2.4: Upper and Lower Bounds. Completeness

A subset $A$ of an ordered field $F$ is said to be bounded below (or left bounded) iff there is $p \in F$ such that

$$(\forall x \in A) \quad p \leq x$$

$A$ is bounded above (or right bounded) iff there is $q \in F$ such that

$$(\forall x \in A) \quad x \leq q$$

In this case, $p$ and $q$ are called, respectively, a lower (or left) bound and an upper (or right) bound, of $A .$ If both exist, we simply say that $A$ is bounded (by $p$ and $q ) .$ The empty set $\emptyset$ is regarded as ("vacuously") bounded by any p and $q$ (cf. the end of Chapter $1, §3 )$.

The bounds $p$ and $q$ may, but need not, belong to $A .$ If a left bound $p$ is itself in $A,$ we call it the least element or minimum of $A,$ denoted min $A$. Similarly, if $A$ contains an upper bound $q,$ we write $q=\max A$ and call $q$ the largest element or maximum of $A .$ However, $A$ may well have no minimum or
maximum.

Note 1. A finite set $A \neq \emptyset$ always has a minimum and a maximum (see Problem 9 of §§ 5-6 )\).

Note 2. A set $A$ can have at most one maximum and at most one minimum. For if it had $t w o$ maxima $q, q^{\prime},$ then

$$q \leq q^{\prime}$$

(since $q \in A$ and $q^{\prime}$ is a right bound); similarly

$$q^{\prime} \leq q;$$

so $q=q^{\prime}$ after all. Uniqueness of $\min A$ is proved in the same manner.

Note 3. If $A$ has one lower bound $p,$ it has many (e.g., take any $p^{\prime}<p )$.

Similarly, if $A$ has one upper bound $q,$ it has many (take any $q^{\prime}>q )$.

Geometrically, on the real axis, all lower (upper) bounds lie to the left (right) of $A ;$ see Figure $1 .$

Examples

(1) Let

$$A=\{1,-2,7\}.$$

Then $A$ is bounded above $($ e.g. $,$ by $7,8,10, \dots)$ and below $($ e.g. $,$ by $-2,-5,-12, \dots )$.

We have $\min A=-2, \max A=7$.

(2) The set $N$ of all naturals is bounded below (e.g., by $1,0, \frac{1}{2},-1, \ldots$) and $1=\min N;$ N has no maximum, for each $q \in N$ is exceeded by some $n \in N$ (e.g. $, n=q+1$).

(3) Given $a, b \in F(a \leq b),$ we define in $F$ the open interval

$$(a, b)=\{x | a<x<b\};$$

the closed interval

$$[a, b]=\{x | a \leq x \leq b\};$$

the half-open interval

$$(a, b]=\{x | a<x \leq b\};$$

and the half-closed interval

$$[a, b)=\{x | a \leq x<b\}.$$

Clearly, each of these intervals is bounded by the endpoints a and $b ;$ moreover, $a \in[a, b]$ and $a \in[a, b)$ (the latter provided $[a, b) \neq \emptyset,$ i.e., $a<$ $b ),$ and $a=\min [a, b]=\min [a, b) ;$ similarly, $b=\max [a, b]=\max (a, b]$. But $[a, b)$ has no maximum, $(a, b]$ has no minimum, and $(a, b)$ has neither. (Why?)

Geometrically, it seems plausible that among all left and right bounds of $A$ (if any) there are some "closest" to $A,$ such as $u$ and $v$ in Figure $1,$ i.e., a least upper bound $v$ and a greatest lower bound $u .$ These are abbreviated

$$\operatorname{lub} A \text{ and } \mathrm{glb} A$$

and are also called the supremum and infimum of $A,$ respectively; briefly,

$$v=\sup A, u=\inf A$$

However, this assertion, though valid in $E^{1},$ fails to materialize in many other fields such as the field $R$ of all rationals (cf. $§§11-12 ) .$ Even for $E^{1},$ it cannot be proved from Axioms 1 through 9.

On the other hand, this property is of utmost importance for mathematical analysis; so we introduce it as an axiom (for $E^{1} ),$ called the completeness axiom. It is convenient first to give a general definition.

Definition

An ordered field $F$ is said to be complete iff every nonvoid right-bounded subset $A \subset F$ has a supremum $($ i.e., a lub) in $F$.

Note that we use the term "complete" only for ordered fields.

With this definition, we can give the tenth and final axiom for $E^{1}$.

The Completeness Axiom

Definition

The real field $E^{1}$ is complete in the above sense. That is, each right-bounded set $A \subset E^{1}$ has a supremum $(\operatorname{sup} A ) \text { in } E ^ { 1 }$, provided $A \neq \emptyset$.

The corresponding assertion for infima can now be proved as a theorem.

Theorem $\PageIndex{1}$

In a complete field $F$ ( such as $E^{1}$), every nonvoid left-bounded subset $A \subset F$ has an infimum $(i . e .,$a glb$)$.

Proof

Let $B$ be the (nonvoid) set of all lower bounds of $A$ (such bounds exist since $A$ is left bounded $) .$ Then, clearly, no member of $B$ exceeds any member of $A,$ and so $B$ is right bounded by an element of $A .$ Hence, by the assumed completeness of $F, B$ has a supremum in $F,$ call it $p .$

We shall show that $p$ is also the required infimum of $A,$ thus completing the proof.

Indeed, we have

(i) $p$ is a lower bound of $A .$ For, by definition, $p$ is the least upper bound of $B .$ But, as shown above, each $x \in A$ is an upper bound of $B .$ Thus

$$(\forall x \in A) \quad p \leq x$$

(ii) $p$ is the greatest lower bound of $A .$ For $p=\sup B$ is not exceeded by any member of $B .$ But, by definition, $B$ contains all lower bounds of $A ;$ so $p$ is not exceeded by any of them, i.e.,

$$p=\mathrm{g} 1 \mathrm{b} A=\mathrm{inf} A$$

Note 4. The lub and glb of $A$ (if they exist) are unique. For inf $A$ is, by definition, the maximum of the set $B$ of all lower bounds of $A,$ and hence unique, by Note $2 ;$ similarly for the uniqueness of sup $A .$

Note 5. Unlike min $A$ and max $A,$ the glb and lub of $A$ need not belong to A. For example, if $A$ is the interval $(a, b)$ in $E^{1}(a<b)$ then, as is easily seen,

$$a=\inf A \text{ and } b=\sup A$$

though $a, b \notin A .$ Thus sup $A$ and inf $A$ may exist, though max $A$ and min $A$ do not.

On the other hand, if

$$q=\max A(p=\min A)$$

then also

$$q=\sup A(p=\inf A) . \quad(\mathrm{Why} ?)$$

Theorem $\PageIndex{2}$

In an ordered field $F,$ we have $q=\sup A(A \subset F)$ iff

(i) $(\forall x \in A) \quad x \leq q$ and

(ii) each field element $p<q$ is exceeded by some $x \in A ;$ i.e.,

$$(\forall p<q)(\exists x \in A) \quad p<x.$$

Equivalently,

(ii') $$(\forall \varepsilon>0)(\exists x \in A) \quad q-\varepsilon<x ; \quad(\varepsilon \in F)$$

Similarly, $p=\inf A$ iff

$$(\forall x \in A) \quad p \leq x \quad \text{ and } \quad(\forall \varepsilon>0)(\exists x \in A) \quad p+\varepsilon>x.$$

Proof

Condition (i) states that $q$ is an upper bound of $A,$ while (ii) implies that no smaller element $p$ is such a bound (since it is exceeded by some $x$ in A). When combined, (i) and (ii) state that $q$ is the least upper bound.

Moreover, any element $p<q$ can be written as $q-\varepsilon(\varepsilon>0) .$ Hence (ii) can be rephrased as $\left(\mathrm{ii}^{\prime}\right) .$

The proof for inf $A$ is quite analogous. $\square$

Corollary $\PageIndex{1}$

Let $b \in F$ and $A \subset F$ in an ordered field $F .$ If each element $x$ of $A$ satisfies $x \leq b(x \geq b),$ so does sup $A$, respectively), provided it exists in $F .$

In fact, the condition

$$(\forall x \in A) \quad x \leq b$$

means that $b$ is a right bound of $A .$ However, sup $A$ is the least right bound, so sup $A \leq b ;$ similarly for inf $A .$

Corollary $\PageIndex{2}$

In any ordered field, $\emptyset \neq A \subseteq B$ implies

$$\sup A \leq \sup B \text{ and } \inf A \geq \inf B$$

as well as

$$inf A \leq \sup A$$

provided the suprema and infima involved exist.

Proof

Let $p=\inf B$ and $q=\sup B$.

As $q$ is a right bound of $B$,

$$x \leq q \text{ for all } x \in B.$$

But $A \subseteq B,$ so $B$ contains all elements of $A .$ Thus

$$x \in A \Rightarrow x \in B \Rightarrow x \leq q$$

so, by Corollary $1,$ also

$$\sup A \leq q=\sup B,$$

as claimed.

Similarly, one gets inf $A \geq \inf B$.

Finally, if $A \neq \emptyset,$ we can fix some $x \in A .$ Then

$$\inf A \leq x \leq \sup A$$

and all is proved. $\square$