Why Boundedness Is Necessary

When Points Can Escape

The Heine–Borel Theorem says that a subset of \(\mathbb{R}\) is compact if and only if it is closed and bounded. The previous tutorial examined what these two conditions say about sets and sequences. Here we focus on why boundedness is necessary: if a set extends arbitrarily far along the real line, we can cover it with open sets that are each bounded but that cannot be reduced to finitely many sets.

Recall that an open cover of \(E\) is a family of open sets whose union contains \(E\). Compactness requires every open cover to have a finite subcover. An unbounded set creates a direct obstacle to this requirement: finitely many bounded sets cannot cover an unbounded set. The earlier Theorem (Subsets and Finite Unions of Bounded Sets) gives the set-theoretic fact behind this argument. We will use it to make the open-cover obstruction precise.

Proof. Suppose, to the contrary, that \(\mathcal{U}\) is an open cover of \(E\), every member of \(\mathcal{U}\) is bounded, and finitely many members \(U_1,\ldots,U_m\) cover \(E\). By the Theorem (Subsets and Finite Unions of Bounded Sets), the finite union \(\bigcup_{j=1}^{m}U_j\) is bounded. Since \(E\subseteq\bigcup_{j=1}^{m}U_j\), the subset part of that same theorem implies that \(E\) is bounded. This contradicts the assumption that \(E\) is unbounded. Therefore the cover has no finite subcover. \(\square\)

This theorem gives more than a single counterexample: any open cover of an unbounded set made up of bounded open sets is guaranteed to fail the finite-subcover test. Such covers are easy to construct. For example, the intervals \((-n,n)\), for positive integers \(n\), are bounded and open, and their union is all of \(\mathbb{R}\). They therefore cover every subset of \(\mathbb{R}\).

Worked Examples: Covers That Cannot Be Reduced

An open cover does not have to consist of intervals centered at the origin. The next example uses a different family of bounded intervals and a set that extends in both directions. It illustrates the general obstruction even when the points of the set do not lie on just one half-line.

Closedness Cannot Replace Boundedness

The last example is closed as well as unbounded, but the same issue occurs for many other closed unbounded sets. The Heine–Borel Theorem therefore cannot be weakened by dropping boundedness: closedness alone does not imply compactness. In fact, the result above shows precisely what fails for a closed unbounded set. It has a bounded open cover for which no finite subcover exists.

There is also a useful way to see that unboundedness persists even inside a particularly well-separated part of a set. This strengthens the contrast: an unbounded set is not merely able to contain points far apart; it contains a closed, unbounded subset whose distinct points stay more than one unit apart.

Proof. Since \(E\) is unbounded, choose \(x_1\in E\) with \(|x_1|>1\). Once \(x_n\) has been chosen, unboundedness lets us choose \(x_{n+1}\in E\) such that \(\displaystyle |x_{n+1}|>\max\{|x_n|+1,n+1\}\). It follows inductively that \(|x_n|>n\), and that the magnitudes increase by more than \(1\) at each step. Let \(D=\{x_n:n\geq1\}\). Since \(|x_n|>n\), the set \(D\) is unbounded.

If \(m>n\), then the increase in magnitude from \(x_n\) to \(x_m\) is greater than \(m-n\), and hence greater than \(1\). The reverse triangle inequality gives \(\displaystyle |x_m-x_n|\geq\big||x_m|-|x_n|\big|>1.\) Thus distinct points of \(D\) are separated by more than \(1\).

It remains to prove that \(D\) is closed. Take any convergent sequence \((d_j)\) in \(D\), with limit \(d\). A convergent sequence is bounded, so there is some \(B>0\) such that \(|d_j|\leq B\) for every \(j\). Because \(|x_n|>n\), only finitely many points of \(D\) can satisfy \(|x_n|\leq B\). The sequence \((d_j)\) therefore takes its values in a finite subset of \(D\).

If this finite subset has more than one point, let \(\delta>0\) be the smallest distance between any two distinct points in it. Such a positive minimum exists because there are only finitely many pairs, and the points are distinct. Since a convergent sequence is Cauchy, there is an index \(J\) such that \(|d_j-d_k|<\delta\) whenever \(j,k\geq J\). No two distinct values from the finite subset are this close, so all terms after \(J\) are equal. The sequence is eventually constant, and its limit is that constant value in \(D\). If the finite subset has only one point, the sequence is constant from the start and its limit is again in \(D\). We have shown that every convergent sequence in \(D\) has its limit in \(D\). The Sequential Criterion for Closedness implies that \(D\) is closed. \(\square\)

This theorem gives a second useful picture of the obstruction: every unbounded set contains a closed part that still stretches indefinitely. The set \(D\) is not compact, by the Heine–Borel Theorem, because it is unbounded. Its separation also makes clear that adding more and more points at distant locations cannot be prevented simply by making those points isolated.

Why the Condition Matters

Compactness rules out the possibility that an open cover needs points from arbitrarily far along the real line. When every set in a cover is bounded, finitely many selected members stay within some bounded region: this is exactly the finite-union property for bounded sets. If the set being covered is unbounded, those finitely many members must miss some of its points.

A common pitfall is to test boundedness using only a few points or a finite portion of a set. Boundedness requires one number \(M\) such that \(|x|\leq M\) for every point \(x\) in the entire set. No matter how far a finite collection of points extends, an unbounded set has points beyond them. Another pitfall is to think that closedness prevents this escape. Closedness controls limits of points that converge to a real number; it does not stop points from moving farther and farther away without converging in \(\mathbb{R}\).