Suprema and Infima

Throughout, $\mathbb{R}$ denotes the complete ordered field (Complete ordered field (least-upper-bound property)) and $S\subseteq \mathbb{R}$ is a subset of it.

The notions upper bound and bounded above are already fixed by Complete ordered field (least-upper-bound property) and are only recalled here, never redefined: $u\in \mathbb{R}$ is an upper bound of $S$ if $s\le u$ for all $s\in S$, and $S$ is bounded above if it has at least one upper bound. The dual notions are:

• $\mathrm{\ell }\in \mathbb{R}$ is a lower bound of $S$ if $\mathrm{\ell }\le s$ for all $s\in S$.
• $S$ is bounded below if it has at least one lower bound.
• $S$ is bounded if it is both bounded above and bounded below, that is, if there are $\mathrm{\ell },u\in \mathbb{R}$ with $\mathrm{\ell }\le s\le u$ for every $s\in S$.

• A bound is an element of $\mathbb{R}$ and is not required to lie in $S$. A bound that does lie in $S$ is a maximum or a minimum (Maximum and minimum of a set), and that is a strictly stronger condition (FALSE: the supremum of a set belongs to the set).
• Bounds come in half-lines: if $\mathrm{\ell }$ is a lower bound of $S$ then so is every ${\mathrm{\ell }}^{\mathrm{\prime }\mathrm{\prime }}\le \mathrm{\ell }$, and if $u$ is an upper bound then so is every $u^{\mathrm{\prime }\mathrm{\prime }}\ge u$. Consequently a set that has one bound of a given kind has infinitely many, and the interesting question is whether the collection of them has a best element, which is what a supremum (Complete ordered field (least-upper-bound property)) or an infimum (Greatest lower bound (infimum)) is.
• Bounded above and bounded below are independent conditions. The set $A=\{n\cdot 1_{\mathbb{R}}:n\ge 1\}$ of canonical naturals of $\mathbb{R}$ is bounded below by $0$ (Canonical naturals are positive and strictly increasing) and is not bounded above (Every complete ordered field is Archimedean); its reflection $-A$ is bounded above and not bounded below (Reflection through zero exchanges upper and lower bounds).
• The empty set is bounded, and vacuously so: every real number is both an upper bound and a lower bound of $\mathrm{\varnothing }$, since the defining condition quantifies over no elements. Having bounds is therefore much weaker than having a least upper bound or a greatest lower bound (FALSE: every subset of $\mathbb{R}$ has a supremum).

Let $S\subseteq \mathbb{R}$ and $\mathrm{\ell }\in \mathbb{R}$. Then $\mathrm{\ell }$ is a greatest lower bound, or infimum, of $S$ if both of the following hold:

• $\mathrm{\ell }$ is a lower bound of $S$ (Lower bound, bounded below, bounded set), that is, $\mathrm{\ell }\le s$ for every $s\in S$;
• ${\mathrm{\ell }}^{\mathrm{\prime }}\le \mathrm{\ell }$ for every lower bound ${\mathrm{\ell }}^{\mathrm{\prime }}$ of $S$.

Written out in one line:

$\mathrm{\ell }\text{ is an infimum of }S⟺[(\mathrm{\forall }s\in S)\mathrm{\ell }\le s]\text{ and }[(\mathrm{\forall }{\mathrm{\ell }}^{\mathrm{\prime }}\in \mathbb{R})((\mathrm{\forall }s\in S){\mathrm{\ell }}^{\mathrm{\prime }}\le s)\Rightarrow {\mathrm{\ell }}^{\mathrm{\prime }}\le \mathrm{\ell }]\mathrm{.}$

An infimum, when it exists, is unique (Suprema and infima are unique ↗), so we may write $\inf S$ for it.

• This is the exact dual of the least upper bound (supremum) of Complete ordered field (least-upper-bound property): reverse every inequality and swap "least" for "greatest". The two notions are related by reflection through $0$ (Reflection through zero exchanges upper and lower bounds).
• Existence is deliberately not part of the definition. That every nonempty subset of $\mathbb{R}$ which is bounded below actually has an infimum is a theorem, Every nonempty set bounded below has an infimum, derived from the least-upper-bound property; it is not an axiom and it is not free.
• As with a supremum, an infimum need not belong to $S$; when it does, it is the minimum of $S$ (Maximum and minimum of a set).
• The usable form of the definition in later arguments is the epsilon characterisation Epsilon characterisation of the infimum: $\mathrm{\ell }=\inf S$ exactly when $\mathrm{\ell }$ is a lower bound that cannot be raised by any positive amount without losing that property.

Let $S\subseteq \mathbb{R}$ and $m\in \mathbb{R}$.

• $m$ is a maximum (or greatest element) of $S$ if $m\in S$ and $s\le m$ for every $s\in S$.
• $m$ is a minimum (or least element) of $S$ if $m\in S$ and $m\le s$ for every $s\in S$.

A set has at most one maximum: if $m_1$ and $m_2$ are both maxima then $m_1\in S$ gives $m_1\le m_2$ and $m_2\in S$ gives $m_2\le m_1$, so $m_1=m_2$ by antisymmetry of the order, which is immediate from the trichotomy axiom of an ordered field (Ordered field, Complete ordered field (least-upper-bound property)). The same argument applies to minima, so we may write $\max S$ and $\min S$.

• A maximum is precisely an upper bound of $S$ (Complete ordered field (least-upper-bound property)) that happens to lie in $S$; a minimum is a lower bound of $S$ (Lower bound, bounded below, bounded set) that lies in $S$. In particular a set with a maximum is bounded above and a set with a minimum is bounded below.
• The empty set has neither a maximum nor a minimum, because the requirement $m\in S$ cannot be met.
• The membership requirement $m\in S$ is exactly what separates a maximum from a supremum, and it is the theme of this page. A supremum is a bound on the set and is not asked to belong to it; a maximum is an element of the set. The two agree exactly when the supremum happens to be attained (The supremum is attained exactly when a maximum exists), and they genuinely differ in general (FALSE: the supremum of a set belongs to the set).
• Every nonempty finite subset of $\mathbb{R}$ has both a maximum and a minimum (Every nonempty finite set of reals has a maximum and a minimum), which is what licenses the notation $\max \{a_1,\dots ,a_n\}$. Infinite sets need not: the failure of attainment is an infinitary phenomenon.

Let $S\subseteq \mathbb{R}$. If $u_1$ and $u_2$ are both least upper bounds of $S$ (Complete ordered field (least-upper-bound property)), then $u_1=u_2$. If ${\mathrm{\ell }}_1$ and ${\mathrm{\ell }}_2$ are both greatest lower bounds of $S$ (Greatest lower bound (infimum)), then ${\mathrm{\ell }}_1={\mathrm{\ell }}_2$.

Consequently a set has at most one supremum and at most one infimum, and the notations $\sup S$ and $\inf S$ denote single, well-determined real numbers whenever they exist.

For $S\subseteq \mathbb{R}$ write $-S:=\{-s:s\in S\}$. Then $-(-S)=S$, and $S$ is nonempty if and only if $-S$ is nonempty. Moreover, for all $u,\mathrm{\ell }\in \mathbb{R}$:

1. $u$ is an upper bound of $S$ if and only if $-u$ is a lower bound of $-S$;
2. $\mathrm{\ell }$ is a lower bound of $S$ if and only if $-\mathrm{\ell }$ is an upper bound of $-S$.

Consequently $S$ is bounded above if and only if $-S$ is bounded below, $S$ is bounded below if and only if $-S$ is bounded above, and $S$ is bounded if and only if $-S$ is bounded (Lower bound, bounded below, bounded set).

Let $S\subseteq \mathbb{R}$ be nonempty and bounded below. Then $S$ has a greatest lower bound in $\mathbb{R}$ (Greatest lower bound (infimum)), and it is given by

$\inf S=-\sup (-S),\text{where }-S=\{-s:s\in S\}\mathrm{.}$

In particular the complete ordered field $\mathbb{R}$ has the greatest-lower-bound property, which is therefore not an extra axiom: it is a consequence of the least-upper-bound property.

• The theorem is not a restatement of the least-upper-bound property: it is proved from it, by transporting the problem across the order-reversing bijection $x\mapsto -x$ of Reflection through zero exchanges upper and lower bounds. Nothing about $\mathbb{R}$ beyond the complete-ordered-field axioms is used.
• The hypotheses are both needed. The empty set is bounded below by every real and has no greatest lower bound, and a set unbounded below has no lower bound at all; the dual failures for suprema are recorded in FALSE: every subset of $\mathbb{R}$ has a supremum.
• The identity $\inf S=-\sup (-S)$ is the standard device for turning any statement about suprema into its dual; Epsilon characterisation of the infimum is the first application on this page.

Let $S\subseteq \mathbb{R}$ be nonempty and bounded above, and let $u$ be an upper bound of $S$ (Complete ordered field (least-upper-bound property)). Then

$u=\sup S⟺\text{for every }\epsilon >0\text{ there exists }s\in S\text{ with }u-\epsilon <s\mathrm{.}$

In words: among the upper bounds of $S$, the supremum is exactly the one that cannot be lowered by any positive amount and still bound $S$.

Let $S\subseteq \mathbb{R}$ be nonempty and bounded below, and let $\mathrm{\ell }$ be a lower bound of $S$ (Lower bound, bounded below, bounded set). Then

$\mathrm{\ell }=\inf S⟺\text{for every }\epsilon >0\text{ there exists }s\in S\text{ with }s<\mathrm{\ell }+\epsilon \mathrm{.}$

In words: among the lower bounds of $S$, the infimum is exactly the one that cannot be raised by any positive amount and still bound $S$ from below.

Let $S\subseteq \mathbb{R}$ be nonempty.

1. If $S$ has a maximum (Maximum and minimum of a set), then $\sup S$ exists and $\sup S=\max S$.
2. If $\sup S$ exists and $\sup S\in S$, then $S$ has a maximum and $\max S=\sup S$.

Hence, for a set whose supremum exists, the supremum is attained (belongs to the set) precisely when the set has a maximum, and then the two agree.

• The dual statement, that $\inf S$ is attained exactly when $S$ has a minimum and then $\inf S=\min S$, is not proved above. It follows by reflection: $m=\max S$ if and only if $-m=\min (-S)$, and $\inf X=-\sup (-X)$ (Reflection through zero exchanges upper and lower bounds, Every nonempty set bounded below has an infimum), so applying the two claims proved here to $-S$ and negating gives the minimum form.
• Claim 1 needs no completeness assumption: a set with a maximum has a supremum for free, since the maximum is already the least upper bound. Only claim 2 presupposes that $\sup S$ exists, which for a nonempty set bounded above is guaranteed by the least-upper-bound property (Complete ordered field (least-upper-bound property)).
• The converse of "the supremum exists" is not "the maximum exists": the set $\{x\in \mathbb{R}:0<x<1\}$ has supremum $1$ and no maximum (FALSE: the supremum of a set belongs to the set). What forces attainment is being nonempty and finite (Every nonempty finite set of reals has a maximum and a minimum); finiteness alone does not, since $\mathrm{\varnothing }$ is finite and has no maximum (Maximum and minimum of a set).

For every $n\in \mathbb{N}$ and all $a_0,a_1,\dots ,a_n\in \mathbb{R}$, the set $\{a_0,a_1,\dots ,a_n\}$ has a maximum and a minimum (Maximum and minimum of a set).

What is proved below is exactly the displayed statement, by induction on $n$.

The usual reading, that every nonempty finite subset of $\mathbb{R}$ has a maximum and a minimum, follows once one identifies the nonempty finite subsets of $\mathbb{R}$ with the sets listable as $\{a_0,\dots ,a_n\}$. That identification is recorded as a stipulation in the Given below, because this page has no definition of finiteness to prove it against. It is discharged, not merely assumed: The nonempty finite subsets of $\mathbb{R}$ are exactly the listable ones ↗ proves that the two descriptions of a nonempty finite subset of $\mathbb{R}$ agree. That lemma is recorded in justified_by rather than in deps, since it is about the sets this lemma quantifies over and therefore depends on this one. This is what licenses the notation $\max \{a_1,\dots ,a_n\}$ and $\min \{a_1,\dots ,a_n\}$ for finite sets of real numbers from this page onwards.

• Where the stipulation is discharged. Finiteness itself is defined later, in Finite, countably infinite, countable, uncountable ↗, as equinumerosity with a von Neumann natural; with that definition in hand The nonempty finite subsets of $\mathbb{R}$ are exactly the listable ones ↗ proves that a subset of $\mathbb{R}$ is nonempty and finite exactly when it is listable as $\{a_0,\dots ,a_n\}$, which is the Given below. So nothing on this page rests on an assumption that is never paid for; it is paid for later, and the payment is recorded in justified_by.
• Only the total order is used, never completeness. The base case needs reflexivity, the inductive step needs totality and transitivity, and the induction itself runs over $\mathbb{N}$. The same induction works in any totally ordered field; what is recorded here is its specialisation to $\mathbb{R}$.
• Nonemptiness is essential: $\mathrm{\varnothing }$ is finite and has no maximum (Maximum and minimum of a set). Finiteness is essential too: $\{x\in \mathbb{R}:0<x<1\}$ is bounded and has no maximum (FALSE: the supremum of a set belongs to the set).
• Combined with claim 1 of The supremum is attained exactly when a maximum exists, this says every nonempty finite subset of $\mathbb{R}$ has a supremum, and that the supremum is attained, because it equals the maximum. The infimum half is not part of The supremum is attained exactly when a maximum exists, which speaks only of maxima and suprema; it follows from the minimum proved here together with the reflection identity $\inf X=-\sup (-X)$ (Reflection through zero exchanges upper and lower bounds, Every nonempty set bounded below has an infimum).

Let $S$ and $T$ be subsets of $\mathbb{R}$ with $\mathrm{\varnothing }\mathrm{\ne }S\subseteq T$, and suppose $T$ is bounded above. Then $S$ is nonempty and bounded above, both $\sup S$ and $\sup T$ exist, and

$\sup S\le \sup T\mathrm{.}$

• The hypothesis that the larger set is bounded above cannot be weakened to the smaller one being bounded above: $\sup T$ has to exist for the inequality to mean anything.
• The dual statement, $\inf T\le \inf S$ for $\mathrm{\varnothing }\mathrm{\ne }S\subseteq T$ with $T$ bounded below, follows by applying this lemma to $\mathrm{\varnothing }\mathrm{\ne }-S\subseteq -T$, which gives $\sup (-S)\le \sup (-T)$, and then negating and using $\inf X=-\sup (-X)$ (Reflection through zero exchanges upper and lower bounds, Every nonempty set bounded below has an infimum).

Let $S\subseteq \mathbb{R}$ be nonempty and bounded above and let $a\in \mathbb{R}$. Write $a+S:=\{a+s:s\in S\}$. Then $a+S$ is nonempty and bounded above, and

$\sup (a+S)=a+\sup S\mathrm{.}$