Compact Rectangles and Boxes

From Intervals to Boxes

A closed interval in the real line is compact: every open cover has a finite subcover. The usefulness of this fact is not limited to one dimension. Once finite products of compact spaces are known to be compact, it gives compactness of closed rectangles and higher-dimensional boxes. We will first prove the interval result directly, then explain exactly how the product theorem applies and how the resulting product topology relates to the usual Euclidean topology.

Unless stated otherwise, \(\mathbb{R}\) has its standard topology, and \(\mathbb{R}^n\) has the topology induced by the Euclidean metric. An interval \([a,b]\), where \(a\le b\), has the subspace topology inherited from \(\mathbb{R}\). A family of open sets is an open cover of a space if its union contains the whole space. A space is compact if every open cover has a finite subcover.

Proof. Let \(\mathcal{U}\) be any open cover of \([a,b]\). If \(a=b\), some member of \(\mathcal{U}\) contains \(a\), and that one set covers \([a,b]=\{a\}\). Now assume \(a<b\). Define \(S\) to be the set of \(x\in[a,b]\) for which \([a,x]\) is covered by finitely many members of \(\mathcal{U}\).

Some member \(U\in\mathcal{U}\) contains \(a\). Because \(U\) is open in the subspace \([a,b]\), there is \(\delta>0\) such that \((a-\delta,a+\delta)\cap[a,b]\subseteq U\). Taking \(x=a+\min(\delta/2,(b-a)/2)\), we have \(a<x\le b\) and \([a,x]\subseteq U\). Thus \(x\in S\), so \(S\) is nonempty. Since \(S\subseteq[a,b]\), it is bounded above. Let \(s=\sup S\).

Choose a member \(V\in\mathcal{U}\) containing \(s\). Openness in the subspace gives \(\eta>0\) such that \((s-\eta,s+\eta)\cap[a,b]\subseteq V\). If \(s<b\), the definition of supremum gives an \(x\in S\) with \(s-\eta/2<x\le s\). Finitely many members of \(\mathcal{U}\) cover \([a,x]\). Also \(V\) covers \([x,\min(b,s+\eta/2)]\), since each point of this interval lies in \((s-\eta,s+\eta)\cap[a,b]\). Therefore finitely many members cover \([a,\min(b,s+\eta/2)]\). Its right endpoint is greater than \(s\), contradicting that \(s\) is an upper bound for \(S\). Hence \(s=b\).

Finally choose \(V\in\mathcal{U}\) containing \(b\), and choose \(\eta>0\) with \((b-\eta,b+\eta)\cap[a,b]\subseteq V\). Since \(s=b=\sup S\), there is \(x\in S\) with \(b-\eta/2<x\le b\). A finite collection covers \([a,x]\), and \(V\) covers \([x,b]\). Together they cover \([a,b]\). Thus every open cover has a finite subcover, as required. \(\square\)

Applying the Finite-Product Theorem

We use the finite-product compactness theorem established in the earlier material on compactness of finite products: if \(X_1,\ldots,X_n\) are compact spaces, then their product \(X_1\times\cdots\times X_n\), with the product topology, is compact. In particular, the result applies to any finite number \(n\ge1\) of compact intervals. The theorem includes products with singleton factors; no nondegeneracy of the intervals is required.

Proof. By the closed-interval theorem, each factor \([a_i,b_i]\) is compact. Applying the finite-product compactness theorem to these \(n\) factors proves that their product with the product topology is compact. \(\square\)

To interpret this as a statement about familiar Euclidean rectangles and boxes, we must check that the product topology on the factors agrees with the subspace topology inherited from \(\mathbb{R}^n\). The product topology has basic open sets of the form \(\prod_{i=1}^n U_i\), where each \(U_i\) is open in its interval factor. The Euclidean subspace topology on a box consists of intersections of the box with Euclidean-open subsets of \(\mathbb{R}^n\).

Proof. First let \(\prod_{i=1}^n U_i\) be a basic product-open set, with \(U_i\) open in \([a_i,b_i]\). For each \(i\), there is an open set \(V_i\subseteq\mathbb{R}\) such that \(U_i=V_i\cap[a_i,b_i]\). The product \(\prod_{i=1}^n V_i\) is open in Euclidean \(\mathbb{R}^n\): around any of its points, each coordinate has an open interval contained in the corresponding \(V_i\), and sufficiently small Euclidean balls stay inside all these intervals. Thus \(\prod_i U_i=B\cap\prod_i V_i\) is open in the subspace topology.

Conversely, let \(O\) be open in Euclidean \(\mathbb{R}^n\), and let \(x=(x_1,\ldots,x_n)\in O\cap B\). There is an \(\varepsilon>0\) such that the Euclidean ball of radius \(\varepsilon\) about \(x\) lies in \(O\). For each coordinate, take \(U_i=(x_i-\varepsilon/(2\sqrt{n}),x_i+\varepsilon/(2\sqrt{n}))\cap[a_i,b_i]\). These are open in their interval factors and contain \(x_i\). Every \(y\in\prod_i U_i\) satisfies \(\|y-x\|^2<n(\varepsilon/(2\sqrt{n}))^2=\varepsilon^2/4<\varepsilon^2\). Hence \(\prod_i U_i\subseteq O\cap B\). Every point of \(O\cap B\) therefore has a product-open neighborhood within \(O\cap B\), proving that \(O\cap B\) is product-open. The topologies agree. \(\square\)

Combining the corollary and proposition shows that a closed Euclidean rectangle or box is compact in its usual Euclidean subspace topology. The role of the topology matters: the product theorem applies to the product topology, and the proposition verifies that this is precisely the topology the box inherits from Euclidean space.

Why the Closed Endpoints Matter

Compactness is about arbitrary open covers, not just the fact that a set is bounded or visually finite in extent. A common mistake is to say that \((0,1)\) is compact because it is a bounded interval. This confuses the open interval with the closed interval \([0,1]\): the open interval is not compact.

Indeed, for \(n\in\mathbb{N}\), let \(U_n=(1/n,1)\), regarded as open in the subspace \((0,1)\). These sets cover \((0,1)\): given \(x\in(0,1)\), choose \(n\in\mathbb{N}\) with \(n>1/x\), so \(1/n<x<1\), and therefore \(x\in U_n\). But any finite collection \(U_{n_1},\ldots,U_{n_k}\) has union \(U_N\), where \(N=\max\{n_1,\ldots,n_k\}\), because \(1/N\) is the smallest of their left endpoints. This union misses every point \(x\) with \(0<x\le1/N\). Hence no finite subcollection covers \((0,1)\).

The correction is to apply the theorem only to \([a,b]\), including both endpoints. The finite-product theorem then gives compactness of products of those closed intervals. It does not say that an arbitrary bounded rectangle, or a product with an open interval factor, is compact. The open cover above verifies the failure directly.