\leq x_i\leq1\). This proves \(S^n\subseteq C_n\).

To prove closedness, take any \(x\in C_n\setminus S^n\). Then \(\|x\|\ne1\), so \(\delta=|\|x\|-1|>0\). The triangle inequality gives \(\|y\|\leq\|y-x\|+\|x\|\) and \(\|x\|\leq\|x-y\|+\|y\|\); together these imply \(|\|y\|-\|x\||\leq\|y-x\|\). If \(y\in C_n\) and \(\|y-x\|<\delta/2\), then \(\|y\|\ne1\): otherwise \(\delta=|\|x\|-1|=|\|x\|-\|y\||<\delta/2\), a contradiction. Thus the relative open ball \(B(x,\delta/2)\cap C_n\) misses \(S^n\). Every point of \(C_n\setminus S^n\) has such a relative open neighborhood, so \(C_n\setminus S^n\) is open in \(C_n\). Therefore \(S^n\) is closed in \(C_n\). \(\square\)

We also use the closed-subspace theorem: a closed subset of a compact space is compact. The proof shows precisely where closedness matters. If \(F\) is closed in a compact space \(X\), let \(\{V_\alpha\}\) be an open cover of \(F\) in its subspace topology. For each \(\alpha\), write \(V_\alpha=U_\alpha\cap F\) with \(U_\alpha\) open in \(X\). The sets \(U_\alpha\), together with \(X\setminus F\), form an open cover of \(X\). Compactness gives a finite subcover of \(X\). Removing \(X\setminus F\) from this finite list leaves finitely many \(U_\alpha\) that cover \(F\), and their corresponding \(V_\alpha\) cover \(F\). This proves the theorem.

Proof. The cube \(C_n\) is compact by the proposition above, and \(S^n\) is closed in \(C_n\) by the closedness proposition. The closed-subspace theorem therefore implies that \(S^n\), with its subspace topology, is compact. This subspace topology is the standard Euclidean topology on the sphere. \(\square\)

Worked Examples in Low Dimensions

A Common Compactness Pitfall

Common pitfall. A tempting but incorrect argument is: “The cube \(C_n\) is compact, and \(S^n\subseteq C_n\), so \(S^n\) is compact.” Compactness is not inherited by every subset; the closed-subspace theorem requires the subset to be closed.

For a concrete failure, \((-1,1)\) is a subset of the compact interval \([-1,1]\), but it is not compact. Consider the open sets \(U_m=(-1+1/m,\,1-1/m)\) for integers \(m\geq2\). They are open in \(\mathbb{R}\), and their union covers \((-1,1)\): given \(x\in(-1,1)\), choose \(m\) large enough that \(1/m<\min(1+x,1-x)\), which gives \(-1+1/m<x<1-1/m\). But a finite subcollection has a largest index \(M\), and because \(U_m\subseteq U_M\) whenever \(2\leq m\leq M\), that finite subcollection is contained in \(U_M\). The point \(1-1/(2M)\) lies in \((-1,1)\) but is greater than \(1-1/M\), so it is not covered by \(U_M\). There is no finite subcover. In the sphere argument, the extra closedness proof is what makes the compactness conclusion valid.