In this section we prove an important principle about real numbers that is often used as an axiom in place of Axiom C.

Definitions

A set S of real numbers has the upper bound M if x ≤ M for every number x in S; we also say that the set S is bounded above by M. The set S has the lower bound m if x ≥ m for every number x in S; we also say that S is bounded below by m. A set S is bounded if S has an upper and a lower bound. Suppose that f is a function on R¹ whose domain D contains the set S. We denote by f |S the restriction of to the set S; that is, f |S has domain S and f |S(x) f(x) for all x in S. A function f is bounded above, bounded below, or bounded if the set R consisting of the range of f satisfies the corresponding condition.

 Suppose we define

Then f is not bounded on S ={x :0 ≤ x ≤ 1}. See Figure 1.1.

 Figure 1.1

We now prove the fundamental principle on upper and lower bounds (see Figure 1.2).

Figure 1.2 U is the least upper bound for S.

Theorem 1.1

If a nonempty set S ⊂ R¹ has an upper bound M, then it has a least upper bound U; that is, there is a unique number U that is an upper bound for S but is such that no number U^`<U is an upper bound for S. If a nonempty set has a lower bound m, it has a greatest lower bound L.

Proof

The second statement follows by applying the first to the set S^/={x :−x ∈ S}.We prove the first statement using the Nested intervals theorem.

If M ∈ S, then we may take U= M, since in this case every x in S is less than or equal to U, and if U^/<U, then U^/is not an upper bound, since U ∈ S and U>U^/.If M is not in S, let b₁ =M and choose a₁ as any point of S. Now, either (a₁ + b₁)/2 is greater than every x in S, or there is some x in S greater than or equal to (a₁ + b₁)/2. In the first case, if we define a₂ = a₁ and b₂ = (a₁ + b₁)/2, then a₂ ∈ S  and b₂ is greater than every x in S. In the second case, choose for a2 one of the numbers in S that is greater than or equal to (a₁ + b₁)/2 and set b₂ = b₁; then we again have a₂ ∈ S and b₂ greater than every x in S. Continuing in this way, we define an infinite sequence {[a_n,b_n]} of closed intervals such that(1.2)

for each n, and so for each n,(1,3)

From the Nested intervals theorem, it follows that there is a unique number U in all these intervals and

Let x be any number in S. Then x<b_n for each n, so that x ≤ U by the limit of inequalities. Thus U is an upper bound for S. But now let U^/<U and let ε = U – U^/ . Then, since a_n → U, it follows that there is an N such that U^/=U − ε<a_n ≤ U for all n>N. But since all the an ∈ S, it is clear that U^/ is not an upper bound. Therefore, U is unique.

Definitions

The number U in Theorem 1.1, the least upper bound, is also called the supremum of S and is denoted by

l. u .b. S or sup S.

The number L of Theorem 1.1, the greatest lower bound, is also called the infimum of S and is denoted by

g. l .b. S or in f S.

Corollary

If x₀ is the largest number in S, that is, if x₀ ∈ S and x₀ is larger than every other number in S, then x₀ = sup S. If S is not empty and U= sup S, with U not in S, then there is a sequence {x_n} such that each x_n is in S and x_n → U. Also, if ε> 0 is given, there is a_n x in S such that  x>U − ε. Corresponding results hold for in f S.

These results follow from the definitions and from the proof of Theorem 1.1 With the help of this corollary it is possible to show that the Axiom of Continuity is a consequence of Theorem 1.1.

Definitions

Let f have an interval I of R¹ as its domain and a set in R¹ as its range.

We say that f is increasing on I if f(x₂)>f(x₁) whenever x₂ >x₁. The

function f is non decreasing on I if f(x₂) ≥ f(x₁) whenever x₂ >x₁. The

function f is decreasing on I if f(x₂)<f(x₁) whenever x₂ >x₁. The

function f is non increasing on I if f(x₂) ≤ f(x₁) whenever x₂ >x₁.A

function that has any one of these four properties is called monotone

on I.

Monotone functions are not necessarily continuous, as the following step function exhibits:

See Figure 1.3. Also, monotone functions may not be bounded. The function f : x → 1/(1 − x) is monotone on the interval I ={x :0 ≤ x< 1} but is not bounded there.

Figure 1.3 A step function.

If a function f defined on an interval I is continuous on I, then an extension of the Intermediate-value theorem shows that the range of f is an interval. In order to establish this result, we use the following characterization of an interval on R¹.

Theorem 1.2

A set S in R¹is an interval ⇔ (i) S contains more than one point and (ii) for every x₁,x₂ ∈ S the number x is in S whenever x ∈ (x₁,x₂).

Proof

If S is an interval, then clearly the two properties hold. Therefore, we suppose that (i) and (ii) hold and show that S is an interval. We consider two cases.

Case 1: S is bounded. Define a = inf S and b =sup S. Let x be an element of the open interval (a, b). We shall show that x ∈ S.

Since a = inf S, we use the corollary to Theorem 1.1 to assert that there is an x₁ ∈ S with x₁ <x. Also, since b = sup S, there is an x₂ in S with x₂ >x. Hence x ∈ (x₁,x₂), which by (ii) of the hypothesis shows that x ∈ S. We conclude that every element of (a, b) is in S. Thus S is either a closed interval, an open interval, or a half-open interval; its endpoints are a, b.

Case 2: S is unbounded in one direction. Assume that S is unbounded above and bounded below, the case of S bounded above and unbounded below being similar. Let a= inf S. Then S ⊂ [a,∞). Let x be any number such that x>a. We shall show that x ∈ S. As in Case 1, there is a number x₁ ∈ S such that x₁ <x. Since S has no upper bound there is an x₂ ∈ S such that x₂ >x. Using (ii) of the hypothesis, we conclude that x ∈ S. Therefore, S = [a,∞) or S= (a,∞).

We now establish a stronger form of the Intermediate-value theorem.

Theorem 1.3

Suppose that the domain of f is an interval I and f is continuous on I. Assume that f is not constant on I. Then the range of f , denoted by J , is an interval.

Proof

We shall show that J has properties(i) and (ii) of Theorem 1.2 and therefore is an interval. Since f is not constant, its range must have more than one point, and property (i) is established. Now let y₁,y₂ ∈ J . Then there are numbers x₁,x₂ ∈ I such that f(x₁) = y₁ and f(x₂) = y₂. We may assume that x₁ <x₂. The function f is continuous on [x₁,x₂], and so we may apply the Intermediate-value theorem. If c is any number between y₁ and y₂, there is an x₀ ∈ [x₁,x₂] such that f(x₀) = c. Thus c ∈ J , and we have established property (ii). The set J is an interval.

If f is continuous on I ={x : a ≤ x ≤ b}, it is not necessarily the case that J is determined by f(a) and f(b). Figure 1.4 shows that J may exceed the interval [f(a), f(b)]. In fact, I may be bounded and J unbounded, as is illustrated by the function f : x → 1/x with I ={x :0 <x< 1}.The range is J ={y :1 <y< ∞}. The function f : x → 1/(1 + x2) with I ={x :0 <x< ∞} and J ={y :0 <y< 1} is an example of a continuous function with an unbounded domain and a bounded range.

Consider the restriction to [0,∞) of the function f : x → x_n where n is a positive integer. By Theorem1.3, the range of f is an interval, which must be [0,∞), since f(0) = 0, f is increasing, and f(x) →+∞ as x →+∞.

Hence, for each x ≥ 0, there isat least one number y ≥ 0 such that y_n = x. If y^/>y, then (y^/)n >y_n = x. Also, if 0 ≤ y^/<y, then (y^/)n <x.

Thus the solution y is unique. We denote it by x₁/n. Every positive number has an nth root for every positive integer n. The function x → x₁/n, x ≥ 0, can be shown to be continuous on [0,∞).

Figure 1.4

Problems

In Problems1 through 8 find l.u.b. S and g.l.b. S. State whether or not these numbers are in S.

Basic Elements of Real Analysis, Murray H. Protter, Springer, 1998 .Page(44-53)