When Is a Bound an Extreme Value?
An upper bound controls a set from above, but it need not belong to the set. A lower bound controls a set from below, but it too may be outside the set. A maximum or minimum is more specific: it is a bound that the set actually contains. This distinction determines whether a set has a greatest or least element, rather than merely being bounded in that direction.
The two conditions in each definition have different roles. Membership says the proposed extreme value is an element of the set; the inequality says it is at least as large as every element, or at most as large as every element. By the Characterization of a Maximum from “Upper Bounds,” a number \(M\in S\) is a maximum exactly when it is an upper bound for \(S\). The corresponding criterion for a minimum is that it belongs to \(S\) and is a lower bound for \(S\).
The empty set has no maximum or minimum, because neither can satisfy the required membership condition. In contrast, every real number is both a lower and an upper bound for the empty set, since there are no elements to check. Having bounds, therefore, does not by itself guarantee an extreme value.
Worked Example: An Endpoint That Is a Maximum
Let \(S=[-2,5)\). Every \(x\in S\) satisfies \(-2\leq x<5\), so \(5\) is an upper bound. But \(5\notin S\), because the right endpoint is excluded. Thus \(5\) is not a maximum.
The set has no maximum at all. To check this, suppose \(M\in S\). Then \(M<5\), so \(5-M>0\). Set \(x=(M+5)/2\). Since \(M<5\), we have \(M<x<5\); also \(M\geq-2\), so \(x>M\geq-2\). Hence \(x\in S\) and \(x>M\). This shows that no element \(M\) can be the maximum.
The left endpoint \(-2\), however, belongs to \(S\), and every element of \(S\) is at least \(-2\). Therefore \(-2=\min S\). This example has a minimum but no maximum.
Maximum and Minimum Are Unique
A set cannot have two different maximum values: if two elements each dominate every element of the set, each must be at least the other. The same reasoning applies to a minimum. This makes the notation \(\max S\) and \(\min S\) unambiguous whenever the relevant values exist.
Proof. Suppose \(M\) and \(N\) are both maxima of \(S\). Since \(N\in S\) and \(M\) is a maximum, \(N\leq M\). Since \(M\in S\) and \(N\) is a maximum, \(M\leq N\). Antisymmetry of the order gives \(M=N\).
Now suppose \(m\) and \(n\) are both minima of \(S\). Since \(n\in S\) and \(m\) is a minimum, \(m\leq n\). Since \(m\in S\) and \(n\) is a minimum, \(n\leq m\). Again, antisymmetry gives \(m=n\). Thus each type of extreme value is unique if it exists. \(\square\)
Worked Example: Checking Both Extremes in a Finite Set
Let \(S=\{-4,1,3,8\}\). The number \(8\) belongs to \(S\), and direct comparison gives \(-4\leq8\), \(1\leq8\), \(3\leq8\), and \(8\leq8\). Thus \(8\) is a maximum. Similarly, \(-4\in S\), and \(-4\leq-4\), \(-4\leq1\), \(-4\leq3\), and \(-4\leq8\), so \(-4\) is a minimum.
The uniqueness theorem ensures there are no other values that could also be the maximum or minimum. Notice that merely listing bounds would not establish either claim: membership must be checked as well.
Every Nonempty Finite Set Has a Maximum and Minimum
For a finite list of real numbers, one can compare entries and keep the largest and smallest seen so far. The same idea proves a general result about finite sets. The proof uses induction on the number of elements and does not require the elements to be written in any particular order.
Proof. We use induction on the cardinality \(n\) of the set. If \(n=1\), then \(S=\{a\}\) for some real number \(a\). Its only element is both at least and at most every element of \(S\), so it is both the maximum and the minimum.
Now let \(n\geq2\), and suppose the theorem holds for every nonempty set of cardinality \(n-1\). Choose \(a\in S\) and let \(T=S\setminus\{a\}\). Then \(T\) has \(n-1\) elements and is nonempty. By the induction hypothesis, \(T\) has a maximum \(M_T\) and a minimum \(m_T\).
Compare \(a\) and \(M_T\). If \(a\leq M_T\), then every element of \(T\) is at most \(M_T\), and \(a\leq M_T\), so every element of \(S=T\cup\{a\}\) is at most \(M_T\). Also \(M_T\in T\subseteq S\). Thus \(M_T\) is a maximum of \(S\). If \(M_T<a\), every element of \(T\) is at most \(M_T<a\), and \(a\in S\); hence \(a\) is a maximum of \(S\).
For the minimum, if \(m_T\leq a\), every element of \(T\) is at least \(m_T\), and \(a\geq m_T\), so \(m_T\) is a minimum of \(S\). If \(a<m_T\), every element of \(T\) is at least \(m_T>a\), so \(a\) is a minimum of \(S\). In either case, \(S\) has both extrema. Induction proves the theorem for every positive finite cardinality. \(\square\)
In the proof, the comparison is between two real numbers, \(a\) and \(M_T\), or \(a\) and \(m_T\); trichotomy ensures the stated cases cover all possibilities. For two real numbers \(a\) and \(b\), the larger of them is their maximum and the smaller is their minimum. If \(a=b\), the same number serves as both.
Worked Example: Extrema of a Finite Set Given by a Rule
Let \(S=\{2k-7:k\in\mathbb{Z},\ 0\leq k\leq4\}\). The integer \(k\) can be \(0,1,2,3,\) or \(4\), giving \(S=\{-7,-5,-3,-1,1\}\). For each permitted \(k\), \(k\leq4\), so \(2k\leq8\) and \(2k-7\leq1\). Since \(k=4\) is permitted, \(1=2(4)-7\) belongs to \(S\), and therefore \(\max S=1\).
Also \(k\geq0\) gives \(2k-7\geq-7\), and \(k=0\) is permitted, so \(-7=2(0)-7\) belongs to \(S\). Thus \(\min S=-7\). The calculation checks both the bound condition and the membership condition for each extreme.
Extrema of a Union
When two nonempty sets already have maxima and minima, the extrema of their union can be found by comparing the two maxima and the two minima. This does not require listing every element of the union: each set’s extreme value summarizes the comparisons needed within that set.
Proof. Write \(M_S=\max S\) and \(M_T=\max T\). If \(M_S\leq M_T\), then every \(x\in S\) satisfies \(x\leq M_S\leq M_T\), and every \(x\in T\) satisfies \(x\leq M_T\). Thus \(M_T\) is an upper bound for \(S\cup T\). Since \(M_T\in T\subseteq S\cup T\), it is the maximum of the union. If \(M_T<M_S\), the same argument with \(S\) and \(T\) exchanged shows that \(M_S\) is the maximum.
For the minimum, write \(m_S=\min S\) and \(m_T=\min T\). If \(m_S\leq m_T\), every element of \(S\) is at least \(m_S\), and every element of \(T\) is at least \(m_T\geq m_S\). Since \(m_S\in S\subseteq S\cup T\), it is the minimum of the union. If \(m_T<m_S\), exchanging the roles of the sets shows that \(m_T\) is the minimum. These cases establish both claims. \(\square\)
Worked Example: Combining Extrema
Let \(S=[-3,2]\) and \(T=\{4,7\}\). The extrema of \(S\) are \(\min S=-3\) and \(\max S=2\); those of \(T\) are \(\min T=4\) and \(\max T=7\). The smaller of the two minima is \(-3\), and the larger of the two maxima is \(7\). Hence \(\min(S\cup T)=-3\) and \(\max(S\cup T)=7\).
Both are elements of the union: \(-3\in S\) and \(7\in T\). Every element of \(S\) lies between \(-3\) and \(2\), and every element of \(T\) is either \(4\) or \(7\), so all elements of the union lie between \(-3\) and \(7\). This also verifies the result directly.
Having One Extreme Does Not Guarantee the Other
A set can have a minimum without a maximum, a maximum without a minimum, both, or neither. The finite-set theorem guarantees both for every nonempty finite set, but infinite sets need not behave that way. Whether an endpoint is included can decide whether a bound is an extreme value.
Worked Example: Four Different Patterns
The interval \([2,6]\) has minimum \(2\) and maximum \(6\), because both endpoints belong to the interval and bound all its elements. The interval \([2,6)\) has minimum \(2\) but no maximum: \(6\) is an upper bound but is excluded, and for any \(x\in[2,6)\), the number \((x+6)/2\) satisfies \(x<(x+6)/2<6\), so \(x\) cannot be a maximum.
The interval \((2,6]\) has maximum \(6\) but no minimum. For each \(x\in(2,6]\), the number \((2+x)/2\) satisfies \(2<(2+x)/2<x\), so no element can be the minimum. Finally, \((2,6)\) has neither: its endpoints are excluded, and for any \(x\) in the interval, the midpoint between \(x\) and \(6\) is a larger element while the midpoint between \(2\) and \(x\) is a smaller element.
The central question is not simply whether a set has an upper or lower bound. It is whether the relevant bound is attained by an element of the set. This distinction will matter when studying the supremum and infimum: those concepts describe least upper and greatest lower bounds even when the set does not contain them.
Check Your Understanding
Use the definitions and results in this tutorial to answer each question.
- What two conditions must a number satisfy to be the maximum of a set?
- Can the empty set have a maximum, even though every real number is an upper bound for it? Explain.
- Why can a set have at most one minimum?
- State the conclusion of the Extrema of Finite Sets Theorem for a nonempty set with seven elements.
- For \(S=[-1,3)\), identify a minimum and explain why \(3\) is not a maximum.
- If two nonempty sets have maxima \(5\) and \(2\), what is the maximum of their union, and why?