Bounds of a Set

Boundedness is the entry point to almost every finiteness or convergence argument in analysis. The Completeness Axiom asserts that every nonempty subset of $\mathbb{R}$ that is bounded above has a least upper bound — the supremum — and this single axiom powers the proofs of the Bolzano–Weierstrass theorem, the Extreme Value Theorem, and the convergence of monotone bounded sequences, all staples of a first analysis course. In numerical analysis and optimization, feasible regions are routinely required to be bounded (compact, once closedness is added) to guarantee that optimal solutions exist rather than escape to infinity. In probability theory, a random variable is called bounded when its range is a bounded subset of $\mathbb{R}$, a hypothesis needed for many concentration inequalities (e.g., Hoeffding's inequality). In functional analysis, "bounded operator" and "bounded linear functional" generalize this same order-theoretic idea to normed spaces, replacing the absolute-value characterization proved above with an operator-norm analogue. Interval arithmetic in scientific computing, and rigorous numerics more broadly, track upper and lower bounds on quantities through a computation to certify final answers within guaranteed error bars — a direct engineering descendant of the definitions given here.

Ask the tutor: Why does the existence of upper bounds for a set not automatically give a least upper bound in an arbitrary ordered field?

Picture the ordered field as a line. A set $S$ sitting on that line is bounded above if you can find a vertical marker to the right of which nothing in $S$ lives — the marker itself need not touch $S$, and there is nothing stopping you from sliding it further right and still having a valid upper bound. This is the crucial asymmetry to internalize: existence of one bound is all the definition demands, and if one upper bound exists, then automatically infinitely many do (anything larger also works, by Consequences of the Order Axioms — transitivity of $\le$). Boundedness says nothing about tightness; that refinement is the job of the supremum, introduced later. Symmetrically, bounded below means a marker exists with nothing in $S$ to its left. A set is simply bounded when it is squeezed between two such markers — pictorially, trapped inside a closed interval $[m, M]$. The set $S = \{x \in \mathbb{Q} : x^2 < 2\}$ is a perfect illustration of subtlety: it is visibly bounded above (by $2$, by $1.5$, by $1.42$, ...), yet the collection of its upper bounds, itself a subset of $\mathbb{Q}$, has no smallest member — every rational upper bound can be nudged downward to a slightly smaller rational upper bound, forever, because $\sqrt 2 \notin \mathbb{Q}$.

Two immediate order-theoretic facts, both direct from Consequences of the Order Axioms, underlie everything that follows.

This lemma shows that the set of upper bounds of a nonempty $S$, if nonempty, is "upward closed" — it is either empty or a ray $[M_0, \to)$-shaped set with no largest element unless it has a least element (this least element, when it exists, is the supremum, the subject of the next article). It also confirms the intuitive picture from the figure: once one upper bound is found, a whole ray of them exists.

The proof technique here is a direct, elementary chase through the order axioms and the definition of absolute value — no compactness, no completeness, nothing beyond what is already available in a bare ordered field. This is deliberate: boundedness is a concept that makes sense, and is fully provable, without any appeal to completeness. It is existence of least upper bounds that will require completeness, not boundedness itself.

Example 1 (asymmetric bounded set with an irrational-flavored gap, worked entirely inside $\mathbb{Q}$). Let

Claim: $S$ is bounded, with $-3$ a minimum (hence a lower bound achieving equality) and $1$ an upper bound not achieved by any element. *Lower bound. For every $n \ge 0$, $\frac{n}{n+1} \ge 0 > -3$, using that $\frac{n}{n+1} \ge 0$ since numerator and denominator are both nonnegative (with $n+1 > 0$), and basic ordered-field arithmetic. Also $-3 \le -3$ trivially. So every element of $S$ satisfies $x \ge -3$; $-3$ is a lower bound, and since $-3 \in S$, it is in fact the minimum of $S$. Upper bound. We check $\frac{n}{n+1} \le 1$ for all $n \ge 0$: equivalent (multiplying both sides by the positive number $n+1$, which preserves the inequality direction, a consequence of the order axioms) to $n \le n+1$, true since $0 \le 1$ and adding $n$ to both sides preserves order. Also $-3 \le 1$. So $1$ is an upper bound of $S$. Is $1$ the smallest upper bound? No element of $S$ equals $1$: if $\frac{n}{n+1}=1$ then $n = n+1$, i.e. $0=1$, false in any field. So $1 \notin S$, meaning $S$ has no maximum. But is $1$ still the least upper bound? Suppose $M < 1$ were an upper bound. Since $\frac{n}{n+1} \to 1$ in the sense that $1 - \frac{n}{n+1} = \frac{1}{n+1}$ can be made smaller than $1-M>0$ by choosing $n$ large (this uses the Archimedean property, which holds in $\mathbb{Q}$), we could find $n$ with $\frac{n}{n+1} > M$, contradicting that $M$ is an upper bound. Hence no rational number smaller than $1$ is an upper bound — $1$ is in fact the least upper bound, but this argument used the Archimedean property of $\mathbb{Q}$, an extra ingredient beyond bare boundedness; we flag it here only to show boundedness by itself doesn't decide whether the bound is attained or how tight it is. (We are not yet claiming a general theorem that least upper bounds exist — that is precisely the content reserved for <a class="wiki-link" href="/wiki/math-analysis-suprema-and-infima" data-track="wiki_article.body.wiki_link">Suprema and Infima</a>.) Boundedness via absolute value.* Take $K = 3 = \max(|-3|, |1|)$. Every $x \in S$ satisfies $-3 \le x \le 1$, hence $|x| \le 3$ (checking both sides: $x \ge -3 \ge -K$ and $x \le 1 \le K$). This matches the theorem above exactly.

Example 2 (a set that is bounded below but not above, and the empty set edge case). Let $T = \{2^n : n \in \mathbb{Z}_{\ge 0}\} = \{1, 2, 4, 8, 16, \dots\} \subseteq \mathbb{Q}$. Bounded below: $1$ is a lower bound. Proof: $2^n \ge 1$ for all $n \ge 0$. For $n=0$, $2^0=1\ge 1$. Inductively, if $2^n \ge 1$ then $2^{n+1} = 2\cdot 2^n \ge 2 \cdot 1 = 2 \ge 1$ (using that multiplying an inequality by the positive number $2$ preserves it, and $2 \ge 1$ from field axioms plus order). So by induction $2^n \ge 1$ for all $n\ge 0$. Not bounded above: Suppose toward contradiction that $T$ has an upper bound $M \in \mathbb{Q}$. By the Archimedean property of $\mathbb{Q}$ (available since $\mathbb{Q}$, like any ordered field containing $\mathbb{Z}$, is Archimedean in the sense that no rational bounds all natural numbers — this is a standard consequence of order-field axioms applied to $\mathbb{Q}$), there is a natural number $n$ with $n > M$. Since $2^n \ge n+1 > n$ for all $n \ge 0$ (an easy induction: $2^0=1\ge 1$, and $2^{n+1}=2\cdot2^n \ge 2(n+1) = 2n+2 \ge n+2 > n+1$ using $n \ge 0$), we get $2^n > n > M$, so $2^n \in T$ violates $2^n \le M$. Contradiction. Hence $T$ is unbounded above. So $T$ is bounded below but not bounded above, hence (by definition) not a bounded set. This shows the two halves of boundedness are logically independent: a set may satisfy one half and fail the other. Empty set edge case. By contrast, consider $\varnothing \subseteq \mathbb{Q}$. The defining condition "$x \le M$ for all $x \in \varnothing$" is a universally quantified statement over an empty domain, hence vacuously true for every $M \in \mathbb{Q}$. So every rational number is an upper bound of $\varnothing$ (and symmetrically, every lower bound). The empty set is therefore bounded (indeed bounded above by anything, and below by anything), a fact that feels paradoxical until one recalls that vacuous truth is a straightforward consequence of the formal definition of $\forall$ over an empty set, not a special exception carved out for $\varnothing$.