What Is an Upper Bound?
A set can be bounded above even when it has no largest element. To describe this precisely, we ask whether one real number lies at least as far to the right as every element of the set. That number need not belong to the set, and many different numbers may serve as upper bounds.
The definition has a universal condition: the same \(U\) must work for every element of \(S\). Checking that \(U\) is greater than some elements, or even many elements, does not establish that it is an upper bound. The empty set is bounded above: every real number is an upper bound for it, because there are no elements that could violate the condition.
When \(S\) is nonempty, an upper bound must be at least as large as each particular element. But it need not be one of those elements. Also, if \(U\) is an upper bound and \(V\geq U\), then \(V\) is an upper bound too: for every \(x\in S\), \(x\leq U\leq V\). Thus a bounded-above set generally has many upper bounds.
Worked Example: A Set with an Upper Bound but No Maximum
Let \(S=\{4-1/n:n\in\mathbb{N},\ n\geq1\}\). For every such \(n\), \(1/n>0\), so \(4-1/n<4\). Therefore \(4\) is an upper bound for \(S\). It is not an element of \(S\): if \(4-1/n=4\), then \(1/n=0\), which is impossible for a positive integer \(n\).
In fact, \(S\) has no maximum. For each \(n\geq1\), \(n+1>n>0\), so \(1/(n+1)<1/n\). Subtracting these positive reciprocals from \(4\) gives \(4-1/n<4-1/(n+1)\). The element indexed by \(n+1\) is therefore larger than the element indexed by \(n\). No element of \(S\) can be its largest element. This example shows that having an upper bound does not mean having a maximum.
Upper Bounds and Inclusion
If one set is contained in another, every upper bound for the larger set also bounds the smaller set. The reverse need not hold: a number may bound a small set but fail to bound a larger one. This gives a convenient way to transfer known bounds.
Proof. Suppose \(U\) is an upper bound for \(S\). Let \(x\in E\). Since \(E\subseteq S\), we have \(x\in S\), and the definition of upper bound gives \(x\leq U\). This holds for every \(x\in E\), so \(U\) is an upper bound for \(E\). If \(S\) is bounded above, choose one of its upper bounds \(U\); the same \(U\) bounds every subset \(E\). The argument includes \(E=\varnothing\), for which the condition holds vacuously. \(\square\)
Worked Example: Using Inclusion to Find a Bound
Let \(S=\{x\in\mathbb{R}:x^2<16\}\), and let \(E=\{x\in S:x\geq0\}\). If \(x\in S\), then \(x^2<16\). Since \(|x|\geq0\) and \(|x|^2=x^2\), the comparison of absolute values by squares gives \(|x|<4\), and hence \(x<4\). Thus \(4\) is an upper bound for \(S\).
Because \(E\subseteq S\), the theorem shows that \(4\) is also an upper bound for \(E\). Directly, the condition \(x\geq0\) and \(x^2<16\) gives \(0\leq x<4\), confirming the same bound. The inclusion argument is useful when the subset has an awkward description but the larger set has an easy bound.
Combining Upper Bounds
To bound a union, a number must bound every element in either part. Thus it must be an upper bound for each set separately. If the two sets have different upper bounds, the larger of those bounds works for their union. This parallels the finite-union property for bounded sets, but here the conclusion keeps track of the direction of the bound.
Proof. Suppose first that \(U\) is an upper bound for \(S\cup T\). Since \(S\subseteq S\cup T\) and \(T\subseteq S\cup T\), the Upper Bounds Pass to Subsets Theorem shows that \(U\) bounds each of \(S\) and \(T\).
Conversely, suppose \(U\) is an upper bound for both \(S\) and \(T\). If \(x\in S\cup T\), then \(x\in S\) or \(x\in T\). In the first case \(x\leq U\) because \(U\) bounds \(S\); in the second case \(x\leq U\) because \(U\) bounds \(T\). Thus \(U\) bounds the union.
If the union is bounded above, its upper bound also bounds each set by the first part, so both sets are bounded above. If both sets are bounded above, choose upper bounds \(U_S\) and \(U_T\), and let \(U=\max\{U_S,U_T\}\). Every element of either set is at most \(U\), so the first part shows that \(U\) bounds the union. This reasoning also covers an empty set. \(\square\)
Worked Example: Finding a Common Bound for Two Sets
Let \(S=\{x\in\mathbb{R}:x\leq -2\}\) and \(T=\{x\in\mathbb{R}:1\leq x<6\}\). The number \(-2\) is an upper bound for \(S\), since every \(x\in S\) satisfies \(x\leq-2\). The number \(6\) is an upper bound for \(T\), since every \(x\in T\) satisfies \(x<6\), and therefore \(x\leq6\).
The larger of these bounds is \(6\). Every element of \(S\) is at most \(-2\leq6\), and every element of \(T\) is at most \(6\). Hence \(6\) is an upper bound for \(S\cup T\). The gap between the two sets does not affect the argument: a bound controls the elements that belong to the union, not every point between them.
Upper Bounds Under Translation and Scaling
A translation shifts both a set and its upper bounds by the same amount. Multiplication by a positive number preserves the direction of inequalities, so it scales an upper bound in the same way. Multiplication by a negative number reverses inequalities; in that case, an upper bound alone is not enough. A lower bound, meaning a number \(L\) satisfying \(L\leq x\) for every element \(x\) of the set, supplies the needed information.
- For any \(b\in\mathbb{R}\), \(U+b\) is an upper bound for \(\{x+b:x\in S\}\).
- If \(a>0\), then \(aU+b\) is an upper bound for \(\{ax+b:x\in S\}\).
Proof. Let \(x\in S\). Since \(U\) is an upper bound, \(x\leq U\). Adding \(b\) to both sides gives \(x+b\leq U+b\), proving the translation claim.
If \(a>0\), multiplication by \(a\) preserves the inequality, so \(ax\leq aU\). Adding \(b\) gives \(ax+b\leq aU+b\). This holds for each \(x\in S\), so \(aU+b\) bounds the transformed set.
Now suppose \(a<0\) and \(L\leq x\) for every \(x\in S\). Multiplication by the negative number \(a\) reverses the inequality, giving \(ax\leq aL\). Adding \(b\) yields \(ax+b\leq aL+b\). This holds for every element \(ax+b\) of the transformed set, so \(aL+b\) is an upper bound. If \(S=\varnothing\), each transformed set is empty and every real number is an upper bound; the conclusions hold in that case as well. \(\square\)
Worked Example: A Bound After a Transformation
Let \(S=\{x\in\mathbb{R}:2\leq x\leq5\}\), and form \(E=\{-2x+3:x\in S\}\). The number \(2\) is a lower bound for \(S\). Because the scale factor \(-2\) is negative, the negative-scaling part of the theorem gives the upper bound \((-2)(2)+3=-1\) for \(E\).
The inequality can be checked directly. For every \(x\in S\), \(2\leq x\), so multiplying by \(-2\) reverses the comparison and gives \(-2x\leq-4\). Adding \(3\) gives \(-2x+3\leq-1\). The endpoint \(x=2\) belongs to \(S\) and produces \(-1\), so this particular upper bound is also an element of \(E\).
Upper Bound Is Not the Same as Maximum
A maximum of a set is an element of the set that is at least as large as every other element. Therefore every maximum is an upper bound, but an upper bound need not be a maximum because it may not belong to the set. An upper bound that belongs to the set is automatically the maximum, since it is at least every element; the only thing an upper bound can lack is membership.
Proof. If \(m\) is the maximum of \(S\), then by definition \(x\leq m\) for every \(x\in S\), so \(m\) is an upper bound. Conversely, suppose \(m\in S\) and \(m\) is an upper bound. Then every \(x\in S\) satisfies \(x\leq m\). Since \(m\) itself belongs to \(S\), it is an element at least as large as every element of \(S\), and hence is the maximum. \(\square\)
The membership condition matters. For the set \(\{4-1/n:n\geq1\}\), the number \(4\) is an upper bound but not a maximum, because \(4\notin S\). By contrast, if a finite nonempty set is written as \(\{-3,1,7/2,5\}\), then \(5\) belongs to the set and every listed element is at most \(5\); it is both an upper bound and the maximum.
A common mistake is to treat “upper bound” as meaning “largest element.” The definition makes no membership requirement, and it does not assert that an upper bound is the smallest possible one. The next topic, lower bounds, studies the corresponding control from the other direction. For now, the key test is direct: verify \(x\leq U\) for every \(x\) in the set.
Check Your Understanding
Use the definition and the results in this tutorial to answer each question.
- Why is every real number an upper bound for the empty set?
- If \(U\) bounds \(S\) and \(E\subseteq S\), why must \(U\) also bound \(E\)?
- If \(S\) and \(T\) have upper bounds \(U_S\) and \(U_T\), what common bound can be used for \(S\cup T\)?
- What information is needed to obtain an upper bound after multiplying a set by a negative number?
- What two conditions must hold for an upper bound \(m\) to be the maximum of a nonempty set?