# 2.4: Upper and Lower Bounds. Completeness

- Page ID
- 19031

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

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 \(\PageIndex{2}\)

I*n 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. \(\square\)