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# 2.4: Upper and Lower Bounds. Completeness

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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. $$\square$$