Bounded Sets

What It Means for a Set to Be Bounded

A set of real numbers is bounded when all of its elements stay within some fixed distance of zero. The bound need not be the smallest possible one. It only needs to work for every element of the set. For example, a set may fit inside a large interval even if a much smaller interval would also contain it.

The condition is vacuously true for the empty set: for instance, \(R=0\) is a bound for \(\varnothing\), because there are no elements that could violate the condition. Every nonempty bounded set has at least one bound, and any larger nonnegative number is also a bound. Indeed, if \(|x|\leq R\) and \(R\leq R'\), then \(|x|\leq R'\).

The absolute-value bound has an equivalent interval description. By the Absolute-Value Bound Criterion from earlier in the course, \(|x|\leq R\) is equivalent to \(-R\leq x\leq R\) when \(R\geq0\). Thus boundedness means that the set lies inside some closed interval centered at zero. More generally, it means the set lies between some two real numbers, which need not be symmetric about zero.

Proof. Suppose first that \(S\) is bounded, with bound \(R\geq0\). For every \(x\in S\), the Absolute-Value Bound Criterion gives \(-R\leq x\leq R\). Therefore \(S\) lies between \(A=-R\) and \(B=R\), and it has a lower bound \(L=-R\) and an upper bound \(U=R\).

Conversely, suppose there are real numbers \(A\leq B\) such that \(A\leq x\leq B\) for every \(x\in S\). Let \(R=\max\{|A|,|B|\}\), so \(R\geq0\), \(-R\leq A\), and \(B\leq R\). For every \(x\in S\), these inequalities give \(-R\leq x\leq R\), and the Absolute-Value Bound Criterion implies \(|x|\leq R\). Hence \(S\) is bounded.

If \(S\) has a lower bound \(L\) and an upper bound \(U\), let \(R=\max\{|L|,|U|\}\). Then \(-R\leq L\leq x\leq U\leq R\) for every \(x\in S\), so the Absolute-Value Bound Criterion gives \(|x|\leq R\), and \(S\) is bounded. Finally, if \(S\) is bounded, the first paragraph produced both a lower and an upper bound. This proves all the stated equivalences. \(\square\)

The definitions of bounded above and bounded below are useful because later results will study upper bounds in more detail. For now, the equivalence says that neither direction can be ignored: controlling only how far right a set extends does not prevent it from extending arbitrarily far left.

Boundedness Passes to Subsets and Finite Unions

A bound that works for a set also works for every subset, since a subset has no elements that are not already in the original set. For a union of two bounded sets, the two bounds might differ, but the larger one controls both sets. These observations give basic tools for constructing new bounded sets from known ones.

Proof. Let \(S\) be bounded, and let \(E\subseteq S\). Choose \(R\geq0\) such that \(|x|\leq R\) for every \(x\in S\). Every \(x\in E\) also belongs to \(S\), so it satisfies \(|x|\leq R\). Thus \(E\) is bounded. This includes \(E=\varnothing\).

Now suppose \(S\) and \(T\) are bounded. Choose bounds \(R_S\geq0\) and \(R_T\geq0\) for them, and set \(R=\max\{R_S,R_T\}\). If \(x\in S\cup T\), then \(x\in S\) or \(x\in T\). In the first case, \(|x|\leq R_S\leq R\); in the second, \(|x|\leq R_T\leq R\). Every element of the union therefore satisfies \(|x|\leq R\), so the union is bounded. \(\square\)

Applying the two-set result repeatedly shows that the union of any positive finite number of bounded sets is bounded. At each step, a bound for the union so far can be combined with a bound for the next set by taking their maximum. This finite argument does not automatically give a bound for a union of infinitely many bounded sets: the individual bounds might grow without limit.

Finding Bounds Directly

To prove a set bounded, it is often helpful to start from the condition defining membership and derive a uniform inequality for every element. “Uniform” here means that the same bound works for all elements, rather than a different bound being chosen for each one. A bound may be deliberately generous if it makes the reasoning simpler.

An unbounded set, by contrast, has no single bound that works for every element. The positive ray gives a direct test of this idea.

Affine Transformations Preserve Boundedness

A bounded set remains bounded after each of its elements is translated, reflected, or scaled by a fixed real number. These operations can move or stretch the set, but a fixed amount of movement and stretching cannot turn a bounded set into an unbounded one. The Triangle Inequality makes this precise.

Proof. Since \(S\) is bounded, choose \(R\geq0\) such that \(|x|\leq R\) for every \(x\in S\). Let \(y\in aS+b\). By the definition of this set, there is an \(x\in S\) such that \(y=ax+b\). The Triangle Inequality, the absolute value of a product, and the bound on \(x\) give

The number \(|a|R+|b|\) is nonnegative and does not depend on the choice of \(y\). It therefore bounds every element of \(aS+b\), proving that this set is bounded. If \(S\) is empty, then \(aS+b\) is empty and is bounded as well; the same conclusion holds. \(\square\)

What Boundedness Does Not Say

Boundedness is a size condition, not a claim about how many elements a set has or whether it contains its boundary points. The interval \((0,1)\) is bounded because \(|x|\leq1\) for each of its elements, even though neither endpoint belongs to it. The set \([0,1]\) is bounded by the same number and does include both endpoints. A finite set is bounded, but bounded sets need not be finite: an interval contains infinitely many real numbers and may still fit inside a fixed range.

Nor does boundedness require a set to be an interval. The union in the earlier example had a gap and was still bounded. Conversely, being an interval alone does not guarantee boundedness: \((0,\infty)\) is an interval, but the direct argument above shows that it is unbounded. To decide boundedness, look for one real number \(R\geq0\) that works in \(|x|\leq R\) for every element—not just for some elements, or for each element separately with a bound that can vary.