Lower Bounds

What Is a Lower Bound?

An upper bound controls a set from above; a lower bound controls it from below. The number serving as a lower bound does not have to belong to the set. Also, a set may have many lower bounds, including numbers much smaller than all its elements.

The defining condition must hold for every element of the set. Checking it for only a selection of elements is not enough. As with upper bounds, the empty set is bounded below: every real number is a lower bound for it, because there are no elements that could violate the condition.

If \(L\) is a lower bound for \(S\) and \(K\leq L\), then \(K\) is also a lower bound: for every \(x\in S\), \(K\leq L\leq x\). Thus, once a set has one lower bound, it has all smaller real numbers as lower bounds too.

Lower Bounds and Inclusion

A lower bound for a set remains a lower bound when we restrict attention to a subset. The subset has no new elements that could fall below the proposed bound. The converse need not hold: a number may bound a small subset from below while failing to bound the larger set.

Proof. Suppose \(L\) is a lower bound for \(S\), and let \(x\in E\). Since \(E\subseteq S\), we have \(x\in S\). The definition of lower bound therefore gives \(L\leq x\). This holds for every \(x\in E\), so \(L\) is a lower bound for \(E\). If \(S\) is bounded below, choose one of its lower bounds; that same number bounds every subset. The argument also applies when \(E\) is empty, since its lower-bound condition holds vacuously. \(\square\)

Lower Bounds for Unions

To bound a union from below, the proposed number must be at most every element in either set. It must therefore be a lower bound for each set separately. If the two sets have different lower bounds, the smaller of those numbers is a lower bound for their union.

Proof. Suppose first that \(L\) is a lower bound for \(S\cup T\). Since \(S\subseteq S\cup T\) and \(T\subseteq S\cup T\), the Lower Bounds Pass to Subsets Theorem shows that \(L\) bounds both \(S\) and \(T\) from below.

Conversely, suppose \(L\) is a lower bound for each set. If \(x\in S\cup T\), then \(x\in S\) or \(x\in T\). In the first case \(L\leq x\) because \(L\) bounds \(S\); in the second case \(L\leq x\) because \(L\) bounds \(T\). Thus \(L\) is a lower bound for the union.

If the union is bounded below, its lower bound also bounds each of its subsets, so both sets are bounded below. If both sets are bounded below, choose lower bounds \(L_S\) and \(L_T\), and set \(L=\min\{L_S,L_T\}\). Then \(L\leq L_S\) and \(L\leq L_T\). Every element of \(S\) is at least \(L_S\), and hence at least \(L\); every element of \(T\) is at least \(L_T\), and hence at least \(L\). The first part of the theorem shows that \(L\) bounds the union. This reasoning includes empty sets: every real number is a lower bound for an empty set, and the union statement still applies. \(\square\)

Negation Turns Lower Bounds into Upper Bounds

Negating every element of a set reverses its direction on the number line. This provides a useful way to translate a question about lower bounds into one about upper bounds. Write \(-S=\{-x:x\in S\}\) for the set obtained by negating the elements of \(S\).

Proof. Suppose \(L\) is a lower bound for \(S\). For each \(x\in S\), we have \(L\leq x\). Order reversal under negation gives \(-x\leq -L\). Every element of \(-S\) has the form \(-x\) for some \(x\in S\), so every element of \(-S\) is at most \(-L\). Hence \(-L\) is an upper bound for \(-S\).

Conversely, suppose \(-L\) is an upper bound for \(-S\). For each \(x\in S\), the element \(-x\) belongs to \(-S\), so \(-x\leq -L\). Reversing the order by negation gives \(L\leq x\). This holds for every \(x\in S\), so \(L\) is a lower bound. The boundedness conclusion follows by applying the equivalence to the existence of a bound. If \(S\) is empty, then \(-S\) is empty as well, and both statements hold vacuously. \(\square\)

Lower Bounds Under Translation and Scaling

A translation shifts a set and its lower bounds by the same amount. Multiplication by a positive number preserves the direction of an inequality. Multiplication by a negative number reverses it, so an upper bound—not a lower bound—is then what supplies a lower bound for the transformed set.

Proof. For translation, let \(x\in S\). Since \(L\leq x\), adding \(b\) gives \(L+b\leq x+b\). Thus \(L+b\) is a lower bound for the translated set.

If \(a>0\), multiplying \(L\leq x\) by \(a\) preserves the inequality, so \(aL\leq ax\). Adding \(b\) gives \(aL+b\leq ax+b\), as required. If \(a<0\), the upper-bound hypothesis gives \(x\leq U\). Multiplication by \(a\) reverses this comparison, giving \(aU\leq ax\), and adding \(b\) gives \(aU+b\leq ax+b\). Each inequality holds for every \(x\in S\), proving the claims. Finally, when \(a=0\), every transformed element has the form \(0x+b=b\); if \(S\) is empty there are no transformed elements, and otherwise \(b\leq b\). Hence \(b\) is a lower bound in both cases. \(\square\)

A Lower Bound Is Not Necessarily a Minimum

A lower bound is any number at most every element of a set. A minimum is a lower bound that also belongs to the set. Therefore a minimum, when it exists, is a lower bound; a lower bound need not be a minimum. This distinction is the lower-bound counterpart of the distinction between an upper bound and a maximum.

For example, let \(S=\{1+1/n:n\in\mathbb{N},\ n\geq1\}\). Since \(1/n>0\), every element of \(S\) is greater than \(1\), so \(1\) is a lower bound. But \(1\notin S\): an equation \(1+1/n=1\) would imply \(1/n=0\), which is impossible for a positive integer \(n\). The lower-bound condition does not require membership, so this is still a valid lower bound.

A common mistake is to treat every lower bound as the “lowest element” of a set. There may be no lowest element, and there can be many lower bounds that do not belong to the set. For instance, the set above is bounded below by \(1\), \(0\), and \(-20\); each number is at most every element of the set, but none of those facts alone makes a number its minimum. To verify a proposed lower bound \(L\), check \(L\leq x\) for every \(x\in S\). Membership is a separate question.