Suprema and Infima: Examples and Counterexamples

The open unit interval $I=(0,1)=\{x\in \mathbb{R}:0<x<1\}$ has $\sup I=1$ and $\inf I=0$, and neither value belongs to $I$. This is the smallest interesting computation in the subject: both bounds exist by the least-upper-bound property (Complete ordered field (least-upper-bound property)) and its dual (Every nonempty set bounded below has an infimum), and both are missed by the set, so $I$ has neither a maximum (The supremum is attained exactly when a maximum exists) nor a minimum. The maximum is ruled out by the attainment criterion; the minimum is ruled out directly, since a minimum would be a lower bound of $I$ lying in $I$ and therefore at most $\inf I=0$ (Maximum and minimum of a set, Greatest lower bound (infimum)).

Nothing here is asserted by inspection. That $1$ bounds $I$ above is checked from the definition; that no smaller number does is checked with the epsilon characterisation (Epsilon characterisation of the supremum), by exhibiting, for each $\epsilon >0$, an explicit element of $I$ lying strictly above $1-\epsilon$. The witness is $\max \{1\mathrm{/}2,1-\epsilon \mathrm{/}2\}$: the second entry does the approximating, and the first keeps the witness inside $I$ when $\epsilon$ is large. The infimum is handled symmetrically with Epsilon characterisation of the infimum and the witness $\min \{1\mathrm{/}2,\epsilon \mathrm{/}2\}$.

• The two entries of the witness $\max \{2^{-1},1-\epsilon \mathrm{/}2\}$ play different roles, and dropping either breaks the argument. For small $\epsilon$ the approximating entry $1-\epsilon \mathrm{/}2$ is the larger one and does the work; for $\epsilon \ge 2$ that entry falls to $0$ or below and leaves $I$, which is exactly what the guard $2^{-1}$ prevents. Using $1-\epsilon$ itself as the witness fails for the same reason and also fails the strict inequality $1-\epsilon <s$.
• The interval $I$ is the standard witness that a supremum need not belong to its set (FALSE: the supremum of a set belongs to the set); that consequence is recorded separately as A supremum need not belong to its set: $\sup (0,1)=1\mathrm{\notin }(0,1)$.
• Compare $\sup [0,1]=1=\max [0,1]$ and $\inf [0,1]=0=\min [0,1]$: the closed interval $[0,1]$ has the same supremum and the same infimum, and attains both. The value of a supremum therefore carries no information at all about whether it is attained.

The closed unit interval $J=[0,1]=\{x\in \mathbb{R}:0\le x\le 1\}$ has $\sup J=1$ and $\inf J=0$, exactly as the open interval $\sup (0,1)=1$ and $\inf (0,1)=0$, with neither attained does, but here both values belong to $J$: they are the maximum and the minimum (Maximum and minimum of a set).

That is the whole content of this example. The two intervals have the same supremum and the same infimum, and differ only in attainment, so the numerical value of a supremum says nothing about whether the set reaches it. The computation is also easier than for the open interval, because a set with a maximum has that maximum as its least upper bound for free, with no epsilon argument and no appeal to completeness (The supremum is attained exactly when a maximum exists).

• Claim 1 of The supremum is attained exactly when a maximum exists needs no completeness: a set with a maximum has a least upper bound because the maximum already is one. The least-upper-bound property of $\mathbb{R}$ (Complete ordered field (least-upper-bound property)) is what handles sets with no maximum, such as the open interval of $\sup (0,1)=1$ and $\inf (0,1)=0$, with neither attained.
• The infimum was computed here directly from Greatest lower bound (infimum) rather than by reflecting through the origin. Both routes are available; the direct one is shorter when the bound is attained, since leastness of the upper bound and greatestness of the lower bound are then immediate from membership.

Write $n\cdot 1$ for the canonical natural number $\underset{n}{\underbrace{1+\cdots +1}}$ of $\mathbb{R}$ (Canonical naturals are positive and strictly increasing) and $1\mathrm{/}n:=(n\cdot 1{)}^{-1}$. The set $H=\{1\mathrm{/}n:n\ge 1\}=\{1,1\mathrm{/}2,1\mathrm{/}3,\dots \}$ has $\inf H=0$, which is not attained, and $\sup H=1$, which is attained, being $\max H$ (Maximum and minimum of a set).

The interesting half is the infimum, and it is exactly the Archimedean property (Every complete ordered field is Archimedean) in disguise. That $0$ is a lower bound is immediate from positivity of the canonical naturals. That no positive number is a lower bound is the statement that the reciprocals of the naturals get below every positive $\epsilon$, which is what the Archimedean property asserts once applied to ${\epsilon }^{-1}$. Leastness is then read off from the epsilon characterisation Epsilon characterisation of the infimum. Nothing is asserted by inspection: for each $\epsilon >0$ an explicit index is produced.

• The set $H$ is bounded, with $0\le h\le 1$ for every $h\in H$, and its members are pairwise distinct, since $n\mapsto 1\mathrm{/}n$ is strictly decreasing on $\{1,2,3,\dots \}$: for $n<m$ the map $n\mapsto n\cdot 1$ is strictly increasing by [L1], so $0<n\cdot 1<m\cdot 1$, and inversion reverses that by [L3], giving $1\mathrm{/}m<1\mathrm{/}n$. It is the standard example showing that a bounded set with infinitely many members can attain one of its two bounds and miss the other. ("Infinite" is used here in its everyday sense: no definition of finiteness is in scope on this page, and nothing above or below depends on one.)
• Positivity of every element is what makes $0$ a lower bound, and the Archimedean property is what makes it the greatest one. In a non-Archimedean ordered field the argument breaks at exactly one point, step 1.4: it is the Archimedean property (Every complete ordered field is Archimedean) that supplies, for a given $\epsilon >0$, an index $n_{\epsilon }$ with ${\epsilon }^{-1}<n_{\epsilon }\cdot 1$, and hence with $1\mathrm{/}{n}_{\epsilon }<\epsilon$ in step 2.1. And Not every ordered field is Archimedean exhibits an ordered field where no such natural exists. What that item establishes is the failure of the Archimedean property there; it says nothing about $H$ or its greatest lower bound, and this page does not compute one. So the value $\inf H=0$ is a statement about $\mathbb{R}$, not a formal manipulation.
• Read as a sequence rather than a set, $H$ is the classical null sequence; the rational form of the same fact is The sequence $1\mathrm{/}n$ is null.

Let $\iota :\mathbb{Q}\to \mathbb{R}$ be the canonical embedding of the rationals as an ordered subfield of $\mathbb{R}$ (The unique embedding of ℚ into an ordered field) and put $S=\{q\in \mathbb{Q}:q>0\text{ and }{q}^2<2\}\mathrm{.}$ Viewed inside $\mathbb{R}$, the set $S$ has a supremum, namely $\sup S=\sqrt{2}$, the unique nonnegative real whose square is $2$ (Square roots exist: a unique $\sqrt{a}\ge 0$ with $(\sqrt{a}{)}^2=a$; the positives are $\{x^2:x\mathrm{\ne }0\}$). Viewed inside $\mathbb{Q}$, the same set has no supremum at all: it has upper bounds in $\mathbb{Q}$, but none of them is least.

This is the reason $\mathbb{R}$ exists. The rationals form an ordered field in which a perfectly ordinary bounded set fails to have a least upper bound, because the number that ought to be that bound is irrational (FALSE: some rational number squares to 2). The least-upper-bound property (Complete ordered field (least-upper-bound property)) is precisely the repair, and the value it supplies here is $\sqrt{2}$.

The order-theoretic defect exhibited below is the exact counterpart of the metric defect recorded in FALSE: the rationals are complete, where a Cauchy sequence of rationals fails to have a rational limit. The two are different statements about the same hole in $\mathbb{Q}$, and this library repairs it twice, once by Dedekind cuts and once by Cauchy sequences.

• The set $S$ is bounded above in $\mathbb{Q}$, by $2$, as step 1.5 checks. Both hypotheses of the least-upper-bound property are therefore satisfied inside $\mathbb{Q}$, and the conclusion still fails. The property is a genuine assumption about the field, not a consequence of the order axioms alone.
• The proof of leastness uses density twice, once to find an element of $S$ close below $\sqrt{2}$ and once to squeeze a rational between $\sqrt{2}$ and a putative rational least upper bound. Both uses go through ℚ is dense in every Archimedean ordered field, which is itself a consequence of the Archimedean property (Every complete ordered field is Archimedean).
• The same argument runs with $2$ replaced by any positive rational $c$ that is not the square of a rational, once its two numerical steps are readjusted: step 1.2 must exhibit a positive rational whose square lies below $c$, and step 1.5 a positive rational whose square lies above it, neither of which is the constant $1$ or $2$ in general. Both exist for every positive rational $c$, and uniformly, so nothing here depends on $c$ being convenient: take $p:=c\mathrm{/}(1+c)$ and $w:=1+c$, both positive rationals. Then $p^2<c$, since $p^2<c$ is equivalent to $c^2<c(1+c{)}^2$ and hence, dividing by $c>0$, to $c<(1+c{)}^2=1+2c+c^2$, which holds because $1+c+c^2>0$; and $w^2=1+2c+c^2>c$ for the same reason. Everything after those two steps is unchanged, so the failure is pervasive rather than a curiosity attached to $\sqrt{2}$.

Take $S=(0,1)$ and $T=(2,3)$ and form the sumset $S+T=\{s+t:s\in S,t\in T\}$. Then $S+T=(2,4),\sup (S+T)=4=1+3=\sup S+\sup T,$ which is the identity of Supremum of a sumset: $\sup (S+T)=\sup S+\sup T$ on a concrete pair of intervals.

Two things are actually computed here, and it is worth keeping them apart. The set identity $S+T=(2,4)$ is not a supremum statement at all: the inclusion $\subseteq$ comes from adding inequalities, and the inclusion $\supseteq$ needs a construction, namely a decomposition $x=s_x+t_x$ of each $x\in (2,4)$ with $s_x\in S$ and $t_x\in T$. The supremum statement is then supplied by Supremum of a sumset: $\sup (S+T)=\sup S+\sup T$ together with $\sup S=1$ ($\sup (0,1)=1$ and $\inf (0,1)=0$, with neither attained) and $\sup T=3$, the latter obtained from $\sup S$ by translation (Supremum of a translate: $\sup (a+S)=a+\sup S$) rather than by a second epsilon argument.

The value $4$ is not attained: it fails the strict inequality $x<4$ that defines $(2,4)$, so $S+T$ has no maximum (The supremum is attained exactly when a maximum exists, Maximum and minimum of a set). The identity $\sup (S+T)=\sup S+\sup T$ therefore holds with the supremum on the left unattained, just as $\sup S=1$ is unattained in $\sup (0,1)=1$ and $\inf (0,1)=0$, with neither attained.