Supremum versus maximum

Find least upper and greatest lower bounds, relate them to attained extrema, and handle unbounded or incomplete-domain cases.

Upper Bounds and Supremum

A number \(x\) is an upper bound for a set \(A\) when every element of \(A\) is at most \(x\). Equivalently, \(x\) sits above the whole set: for every \(a\in A\), we have \(a\leq x\).

Collect all upper bounds of \(A\) into one set, denoted \(U A\). When \(A\) is bounded above and this candidate set has a smallest member, that smallest upper bound is the supremum: \[\sup(A)=\min(U A).\] This is the least upper bound of \(A\).

For \(A=\{1,2,3\}\subseteq\mathbb R\), any number \(x\geq3\) is an upper bound, because it is at least as large as \(1\), \(2\), and \(3\). No number below \(3\) can be an upper bound, since it would fail to bound the element \(3\). Thus \(U A=[3,\infty)\).

Therefore \[\min(U A)=\min[3,\infty)=3,\] so \(\sup(A)=3\). In this finite example the supremum is also the maximum. More generally, whenever a set has a maximum, that maximum equals its supremum; in particular, every non-empty finite set satisfies \(\max(A)=\sup(A)\).

Lower Bounds and Infimum

A lower bound is the mirror image of an upper bound. For a set \(A\), a number \(x\) is a lower bound when it sits below the whole set: every element of \(A\) is greater than or equal to \(x\), or equivalently \(x\leq a\) for every \(a\in A\).

We write \(L A\) for the set of all lower bounds of \(A\). Once those lower bounds are collected, the infimum of \(A\) is the largest element of that lower-bound set: \[\inf(A)=\max(L A).\] This is the greatest lower bound.

For \(A=\{1,2,3\}\subseteq\mathbb R\), any number \(x\leq1\) lies below all three elements of \(A\). Thus the lower bounds form \(L A=(-\infty,1]\). The greatest number in that interval is \(1\), so \[\max(L A)=\max(-\infty,1]=1,\] and therefore \(\inf(A)=1\).

This also explains the relation to minima: if a set has a minimum, that minimum is automatically its greatest lower bound. In particular, every non-empty finite set of real numbers satisfies \(\min(A)=\inf(A)\).

Infimum of a Finite Set

For a finite set, the fastest way to find the infimum is often to write out every element and then take the minimum. Consider \[A=\left\{\dfrac{(-1)^n}{n+1}:n\in\{0,1,2,3,4\}\right\}.\]

Substituting the five allowed values gives \[\dfrac{(-1)^0}{0+1}=1,\quad \dfrac{(-1)^1}{1+1}=-\dfrac12,\quad \dfrac{(-1)^2}{2+1}=\dfrac13,\quad \dfrac{(-1)^3}{3+1}=-\dfrac14,\quad \dfrac{(-1)^4}{4+1}=\dfrac15.\]