Worked Compactness Decisions In Rn

A Reliable Strategy for Compactness Decisions

In Euclidean space, compactness can often be decided by checking two simpler properties. The Heine–Borel theorem says that a subset of \(\mathbb{R}^n\), with its usual Euclidean topology, is compact if and only if it is closed and bounded. The challenge in problems involving unions, products, or sequences of sets is to check those two properties separately: closedness and boundedness do not always behave in the same way.

Throughout, boundedness is with respect to the Euclidean norm. A set \(A\subseteq\mathbb{R}^n\) is bounded if there is a real number \(R\geq 0\) such that \(\|a\|\leq R\) for every \(a\in A\). The empty set is bounded and compact. We use the usual convention that a sequence of sets \(K_1,K_2,\ldots\) is increasing if \(K_n\subseteq K_{n+1}\) for every \(n\), and decreasing if \(K_{n+1}\subseteq K_n\) for every \(n\).

Unions: Finite and Infinite Families

A finite union of closed sets is closed, and a finite union of bounded sets is bounded. In contrast, an arbitrary union of closed sets need not be closed, and an arbitrary union of bounded sets need not be bounded. These distinctions give the following result.

Proof. For a finite family \(A_1,\ldots,A_m\) of compact sets, Heine–Borel makes each \(A_i\) closed and bounded. A finite union of closed sets is closed: its complement is the finite intersection of their open complements, hence is open. To check boundedness, choose \(R_i\geq0\) with \(\|a\|\leq R_i\) for all \(a\in A_i\). The maximum \(R=\max\{R_1,\ldots,R_m\}\) bounds every point of the union. Heine–Borel now makes the union compact. For any family, including an infinite one, the stated compactness equivalence is exactly Heine–Borel applied to its union. Finally, if every \(A_\alpha\) is bounded, the union is bounded precisely when there is one \(R\geq0\) such that \(\|a\|\leq R\) for every \(\alpha\in J\) and every \(a\in A_\alpha\); this is the definition of boundedness of the union. \(\square\)

For an infinite union, compactness of every individual member does not settle either test for the union. Even an increasing sequence of compact sets can have a nonclosed union or an unbounded union. The next example checks both possibilities explicitly.

Products: Coordinate Tests

For \(A\subseteq\mathbb{R}^m\) and \(B\subseteq\mathbb{R}^n\), their product \(A\times B\) is regarded as a subset of \(\mathbb{R}^{m+n}\), with the Euclidean metric. If both factors are nonempty, closedness and boundedness of the product can be tested factor by factor.

Proof. We use the Euclidean fact that a set is closed if and only if it contains the limits of all its convergent sequences. To prove the forward implication for closedness, choose \(b\in B\), which is possible because \(B\) is nonempty. If a sequence \((a_k)\) in \(A\) converges to \(a\in\mathbb{R}^m\), then \((a_k,b)\) is a sequence in \(A\times B\) converging to \((a,b)\). Closedness of \(A\times B\) gives \((a,b)\in A\times B\), hence \(a\in A\). Therefore \(A\) is closed. Choosing any \(a\in A\) and applying the same argument to sequences in \(B\) proves that \(B\) is closed.

For the reverse implication, suppose \(A\) and \(B\) are closed, and let \((a_k,b_k)\) be a convergent sequence in \(A\times B\) with limit \((a,b)\). Each coordinate sequence converges, so \(a_k\to a\) and \(b_k\to b\). Closedness gives \(a\in A\) and \(b\in B\), and thus \((a,b)\in A\times B\). The sequence criterion proves that \(A\times B\) is closed.

For boundedness, if \(A\) and \(B\) have bounds \(R\) and \(S\), then every \((a,b)\in A\times B\) satisfies \(\|(a,b)\|^2=\|a\|^2+\|b\|^2\leq R^2+S^2\), so the product is bounded. Conversely, fix \(b\in B\). If the product is bounded by \(M\), then for every \(a\in A\), \(\|a\|^2\leq\|(a,b)\|^2\leq M^2\), so \(A\) is bounded; fixing \(a\in A\) proves \(B\) is bounded. Heine–Borel applied to the factors and product now gives the compactness equivalence. \(\square\)

The nonempty hypothesis in the factor-by-factor statement matters for its converse directions. If \(A=\varnothing\), then \(A\times B=\varnothing\) is closed, bounded, and compact whatever \(B\) is. The product gives no information about whether \(B\) is closed or bounded. For compactness, without assuming nonemptiness, the correct statement is: if both factors are compact, their product is compact; and if the product is compact and both factors are nonempty, both factors are compact.

Sequences of Compact Sets

For an increasing sequence, the union is compact exactly when it passes the closedness and boundedness tests. In particular, compactness of each \(K_n\) does not itself supply a common bound or make the union closed. For a decreasing sequence, there is a stronger conclusion when all sets are nonempty.

Proof. Each \(K_n\) is closed, so their intersection is closed: its complement is the union of the open sets \(\mathbb{R}^n\setminus K_n\), hence open. The intersection is a subset of the bounded set \(K_1\), so it is bounded. It remains to prove it is nonempty.

Suppose instead that \(\bigcap_{n\geq1}K_n=\varnothing\). Then the open sets \(\mathbb{R}^n\setminus K_n\) cover \(K_1\). Compactness of \(K_1\) gives a finite subcover, say by \(\mathbb{R}^n\setminus K_{n_1},\ldots,\mathbb{R}^n\setminus K_{n_r}\). Let \(N\) be the largest of the finitely many indices. Since the sets decrease, \(K_N\subseteq K_{n_i}\) for every \(i\). Nonemptiness of \(K_N\) gives a point in \(K_N\), and that point lies in every \(K_{n_i}\). It is therefore in none of the sets \(\mathbb{R}^n\setminus K_{n_i}\), contradicting that these sets cover \(K_1\), which contains \(K_N\). So the intersection is nonempty. It is closed and bounded, and Heine–Borel makes it compact. \(\square\)

Common Mistake: Transferring a Finite-Union Rule to Infinite Unions

A frequent error is to argue that a union of closed sets is closed because each set in the union is closed. That argument is valid for finite unions, not arbitrary ones. For example, each \(K_n=[0,n/(n+1)]\) above is closed, but their union is \([0,1)\). The points \(n/(n+1)\) belong to that union and converge to \(1\), which is excluded, so the union is not closed. The correction is to test the union itself: for compactness in Euclidean space, verify both that the whole union is closed and that one bound works for every point in it.

The same separation of tests is useful for products and sequences. A product of two nonempty Euclidean sets is closed exactly when both factors are closed, and bounded exactly when both factors are bounded. A decreasing sequence of nonempty compact sets has a nonempty compact intersection; an increasing sequence of compact sets, however, needs its own closedness and boundedness checks before its union can be declared compact.