Suprema and Infima

The supremum and infimum are not merely a real-analysis formality; they are the load-bearing beams of huge swaths of mathematics and its applications. In measure theory and probability, the outer measure of a set is defined as an infimum over covering sums, and expectations of unbounded random variables are handled via suprema of truncations. In optimization and operations research, the value of a linear or convex program is frequently defined as an infimum (or supremum) over a feasible region, and questions of whether an optimum is attained (as opposed to merely approached) are exactly questions about whether an infimum equals a minimum. In functional analysis, the operator norm of a bounded linear map is defined as a supremum over the unit ball, and completeness arguments throughout Banach and Hilbert space theory rest on suprema of Cauchy-type quantities. In computer science and algorithm analysis, worst-case and average-case complexity bounds are phrased as suprema/infima over classes of inputs, and semantics of recursive programs in domain theory use least upper bounds of increasing chains (Scott domains). In economics and game theory, the value of a zero-sum game and utility-maximization problems are phrased with suprema and infima over strategy sets, with the minimax theorem itself an equality of an infimum-supremum and supremum-infimum expression. Even in statistics, the supremum norm underlies the Kolmogorov–Smirnov test statistic, defined as the supremum of the absolute difference between empirical and true distribution functions.

Ask the tutor: Why does the set of rationals below $\sqrt2$ fail to have a rational supremum, and how does that failure motivate the real numbers' completeness axiom?

Picture the real line and a bounded set $S$ sitting on it. The upper bounds of $S$ form a ray: once some number $u$ is an upper bound, every larger number is automatically an upper bound too, since it is still $\ge$ every element of $S$. So the collection of all upper bounds of $S$ is itself an interval of the form $[\alpha, \infty)$ if it has a smallest element $\alpha$ — and that smallest element is exactly $\sup S$. The supremum is the left endpoint of the "upper bound ray," the exact point where bounding-from-above starts. This picture immediately suggests two things. First, $\sup S$ need not belong to $S$ itself — it is a property of the boundary between $S$ and its complement of upper bounds, not necessarily a member of $S$. Second, if $S$ has a maximum element $\max S$, then $\max S$ is automatically the supremum, because it is an upper bound (it dominates every element, including itself) and no smaller number can be an upper bound (since $\max S \in S$ violates any bound smaller than itself). The supremum generalizes the maximum to sets that may not attain one, such as $S = \{x \in \mathbb{Q} : x < \sqrt2\}$, which has no largest rational element yet is bounded above.

:::svg[The set of upper bounds of $S=(0,3]$ is exactly the ray $[3,\infty)$, so $\sup S = 3$]

Example 1 — a set with attained infimum but unattained supremum. Let

List the first several terms of the sequence part: for $n=1$, $\frac{-1}{2} = -0.5$; $n=2$, $\frac{2}{3} \approx 0.667$; $n=3$, $\frac{-3}{4} = -0.75$; $n=4$, $\frac{4}{5} = 0.8$; $n=5$, $\frac{-5}{6} \approx -0.833$; $n=6$, $\frac{6}{7}\approx 0.857$; and in general the even-indexed terms $\frac{n}{n+1}$ increase toward $1$ while odd-indexed terms $\frac{-n}{n+1}$ decrease toward $-1$. Together with the isolated point $5$, we claim:

Verifying $\sup S = 5$. We must check both defining properties. Upper bound: every term $\frac{(-1)^n n}{n+1}$ satisfies $\left|\frac{(-1)^n n}{n+1}\right| = \frac{n}{n+1} < 1 < 5$, so every sequence term is $< 5$; and the extra point $5$ satisfies $5 \le 5$. Hence $5$ is an upper bound of $S$. Least upper bound: let $u$ be any upper bound of $S$. Since $5 \in S$, the defining property of upper bound forces $5 \le u$. Thus every upper bound of $S$ is $\ge 5$, so $5$ is the least one. Both properties hold, so $\sup S = 5$, and note $5 \in S$ — the supremum is attained here, coinciding with $\max S$. Verifying $\inf S = -1$. Lower bound: for every $n$, $\frac{(-1)^n n}{n+1} \ge -\frac{n}{n+1} > -1$ (since $\frac{n}{n+1} < 1$ for all finite $n$), and $5 \ge -1$. So $-1$ is a lower bound. Greatest lower bound: suppose $\ell$ is a lower bound of $S$ with $\ell > -1$. Let $\varepsilon = \ell - (-1) = \ell+1 >0$. Choose an odd $n$ so large that $\frac{n}{n+1} > 1-\varepsilon$ (possible since $\frac{n}{n+1}\to 1$; concretely take $n$ odd with $n > \frac{1-\varepsilon}{\varepsilon}$, which exists by the Archimedean property). Then the term $\frac{-n}{n+1} = -\frac{n}{n+1} < -(1-\varepsilon) = -1+\varepsilon = \ell$, contradicting that $\ell$ is a lower bound. Hence no lower bound exceeds $-1$, so $-1$ is the greatest lower bound: $\inf S = -1$. Here the infimum is not attained, since no term of $S$ equals $-1$ exactly — the odd-indexed terms only approach it.

Example 2 — computing sup and inf of a set defined by an inequality. Let

First solve the inequality algebraically. Factor: $x^2 - 3x - 4 = (x-4)(x+1)$. This product is negative exactly when $x$ lies strictly between the roots $-1$ and $4$ (since the parabola opens upward and is negative between its two real zeros). Hence

We claim $\sup T = 4$ and $\inf T = -1$, with neither attained. Upper bound check for $4$: for all $x \in T$, $x < 4$ by definition of $T$, so in particular $x \le 4$; thus $4$ is an upper bound. Least upper bound check: suppose $u < 4$ is also an upper bound. Set $\varepsilon = \frac{4-u}{2} > 0$ and consider $x_0 = 4 - \varepsilon = \frac{4+u}{2}$. Since $u < 4$, we get $-1 < u < x_0 < 4$ (the middle inequality $u < x_0$ holds because $x_0 - u = \frac{4-u}{2} > 0$, and $x_0 < 4$ because $\varepsilon > 0$; also $x_0 > -1$ since $x_0 > u \ge$ any value would need checking, but more directly $x_0 = \frac{4+u}{2}$, and since $u$ is an upper bound of a nonempty set $T \ni 0$, we have $u \ge 0 > -1$, hence $x_0 > \frac{4-1}{2}=1.5 > -1$). So $x_0 \in T$, yet $x_0 > u$, contradicting that $u$ is an upper bound of $T$. Hence no number less than $4$ is an upper bound, so $4$ is the least upper bound: $\sup T = 4$. Since $4 \notin T$ (as $(4-4)(4+1)=0 \not< 0$), the supremum is not attained. Lower bound and infimum, symmetric argument: for all $x \in T$, $x > -1$, so $-1$ is a lower bound. If $\ell > -1$ were also a lower bound, take $x_1 = \frac{\ell + (-1)}{2} = \frac{\ell - 1}{2}$; one checks $-1 < x_1 < \ell$ and $x_1 < 4$ (since $\ell < 4$ was not even needed — $x_1 < \ell < 4$ follows once we confirm $\ell < 4$, which holds because $\ell$ is a lower bound of the nonempty set $T \ni 0$, forcing $\ell \le 0 < 4$), so $x_1 \in T$ while $x_1 < \ell$, contradicting that $\ell$ is a lower bound. Hence $\inf T = -1$, not attained since $-1 \notin T$. By the uniqueness theorem proved above, these are the supremum and infimum of $T$ — there is no ambiguity or alternative candidate value, even though we found them by an inequality-chasing argument rather than by inspection.