4.2 Upper bounds, supremum, and infimum

Motivation

Once a set is ordered, questions about its edges become unavoidable. Does the set contain a largest element? If it does not, can the ambient ordered set still supply a sharp upper boundary? What hypotheses ensure that such a boundary actually exists? These are different questions, and each needs its own quantifiers.

The distinction is foundational for completeness. An order relation tells us how elements that already exist compare. It does not, by itself, guarantee that every bounded set has a least upper bound or a greatest lower bound. Supremum and infimum give precise names to those boundaries when they exist; the completeness of $R$, used later in the course, guarantees their existence for the appropriate nonempty bounded subsets of $R$.

Keep the following hierarchy in view:

• maximum and minimum are elements of the set;
• upper and lower bounds are elements of the ambient ordered set;
• boundedness asserts that at least one suitable bound exists;
• supremum and infimum assert that a best such bound exists.

The order language behind bounds

Definition

Partially ordered set and total order

A partially ordered set is a pair $(X,\le)$ in which the relation $\le$ is reflexive, antisymmetric, and transitive. It is a totally ordered set (or linearly ordered set) if it also satisfies comparability: for every $x,y\in X$, either $x\le y$ or $y\le x$.

The definitions of bounds, supremum, and infimum below make sense in a partially ordered set. The approximation criteria later require a total order.

Definition

Maximum, minimum, upper bound, lower bound, and boundedness

Let $Y\subseteq X$, where $(X,\le)$ is a partially ordered set.

• A maximum of $Y$ is an element $m\in Y$ such that $(\forall y\in Y)\,y\le m$.
• A minimum of $Y$ is an element $n\in Y$ such that $(\forall y\in Y)\,n\le y$.
• An upper bound of $Y$ is an element $u\in X$ such that $(\forall y\in Y)\,y\le u$.
• A lower bound of $Y$ is an element $\ell\in X$ such that $(\forall y\in Y)\,\ell\le y$.
• The set $Y$ is bounded above in $X$ if $(\exists u\in X)(\forall y\in Y)\,y\le u$. It is bounded below in $X$ if $(\exists \ell\in X)(\forall y\in Y)\,\ell\le y$.

Thus “upper bound” names a particular element, whereas “bounded above” is an existence statement about at least one such element. It does not yet say that a least upper bound exists.

The ambient set $X$ is part of every statement. For example, a rational subset may be bounded above in $Q$ yet fail to have a supremum in $Q$, while it has one when regarded as a subset of $R$.

Supremum and infimum

Definition

Supremum and infimum

Let $Y$ be a nonempty subset of a partially ordered set $X$.

• An element $s\in X$ is the supremum of $Y$, written $s=\sup_X(Y)$, if $(\forall y\in Y)\,y\le s$ and, for every upper bound $u\in X$ of $Y$, $s\le u$.
• An element $t\in X$ is the infimum of $Y$, written $t=\inf_X(Y)$, if $(\forall y\in Y)\,t\le y$ and, for every lower bound $\ell\in X$ of $Y$, $\ell\le t$.

When the ambient set is clear, we write simply $\sup(Y)$ and $\inf(Y)$.

There are three logically separate checks. First define what a candidate would have to satisfy. Next verify the relevant boundedness condition. Finally justify that the best bound exists in the ambient set. A general ordered field does not provide this final step: $Q$ is an ordered field but is not complete. For $R$, the later completeness theorem supplies existence whenever $Y$ is nonempty and bounded above (for a supremum), or nonempty and bounded below (for an infimum).

Uniqueness and the link with maxima and minima

Theorem

A subset has at most one supremum and at most one infimum

If $Y\subseteq X$ has a supremum, then that supremum is unique. If it has an infimum, that infimum is unique.

Suppose $s$ and $s'$ are both suprema of $Y$. Because $s$ is an upper bound and $s'$ is below every upper bound, $s'\le s$. Reversing their roles gives $s\le s'$; antisymmetry then gives $s=s'$. Reversing all inequalities proves the infimum statement.

A maximum automatically gives a supremum, but the membership condition is essential. If $m=\max(Y)$, then $m\in Y$ and every $y\in Y$ satisfies $y\le m$, so $m$ is an upper bound. If $u$ is any upper bound, applying its defining inequality to the particular element $m\in Y$ gives $m\le u$. Hence $m$ is the least upper bound and $\sup(Y)=m$. Dually, if $n=\min(Y)$, then $n$ is a lower bound; every lower bound $\ell$ satisfies $\ell\le n$ because $n\in Y$; hence $\inf(Y)=n$. Conversely, a supremum that belongs to $Y$ is its maximum, and an infimum that belongs to $Y$ is its minimum.

Dual approximation criteria

Theorem

Order and epsilon characterizations of supremum and infimum

Let $Y$ be a nonempty subset of a totally ordered set $X$.

• For $s\in X$, $s=\sup_X(Y)$ if and only if $(\forall y\in Y)\,y\le s$ and $(\forall u\in X)(u\lt s\Rightarrow(\exists y\in Y)\,u\lt y\le s)$.
• For $t\in X$, $t=\inf_X(Y)$ if and only if $(\forall y\in Y)\,t\le y$ and $(\forall v\in X)(t\lt v\Rightarrow(\exists y\in Y)\,t\le y\lt v)$.

If $X=F$ is an ordered field, these are respectively equivalent to the same bound condition together with

• $(\forall \varepsilon\in F)(\varepsilon>0\Rightarrow (\exists y\in Y)\,s-\varepsilon\lt y\le s)$;
• $(\forall \varepsilon\in F)(\varepsilon>0\Rightarrow (\exists y\in Y)\,t\le y\lt t+\varepsilon)$.

The upper statement says that no element strictly below $s$ remains an upper bound. The lower statement says that no element strictly above $t$ remains a lower bound. The epsilon versions express the same fact using the additive structure of an ordered field. They characterize a candidate; they do not by themselves create a supremum or infimum for every bounded set.

Why the dual approximation criteria are equivalent

Assume $s=\sup(Y)$ and take $u\lt s$. If $u$ were an upper bound, leastness would give $s\le u$, a contradiction. Hence there is $y\in Y$ for which $y\nleq u$. Total comparability turns this into $u\lt y$, while the upper-bound property gives $y\le s$.

Conversely, suppose the displayed upper condition holds, and let $b$ be any upper bound of $Y$. Total comparability gives either $b\lt s$ or $s\le b$. The first alternative would produce $y\in Y$ with $b\lt y$, contradicting that $b$ is an upper bound. Therefore $s\le b$, so $s=\sup(Y)$.

For the lower condition, assume $t=\inf(Y)$ and take $v\gt t$. If $v$ were a lower bound, greatestness would give $v\le t$, a contradiction. Thus some $y\in Y$ fails $v\le y$; total comparability gives $y\lt v$, and the lower-bound property gives $t\le y$. Conversely, if the displayed lower condition holds and $a$ is any lower bound, total comparability gives either $t\lt a$ or $a\le t$. The first alternative would produce $y\in Y$ with $y\lt a$, contradicting $a\le y$. Hence $a\le t$, so $t=\inf(Y)$.

In an ordered field, substitute $u=s-\varepsilon$ and $v=t+\varepsilon$ to get the epsilon conditions. Conversely, for $u\lt s$ choose $\varepsilon=s-u\gt 0$; for $v\gt t$ choose $\varepsilon=v-t\gt 0$. The epsilon conditions then recover the two order conditions exactly.

A reliable proof workflow

The definition is short, but a rigorous extremal-bound proof should make its logic visible. A useful first move is to name the sets of all bounds:

$$U_X(Y)=\{u\in X:(\forall y\in Y)\,y\le u\}, \qquad L_X(Y)=\{\ell\in X:(\forall y\in Y)\,\ell\le y\}.$$

Then $Y$ is bounded above precisely when $U_X(Y)$ is nonempty, while a supremum exists precisely when $U_X(Y)$ has a minimum. Likewise, bounded below means that $L_X(Y)$ is nonempty, whereas an infimum exists precisely when $L_X(Y)$ has a maximum. This formulation exposes the gap between “there is at least one bound” and “there is a best bound.”

When asked to prove that a proposed element is a supremum, use the following four-step discipline.

1. Fix the universe. State the ambient ordered set and verify that the candidate belongs to it. A real candidate that is not rational cannot serve as a supremum inside $Q$, however natural it may look geometrically.
2. Prove the bound half. Start with an arbitrary element of the subset and prove that it lies below the candidate. One diagram, a list of early terms, or a limiting trend is not a substitute for this universal statement.
3. Prove sharpness. Either take an arbitrary upper bound and show that the candidate lies below it, or show that every strictly smaller ambient element is defeated by an element of the subset. In an ordered field, the latter is usually packaged as an epsilon argument. The witness may depend on the smaller element or on epsilon; it need not be one element that works for all choices.
4. Identify the existence input. A direct proof that one explicit candidate satisfies both defining clauses already proves existence for that set. If no candidate has yet been constructed, boundedness alone is insufficient unless a completeness theorem is available.

For an infimum, reverse every inequality and every directional word. Prove that the candidate lies below every element of the set, then show that every lower bound lies below the candidate. In an epsilon proof, points of the set must be found below $t+\varepsilon$, not above $t-\varepsilon$. Writing both bound conditions before manipulating symbols is a simple way to avoid reversing only half of the argument.

The negations also deserve attention. Saying that $u$ is not an upper bound means

$$(\exists y\in Y)\; y\nleq u,$$

not merely that $u$ is not known to be an upper bound. In a total order this is equivalent to finding $y$ with $u\lt y$. Dually, saying that $v$ is not a lower bound means that some $y\in Y$ satisfies $y\lt v$. These conversions are exactly where total comparability enters the approximation proof.

Nonemptiness is not decorative either. For the empty subset, the statements “every element of the subset lies below this candidate” and “every element lies above this candidate” are both vacuously true. Its upper-bound set and lower-bound set are therefore the whole ambient set. Whether those bound sets have extrema depends on the ambient order, so the standard completeness axiom is deliberately stated only for nonempty subsets.

Duality under order reversal

There is a structural reason that every theorem above comes in a supremum and an infimum version. If the order on $X$ is reversed, upper bounds become lower bounds, least becomes greatest, maximum becomes minimum, and supremum becomes infimum. A proof that uses only order relations can therefore be dualized by reversing all inequalities.

In an ordered field, negation realizes this reversal concretely: from $a\le b$ we obtain $-b\le-a$. Consequently, lower bounds of $B$ correspond to upper bounds of $-B$. The final exercise turns this observation into a full existence proof and identity; it also records exactly where completeness of $R$ is used.

Worked examples

Worked example

A finite set in Z

Let $Y=\{1,2,3\}\subseteq Z$. Its maximum is $3$ and its minimum is $1$. Every integer at least $3$ is an upper bound, and every integer at most $1$ is a lower bound. Since a maximum is a supremum and a minimum is an infimum, $\sup(Y)=3$ and $\inf(Y)=1$.

Worked example

The open interval (0,1)

Let $Y=(0,1)\subseteq R$. The element $1$ is an upper bound. If $u\lt1$, then $y=\max\{(u+1)/2,1/2\}$ lies in $(0,1)$ and satisfies $u\lt y$; hence $u$ is not an upper bound. Therefore $\sup(Y)=1$. Dually, $0$ is a lower bound, and for every $v\gt 0$, the element $y=\min\{v/2,1/2\}$ lies in $(0,1)$ and satisfies $y\lt v$; hence $\inf(Y)=0$.

Neither boundary belongs to $Y$, so the interval has neither a maximum nor a minimum.

Worked example

The infimum of the positive rationals

Let $Q_{\gt 0}=\{q\in Q:q\gt 0\}$. The element $0\in Q$ is a lower bound. Now let $\ell\in Q$ be any lower bound. If $\ell\gt 0$, then $\ell/2\in Q_{\gt 0}$, but the lower-bound condition would require $\ell\le\ell/2$, a contradiction. Thus every lower bound satisfies $\ell\le0$; since $0$ itself is a lower bound, $\inf_Q(Q_{\gt 0})=0$.

There is no minimum: for every $q\in Q_{\gt 0}$, the element $q/2$ is a strictly smaller positive rational. This proves, rather than merely sketches, the difference between an infimum and a minimum.

Common mistakes

Common mistake

Combining definition, boundedness, and existence

Showing that $Y$ has an upper bound proves only that $Y$ is bounded above. It does not identify a least upper bound, and in a noncomplete ambient ordered field it does not even guarantee that a least upper bound exists.

Common mistake

Forgetting the ambient ordered set

The notation $\sup(Y)$ is always relative to an ambient ordered set. A boundary may exist in $R$ but be absent from $Q$. State the ambient set whenever a change of universe could affect existence.

Common mistake

Using the approximation test in a partial order

The definitions of supremum and infimum work in a partially ordered set, but the displayed strict-approximation equivalences use total comparability. Without it, “not below” cannot automatically be converted into “strictly above.”

Completeness uses the exact boundedness hypotheses

The least-upper-bound principle is an existence statement with two hypotheses, and both matter:

Theorem

Least-upper-bound principle

An ordered field $F$ is complete if every nonempty subset $Y\subseteq F$ that is bounded above in $F$ has a supremum in $F$. Dually, every nonempty set bounded below has an infimum. The theorem does not assert a supremum for the empty set, nor for a set with no upper bound in the chosen ambient field.

The ambient field cannot be omitted. Let

$$Y=\{q\in Q:q^2\lt 2\text{ and }q\gt 0\}.$$

This set is nonempty and bounded above in $Q$ (for instance, $2$ is an upper bound), but its sharp boundary is not rational. The same subset, viewed inside $R$, has a supremum. Thus “bounded” and “has a supremum” are always relative to the ambient ordered set.

Suprema of geometric sums and finite joins

The definitions are designed for infinite sets as well as intervals. The following example is especially useful because the set is an infinite image of a finite-sum formula, and because its supremum is not a maximum.

Worked example

A geometric-sum image with supremum 2

Define $f:N\to Q$ by

$$f(n)=\sum_{k=0}^{n}2^{-k}=2-2^{-n}, \qquad Y=f(N).$$

The finite geometric identity gives the displayed formula (or it follows by induction from $f(0)=1$ and $f(n+1)=f(n)+2^{-(n+1)}$). Since $2^{-n}>0$, every $f(n)<2$, so $2$ is an upper bound in $Q$. To prove it is least, let $q<2$ be any rational and write $\varepsilon=2-q=a/b>0$ with positive integers $a,b$. Choose $n>b$. The elementary induction $2^n>n$ gives

$$2^{-n}<\frac1n<\frac1b\le\frac ab=\varepsilon.$$

Therefore $f(n)=2-2^{-n}>2-\varepsilon=q$. Every rational below $2$ is defeated by an element of $Y$, so the order approximation criterion gives $\sup_Q(Y)=2$. Also $f(n+1)>f(n)$ for every $n$, so $Y$ has no maximum; its minimum is $f(0)=1$. This is a concrete reminder that an ordered field can contain a bounded set with a supremum that is not attained, while $Q$ still fails completeness for other sets such as the rational $\sqrt{2}$ cut.

The second pattern uses only the order axioms and works in a partial order. It must not be silently replaced by an argument that assumes every pair is comparable.

Theorem

Binary suprema give suprema of nonempty finite subsets

Let $(S,\le)$ be a partially ordered set in which every pair $\{a,b\}$ has a supremum. Then every nonempty finite subset of $S$ has a unique supremum.

Worked example

The binary-join induction

For a singleton $\{x\}$, the element $x$ is its supremum: it is an upper bound, and every upper bound $u$ satisfies $x\le u$. Suppose a finite set $E\subseteq S$ has supremum $s_E$, and add one new element $a$. By hypothesis, the pair $\{s_E,a\}$ has a supremum $s$. For every $x\in E$, transitivity gives $x\le s_E\le s$, and also $a\le s$, so $s$ is an upper bound of $E\cup\{a\}$.

Now let $u$ be any upper bound of $E\cup\{a\}$. It bounds $E$, hence $s_E\le u$, and it bounds $a$, hence $a\le u$. Thus $u$ is an upper bound of $\{s_E,a\}$, so $s\le u$. This proves that $s$ is the least upper bound. Induction on the cardinality gives existence for every nonempty finite subset, while antisymmetry gives uniqueness. No total-order assumption appears anywhere: binary joins, transitivity, and antisymmetry are the required hypotheses.

This distinction is useful when comparing examples. A finite subset of a total order has a maximum, but a finite subset of a partial order may have a supremum that is not one of its elements. The theorem above concerns existence of a best upper bound, not attainment by a maximum and not completeness of every partial order.

Summary

• Bounds live in the ambient ordered set; maxima and minima must also lie in the subset.
• Boundedness is an existence claim for some bound, not for a best bound.
• A supremum is the least upper bound and an infimum is the greatest lower bound; each is unique if it exists.
• A maximum equals the supremum, and a minimum equals the infimum.
• In total orders, strict order approximation characterizes both notions; in ordered fields this becomes the dual epsilon test.
• Completeness, not the ordered-field axioms alone, guarantees these boundaries for all appropriate nonempty bounded subsets of $R$.

Quick checks

Checkpoint

For $Y=(0,1)$, does Y have a maximum? What are $\sup(Y)$ and $\inf(Y)$?

Keep the inside/outside distinction clear.

Solution · Answer

$Y$ has no maximum and no minimum. Its supremum is $1$, and its infimum is $0$. The two bounds exist in the ambient set $R$ but do not belong to $Y$.

Checkpoint

If A has maximum m, what is $\sup(A)$? What is the dual statement for a minimum n?

Use membership in $A$ when comparing with an arbitrary bound.

Solution · Answer

$\sup(A)=m$: the maximum is an upper bound, and every upper bound $u$ satisfies $m\le u$ because $m\in A$. Dually, if $n=\min(A)$, then $\inf(A)=n$: it is a lower bound, and every lower bound $\ell$ satisfies $\ell\le n$ because $n\in A$.

Exercises

Checkpoint

Let $A=\{1-1/n:n\in\mathbb Z^+\}$. Find $\sup(A)$, $\inf(A)$, and decide whether A has a maximum.

For the supremum, prove both the upper-bound condition and the epsilon approximation condition.

Solution · Guided solution

The set begins

$$\left\{0,\frac12,\frac23,\frac34,\dots\right\}.$$

For every $n\in Z^+$, we have $1-1/n\lt1$, so $1$ is an upper bound. Let $\varepsilon\gt 0$. By the Archimedean property, choose $n\in Z^+$ such that $n\gt 1/\varepsilon$. Then $1/n\lt\varepsilon$, and therefore

$$1-\varepsilon\lt1-\frac1n\in A.$$

The epsilon criterion proves $\sup(A)=1$. The set has no maximum: after the term $1-1/n$, the term $1-1/(n+1)$ is strictly larger. Finally, every term is nonnegative and the term for $n=1$ is $0$; hence $\inf(A)=0$, and $0$ is also the minimum.

Checkpoint

Show that if B is a nonempty subset of R and is bounded below, then $\inf(B)=-\sup(-B)$.

First define $-B$, then verify the hypotheses needed to invoke completeness of $R$.

Solution · Guided solution

Define

$$-B=\{-b:b\in B\}.$$

Because $B$ is nonempty, $-B$ is nonempty. Since $B$ is bounded below, choose a lower bound $\ell\in R$. For every $b\in B$, $\ell\le b$, so $-b\le-\ell$. Thus $-\ell$ is an upper bound of $-B$, and $-B$ is bounded above.

We now use, in advance, the completeness of $R$ proved in the subsequent notes: the nonempty, bounded-above set $-B$ has a supremum. Put $S=\sup(-B)$ and $I=-S$. For every $b\in B$, $-b\le S$, hence $I\le b$; so $I$ is a lower bound of $B$. If $j$ is any lower bound of $B$, then $-b\le-j$ for every $b\in B$, so $-j$ is an upper bound of $-B$. Leastness of $S$ gives $S\le-j$, hence $j\le-S=I$. Therefore $I$ is the greatest lower bound of $B$, and

$$\inf(B)=-\sup(-B).$$

Related notes

Read this after 4.1 Total orders and ordered fields and continue to 4.3 Completeness and gaps in Q.