Dr. Neal, WKU MATH 337 The Real Number System

Sets of Numbers A set S is a well-defined collection of objects, with well-defined meaning that there is a specific description from which we can tell precisely whether or not an element is in S . The collection of “large” numbers does not constitute a well-defined set. The set P of prime numbers is well-defined: P is the collection of integers n that are greater than 1 for which the only positive divisors of n are 1 and n . So we can tell precisely whether or not an object is in P . For instance, 13

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∈ P but 14 ∉ P because 14 = 7 × 2 . The set of rational numbers Q is also well-defined. A rational number is an expression of the form a / b , where a and b are integers with b ≠ 0 . Two rational numbers a / b and c / d are said to be equal if and only if ad = bc . It is an easy exercise to show that the sum and product of two rational numbers are also rational numbers. The set of integers Z is a subset of Q because an integer n can be written as n / 1 . Proposition 1.1. Let x and y be rational numbers with x < y . There exists another rational number z such that x < z < y . Proof. Let z = (x + y) / 2 . Because x , y , and 1/2 are rational, x + y is rational, thus so is 12× ( x + y)= z . Moreover, x = 1

2(x + x ) < 1

2(x + y) < 1

2(y + y) = y ; thus, x < z < y . QED

Existence of Irrational Numbers

The Pythagorean Theorem states: “In a right triangle, the square of the hypotenuse equals the sum of the squares of the other two sides.” So if we let c be the length of the hypotenuse of a right triangle with two other sides of length 1, then c2 = 12 + 12 = 2 . Thus, c is a positive number such that c2 = 2 ; that is, c = 2 , which exists as a physical length. But we have already proven using only facts about integers that 2 cannot be written as a ratio of integers. Thus, c is not a rational number; therefore we call it irrational. The set of real numbers ℜ is now the union of the Rationals and Irrationals.

Ordered Sets Definition 1.1. An order on a set S is a relation ≺ on S that satisfies two properties: (i) (Trichotomy) For all x , y ∈ S , exactly one of the following is true:

x ≺ y or x = y or y ≺ x . (ii) (Transitive) For all x , y ,

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z ∈ S , if x ≺ y and y ≺ z , then x ≺ z . We write

€

x≺ y to denote that either x ≺ y or x = y . Clearly

€

≺ is also transitive. That is, if x≺ y and

€

y≺ z , then

€

x≺ z .

Dr. Neal, WKU In general, x ≺ y is read as “ x precedes y” thus indicating an ordering to the set. For example, let S be the set of standard lower-case English letters. Then a ≺ c , c ≺ f , and a ≺ f . We are most concerned with the ordering on the set of real numbers ℜ , and its various subsets such as the natural numbers ℵ, the integers Z , and the rational numbers Q . The ordering used for these sets is the relation < (less than). For any two real numbers x and y , we say x < y if and only if 0 < y − x (i.e., y − x is positive). Then the relation < satisfies the two properties of an order. With numbers, we read x < y as “ x is less than y” and x≤ y is read as “ x is less than or equal to y .” Definition 1.2. Let E be a non-empty subset of an ordered set S . We say that x is the least (first) element of E provided x ∈ E and x ≺ y for all other y ∈ E . We say that x is the greatest (last) element of E provided x ∈ E and y ≺ x for all other y ∈ E . Proposition 1.2. Let E = { x1 , . . . , xn } be a finite subset of n distinct elements from an ordered set. There exists a first (least) element and a last (greatest) element in E . Proof. If n = 1 , then x1 is both the least and greatest element in E . If n = 2 , then the result follows from the Trichotomy axiom of order. Now assume the result holds for some n ≥ 2 , and assume that x1 is the least element and xn is the greatest element in E . Now consider the set E = { x1 , . . . , xn , xn+1} with a new distinct element xn+1 . If xn+1 ≺ x1 , then xn+1 is the least element and xn is the greatest element in E . If x1 ≺ xn+1 ≺ xn , then x1 is the least element and xn is the greatest element in E . If xn ≺ xn+1 , then x1 is the least element and xn+1 is the greatest element in E . With all cases being exhausted, we see that the result holds for a set of cardinality n + 1 provided it holds for a set of cardinality n . By induction, the proposition holds for all n ≥ 1. QED

Upper and Lower Bounds Definition 1.3. Let E be a subset of an ordered set S . (a) We say that E is bounded above if there exists an element β in S such that x≺ β for all x ∈E . Such an element β is called an upper bound of E . (b) An element β in S is called the least upper bound of E or supremum of E , denoted by lubE or supE , if (i) β is an upper bound of E and (ii) if λ is a different upper bound of E , then β≺ λ . Note: Suppose β = supE and λ ≺ β . Then λ cannot be an upper bound of E . Thus there must be an element x ∈E such that λ ≺ x . Example 1.1. Let E = 9 + 1

2, 9 + 2

3, 9 + 3

4, . . . ,9 + n

n + 1, . . .{ } ⊆ ℜ . We claim that

supE = 10 .

Dr. Neal, WKU First, 9 + n

n +1< 9 + n + 1

n + 1= 10 for all integers n ≥ 1; thus, 10 is an upper bound of E .

Next, suppose λ is a different upper bound of E but λ < 10 . Obviously 9 < λ or else it would not be an upper bound of E . Then 9 < λ < 10 , which makes 0 < λ − 9 < 1. Because lim

n→∞

nn + 1

= 1 , we can choose an integer n large enough such that

λ − 9 < nn + 1

< 1 . But then λ < 9 + nn + 1

< 10 , which contradicts the fact that λ is an

upper bound of E . So we must have 10 < λ which makes 10 = supE . Proposition 1.3. Let β be an upper bound of a subset E of an ordered set S . If β ∈E , then β = supE . Proof. Let λ be a different upper bound of E . Then either λ ≺ β or β ≺ λ due to the Trichotomy. But if λ ≺ β , then λ would not be an upper bound of E because β ∈E . So we must have β ≺ λ . Therefore, β is the least upper bound of E . QED Definition 1.4. Let E be a subset of an ordered set S . (a) We say that E is bounded below if there exists an element α in S such that α≺ x for all x ∈E . Such an element α is called a lower bound of E . (b) An element α in S is called the greatest lower bound of E or infimum of E , denoted by glbE or infE , if (i) α is a lower bound of E and (ii) if λ is a different lower bound of E , then λ≺α . Note: Suppose α = inf E and α≺ λ . Then λ cannot be a lower bound of E . Thus there must be an element x ∈E such that x≺ λ . Example 1.2. Let E = 1 / n n ∈ℵ{ } = {1, 1/2, 1/3, 1/4, . . . } ⊆ ℜ . Then supE = 1 , which is in E (see Proposition 1.3), but inf E = 0 which is not in E . Proposition 1.4. Let α be a lower bound of a subset E of an ordered set S . If α ∈E , then α = inf E .

(We leave the proof as an exercise.)

The Least Upper Bound Property Definition 1.5. An ordered set S is said to have the least upper bound property if the following condition holds: Whenever E is a non-empty subset of S that is bounded above, then supE exists and is an element in S .

Dr. Neal, WKU Proposition 1.5. The set of integers Z has the least upper bound property. Proof. Let E be a non-empty subset of the integers Z that is bounded above. Then there exists an integer N such that n ≤ N for all n ∈ E . Because E is non-empty, there is an integer n ∈ E with n ≤ N . By Prop. 1.3, if n is an upper bound of E , then n = supE which exists as an element in Z . If n is not an upper bound of E , then by Prop. 1.2. we pick the greatest element in E from among the finite number of integers in E that are from n to N . By Prop. 1.3, this greatest element in E is supE and is an integer because it was chosen from E . QED In the proof of Proposition 1.5, we found the least upper bound of the non-empty subset E and in fact supE is an element of E . If E were bounded below, then we also would have inf E ∈E . We therefore can state: Proposition 1.6. (The Well-Ordering Principle of Z ). (a) Let E be a non-empty subset of the integers that is bounded above. Then E has a greatest element within the set E . (b) Let E be a non-empty subset of the integers that is bounded below. Then E has a least element within the set E . The Well-Ordering Principle is essential for the study of limits of sequences, and we often shall use it implicitly in the manner illustrated in the following example: Example of the Well-Ordering Principle: Let {an}n =1

∞ be a sequence of real numbers that increases to infinity. Then we can choose the smallest integer N such that aN ≥ 100 . Because the sequence is increasing to infinity, we have a1 < a2 < a3 < . . .< an < . . . and at some point we must have 100 ≤ ak < ak+1 < ak+2 < . . . . The Well-Ordering Principle is then applied to the set of indices, not to the actual sequence values. Let E = {k : 100 ≤ ak} , which is the set of indices k for which ak ≥ 100 . Then E is a non-empty subset of the integers that is bounded below by 1 (the first index). By the Well-Ordering Principle, E has a least element within E . Thus, there exists a smallest index N such that aN ≥ 100 . Example 1.3. The set of rational numbers Q does not have the least upper bound property. As an example, let E ⊆Q be defined as follows:

E = 1. 4, 1. 41, 1. 414, 1. 4142, 1.41421, . . .{ }

= 1 +a1

101 + . . .+ ak10k

k ≥ 1, where ai is the ith decimal place value in 2⎧ ⎨ ⎩

⎫ ⎬ ⎭

.

Then E is a non-empty subset of the rational numbers and E is bounded above by 2 . In fact 2 = supE , but 2 is not a rational number. Thus, Q does not have the least upper bound property.

Dr. Neal, WKU

The Greatest Lower Bound Property Definition 1.6. An ordered set S is said to have the greatest lower bound property if the following condition holds: Whenever E is a non-empty subset of S that is bounded below, then inf E exists and is an element in S . Example 1.4. The set of integers Z has the greatest lower bound property, but the set of rational numbers Q does not. These results follow from Proposition 1.5, Example 1.3, and the following theorem: Theorem 1.1. Let S be an ordered set. Then S has the greatest lower bound property if and only if S has the least upper bound property. Partial Proof. Assume S has the least upper bound property and let E be a non-empty subset of S that is bounded below. Let L be the set of all elements in S that are lower bounds of E . Then L ≠ ∅ because E is bounded below. Because E is non-empty, there exists an element x0 in E . By definition of L , x0 is an upper bound of L ; so L is bounded above. By the least upper bound property of S , α = supL exists and is an element of S . We claim that α = inf E .

S

E (All lower bounds of E)

supL = α = inf E

L

First, let x ∈E . By definition of L , x is an upper bound of L ; thus, α≺ x because α is the least upper bound of L . Thus α is also a lower bound of E . But suppose λ is another (different) lower bound of E . Then λ ∈ L . But then λ≺α because α = supL . But because λ is different than α , we must have λ≺α due to the properties of an order. Thus, α is in fact the greatest lower bound of E and it exists in S . Thus, S has the greatest lower bound property. (We leave the proof of the converse as an exercise.)

Decimal Expansions: Rational vs. Irrational A commonly used result is that a rational number has a decimal expansion that terminates, such as 4.0 (an integer) or 2.375 (a regular number), or has a decimal expansion that recurs such as 5.342 . This property characterizes rational numbers and therefore provides another distinction between rational and irrational numbers. Theorem 1.2. A real number is rational if and only if it has a decimal expansion that either terminates or recurs.

Dr. Neal, WKU Proof. Assume first that we have a real number x with a decimal expansion that either terminates or recurs. If it terminates, then x has the form x = ± (n. a1a2 . . .ak ) , where n ,a1, . . . , ak are all non-negative integers, and 0 ≤ ai ≤ 9 for 1 ≤ i ≤ k . Thus,

x = ± n +a1101

+ . . . + ak10k

⎛ ⎝ ⎜

⎞ ⎠ ⎟ is the sum of rational numbers which is rational. (If all ai are

0, then x = ±n is an integer and thus is a rational number.) If the decimal expansion is recurring, then x has the form x = ±n. a1 . . .aj aj+1 . . .aj+k aj+1 . . . aj+ k a j+1 . . . aj+k . . .

= ± n +a1

101 + . . .+aj

10 j+

110 j

aj+110

+ . . . +aj+k10k

⎛

⎝ ⎜

⎞

⎠ ⎟ +

110 j

aj+1

10k+1 + . . .+aj+k102k

⎛

⎝ ⎜

⎞

⎠ ⎟ + . . .

⎛

⎝ ⎜

⎞

⎠ ⎟

= ± n +a1

101 + . . +aj

10 j+aj+110 j

110

+1

10k+1 +1

102k+1 +. .⎛ ⎝ ⎜

⎞ ⎠ ⎟ + . . +

aj+k10 j

110k

+1

102k +1

103k +. .⎛ ⎝ ⎜

⎞ ⎠ ⎟

⎛

⎝ ⎜

⎞

⎠ ⎟

= ± n +a1

101 + . . .+aj

10 j+aj+1

10 j+11

10k⎛ ⎝ ⎜

⎞ ⎠ ⎟

i = 0

∞∑

i+ . . .+

aj+k10 j+k

110k⎛ ⎝ ⎜

⎞ ⎠ ⎟

i= 0

∞∑

i⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

= ± n +a1

101 + . . .+aj

10 j+

aj+110 j+1 + . . . +

aj+k10 j+ k

⎛

⎝ ⎜

⎞

⎠ ⎟

110k⎛ ⎝ ⎜

⎞ ⎠ ⎟

i= 0

∞∑

i⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

= ± n +a1

101 + . . .+aj

10 j+

110 j

aj+1101 + . . . +

aj+k10k

⎛

⎝ ⎜

⎞

⎠ ⎟ ×

10k

10k −1

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ , which is rational .

Conversely, let x = c / d be a rational number. We must obtain its decimal expansion. By dividing d into c , we may assume that x is of the form x = ± (n + a / b) , where n ,a ,b are non-negative integers with b ≠ 0 and 0 ≤ a < b . If a = 0 , then x = ±n is a terminating decimal expansion. If a ≠ 0 , then let r0 = a and divide b into 10r0 to obtain

10r0 = a1 b + r1 where 0 ≤ r1 < b and 0 ≤ a1 ≤ 9 because 0 ≤ r0 < b .

Then divide b into 10r1 to obtain

10r1 = a2 b + r2 where 0 ≤ r2 < b and 0 ≤ a2 ≤ 9 because 0 ≤ r1 < b .

Continue dividing b into 10r j to obtain

10r j = aj+1 b + r j+1 where 0 ≤ r j+1 < b and 0 ≤ aj+1 ≤ 9 because 0 ≤ r j < b .

Because all remainders ri must be from 0 to b − 1 , there can be at most b distinct remainders. At some point, we will have 10rk = ak+1 b + rk+1 , where rk+1 = ri for some previous remainder ri . Because the next step of dividing 10ri by b yields the same quotient ai+1 and remainder as before, the pattern recurs for the successive divisions. We now claim that x = ± (n. a1a2 .. . ai ai+1. .ak+1 ai+1. . ak+1 ai+1. . ak+1 . . . ) To see this result, we first use 10a = 10r0 = a1 b + r1 . Dividing by 10b gives

Dr. Neal, WKU

ab

=a110

+110

r1b

⎛ ⎝ ⎜

⎞ ⎠ ⎟ .

But then 10r1 = a2 b + r2 , which gives r1

b=a210

+110

r2b

⎛ ⎝ ⎜

⎞ ⎠ ⎟ ; hence,

ab

=a110

+110

a210

+110

r2b

⎛ ⎝ ⎜

⎞ ⎠ ⎟

⎛ ⎝ ⎜

⎞ ⎠ ⎟ =

a110

+a2102

+1102

r2b

⎛ ⎝ ⎜

⎞ ⎠ ⎟ .

Continuing, we obtain

ab

=a110

+a2102

+ . . .+ ai10i

+ai+110i+1

+. . . + ak+110k+1 +

110k+1

rk+1b

⎛ ⎝ ⎜

⎞ ⎠ ⎟ ,

where rk+1 = ri . But then we have

ab

=a110

+a2

102 + . . .+ ai10i

+ai+1

10i+1 +. . . +ak+1

10k+1 +ai+1

10k+2 +1

10k+2ri+1b

⎛ ⎝ ⎜

⎞ ⎠ ⎟

=a110

+a2

102 + . . .+ ai10i

+ai+1

10i+1 +. . . +ak+1

10k+1 +ai+1

10k+2 +ai+ 2

10k+3 +1

10k+3ri+3b

⎛ ⎝ ⎜

⎞ ⎠ ⎟

=a110

+a2

102 + . . .+ ai10i

+ai+1

10i+1 +. . . +ak+1

10k+1 +ai+1

10k+2 +ai+ 2

10k+3 +. . . +ak+1

102(k+1)−i

+1

102(k+1)−irk+1b

⎛ ⎝ ⎜

⎞ ⎠ ⎟

By recursion, we obtain a recurring decimal expansion of x that terminates if b ever divides evenly into a remainder. QED Example 1.5 (a) Find the rational form of 5.86342342342 . . . (b) Find the decimal expansion of 3/8. (c) Find the decimal expansion of 3/14. Solution. (a) By Theorem 1.2, we have

5.86342 = 5 + 810

+6100

+1100

× 0.342 × 103

103 −1

= 5 + 810

+6100

+1100

×342999

= 5 + 810

+6100

+34299900

=162712775

.

(b) For a / b = 3 / 8 , we first divide 8 into 10 × 3 to obtain the next remainder r1 , then continue dividing 8 into 10ri until we obtain a 0 remainder or a repeated remainder. The decimal terms of 3/8 come from the successive quotients:

Dr. Neal, WKU

30 = 3 × 8 + 660 = 7 × 8 + 440 = 5 × 8 + 0 0 = 0 × 8 + 0

Thus, 3/8 = 0.375.

(c) For a / b = 3 / 14 , we first divide 14 into 10 × 3 to obtain the next remainder r1 , then continue dividing 14 into 10ri until we obtain a 0 remainder or a repeated remainder. The decimal terms of 3/14 come from the successive quotients:

30 = 2 ×14 + 220 = 1 ×14 + 660 = 4 × 14 + 440 = 2 × 14 +12120 = 8 × 14 + 880 = 5 × 14 +10100 = 7 ×14 + 2 ← repeated remainder20 = 1 ×14 + 6

↑ repeated quotient

Thus, 3/14 = 0.2142857142857...

Note: The decimal expansion of a rational number is not unique. An expansion that terminates also can be written as an expansion that ends in a string of all 9's. Specifically, the decimal x = n .a1a2 .. . ak , where 1 ≤ ak ≤ 9 , can be re-written as x = n .a1a2 .. . (ak −1) 999999 . . . . This result follows from the geometric series

9 110⎛ ⎝ ⎜

⎞ ⎠ ⎟ i

i= k+1

∞∑ = 9 × (1 / 10)

k+1

1 −1 / 10= 9 × (1 / 10)

k+1

9 / 10=

110⎛ ⎝ ⎜

⎞ ⎠ ⎟ k

.

For example, with k = 3 we have 0.000 9999 . . .= 0.001 . As another example, we have 2.4837 = 2.48369999 .

By convention, we always shall assume that a

rational number never ends in a string of all 9's.

Dr. Neal, WKU

An immediate consequence to Theorem 1.2 is the following result: Corollary 1.1. A number is irrational if and only if it has a decimal expansion that neither terminates nor recurs. We have already seen that 2 is irrational. An example of another irrational number is

x = 0.123456789112233445566778899111222333 . . .

Another Construction of the Reals Theorem 1.2 and Corollary 1.1 give a method of constructing the real numbers based on the natural numbers. A brief outline of the steps follows: I. Let 0 denote no length and let 1 denote a fixed unit of length. II. By extending the initial length of 1 by incremental successive lengths of 1, we obtain the natural numbers ℵ = {1, 2, 3, . . .}. Including 0 gives us the whole numbers W . III. By considering all quotients a / b of natural numbers, we obtain the positive rational numbers Q+ . Also allowing a = 0 gives us the non-negative rational numbers.

IV. Every element in Q+ can be written uniquely in the form n + ai10ii= 1

∞∑ , where n ∈W ,

0 ≤ ai ≤ 9 for all i , the sequence {ai} either terminates by ending in all 0's or is recurring, and the sequence {ai} does not end in a string of all 9's. V. Let the non-negative irrational numbers I+ be defined by all values of the form

n + ai10ii= 1

∞∑ , where n ∈W , 0 ≤ ai ≤ 9 for all i , and the sequence {ai} neither terminates

nor recurs. (In particular, irrational numbers also cannot end in a string of all 9's.) VI. Let the positive real numbers be defined by R+ = Q+ ∪ I+ .

VII. Negative numbers can be written in the form −n ± ai10ii=1

∞∑ .

A real number is therefore either rational or irrational. By comparing the decimal expansions of any two real numbers x and y , we can see that either x < y , x = y , or y < x . Thus, the Reals are ordered. The next results prove that both the rational numbers and the irrational numbers are dense within the Reals.

Dr. Neal, WKU Theorem 1.3. Let x and y be any two real numbers with 0 ≤ x < y . (a) There exists a rational number z such that x < z < y . (b) There exists an irrational number w such that x < w < y . Proof. Let x = m. a1a2a3 . . . and y = n. b1b2b3 . . . be the decimal forms of x and y where 0 ≤ m ≤ n . Suppose first that m < n . Because x cannot end in a string of all 9's, let ai be the first decimal place value such that ai < 9 . Then change ai to ai + 1 and let z = m. a1a2a3 . . .(ai + 1) , which is rational, and let w = m. a1a2a3 . . .(ai + 1)c1c2c3 . . . , where the sequence {ci} neither terminates nor recurs, so that w is irrational. We then have x < z < y and x < w < y . Next assume that m = n . Because x < y , we must have a1 ≤ b1 . We now choose the smallest index i such that ai < bi . Then we must have aj = bj for 1 ≤ j < i or else x would be larger than y . Now because x cannot end in a string of all 9's, choose the least k ≥ i such that ak < 9 . Now let z = m. a1a2a3 . . .ai . . .(ak + 1) . Then z is rational and x < z < y . Finally, let w = m. a1a2a3 . . .ai . . . (ak + 1)c1c2c3 . . . where the sequence {ci} neither terminates nor recurs, so that w is irrational. Then x < w < y . QED Corollary 1.2. Let x and y be any two real numbers with x < y . (a) There exists a rational number z such that x < z < y . (b) There exists an irrational number w such that x < w < y . Proof. If 0 ≤ x < y , then the result is proved in Theorem 1.3. If x < 0 < y , then by Theorem 1.3, we can find such z and w such that x < 0 < z < y and x < 0 < w < y . If x < y ≤ 0 , then 0 ≤ −y < −x . So we can apply Theorem 1.3 to find a rational z and an irrational w such that − y < z < −x and − y < w < −x . But then − z is rational and −w is irrational with x < − z < y and x < −w < y . QED Example 1.6. Apply Theorem 1.3 to x = 8.9942 . . . and y = 9.01 assuming x is rational and assuming x is irrational. Solution. Let z = 8.995 , which is rational. Then x < z < y regardless of whether x is rational or irrational. If x is irrational, then let w = x + 0.001 , which is irrational. Then x < w < y . However if x is rational, then let w = 8.995 + 2 − 1( ) / 1000 = 8.9954142 . . . , which is irrational. (This expression simply appends the decimal digits of 2 to the end of 8.995.) Then x < w < y .

We continue with two of the most important property of real numbers:

Dr. Neal, WKU Theorem 1.4. The set of real numbers has the greatest lower bound property. Proof. Let E a non-empty subset of ℜ that is bounded below. Let r ∈ℜ be a lower bound of E and let F be the set of all integers k such that k ≤ r . Then r is an upper bound of F ; thus, by Prop. 1.6, we can choose a greatest element n in F . Then n ≤ r , so n is also a lower bound of E . By Prop. 1.4, if n ∈ E , then n = inf E . Otherwise, we continue: Choose the greatest integer a1 from 0 to 9 such that n + a1 / 10 is a lower bound of E . If n + a1 / 10 ∈ E , then n + a1 / 10 = inf E . Otherwise, choose the greatest integer a2 from 0 to 9 such that n + a1 / 10 + a2 / 10

2 is a lower bound of E . Continuing, we thereby obtain a real number α = n + ai / 10

ii=1∞∑ with α ≤ n + 1.

We claim α = inf E . So let x ∈E . By construction, n + ai / 10i

i=1k∑ ≤ x for all k ≥ 1 . In

particular, n ≤ x . If n + 1 ≤ x , then α ≤ n + 1 ≤ x . But suppose x < n + 1 with x = n + bi / 10

ii=1∞∑ where the string {bi} does not end in all 9's. If we had x < α , then

we must have bk < ak for some k and we can choose the smallest integer k for which this occurs. But then we would have x < n + ai / 10

ii=1k∑ , which is a contradiction.

Thus, α ≤ x ; that is, α is a lower bound of E . Next, suppose λ is a different lower bound of E . We must show that λ < α . So write λ = m + ci / 10

ii=1∞∑ Then m ≤ λ , so m is also a lower bound of E . By the choice

of n above, we must have m ≤ n . If m < n , then λ < α . So suppose m = n . Now if ak < ck for some k and we choose the smallest k where this occurs, then we would

have n +i= 1

k∑ ci / 10

i ≤ λ ≤ x for all x ∈E . But then this ck contradicts the choice of ak .

So we must have ck ≤ ak for all k . That is, λ ≤ α . But because λ is different that α , we must have λ < α . That is, α = inf E . QED

From Theorems 1.1 and 1.4, we obtain: Corollary 1.2. The set of real numbers has the least upper bound property. That is, any non-empty subset of ℜ that is bounded above has a least upper bound in ℜ . An important consequence of the glb and lub properties of the real line is the fact that the intersection of a nested decreasing sequence of closed intervals is non-empty:

[a1, b1]⊇ [a2 , b2 ] ⊇ [a3, b3 ] ⊇ . . .⊇ [sup{ai}, inf{bi}]

a1 ≤ a2 ≤ a3 ≤ . . . ≤ ai ≤ . . . ≤ sup{ai} ≤ inf{bi} ≤ . . . ≤ bi ≤. . . ≤ b3 ≤ b2 ≤ b1

Theorem 1.5. (Nested Interval Property) Let [ai , bi ] be a nested decreasing sequence of

closed intervals. Then [sup{ai}, inf{bi}]⊆ [ai, bi ]i=1

∞

∩ . In particular, [ai, bi ]i=1

∞

∩ ≠∅ .

Dr. Neal, WKU Proof. Because the intervals are nested decreasing, we have ai ≤ ai+1 ≤ bi+1 ≤ bi for i ≥ 1. So the set of points {ai} is bounded above by each bi and the set of points {bi} is bounded below by each ai ; hence a = sup{ai} and b = inf{bi} exist by the least upper bound and greatest lower bound properties of ℜ . Because each bi is an upper bound of {ai} and a is the least upper bound of {ai}, we have a ≤ bi for all i ≥ 1. But then a becomes a lower bound of the {bi} ; so a ≤ b because b is the greatest lower bound of the {bi} . For all i ≥ 1, we then have

ai ≤ a ≤ b ≤ bi ; thus, [a, b]⊆ [ai, bi ]i=1

∞

∩ . In particular, [ai, bi ]i=1

∞

∩ contains a and b . QED

Exercises 1. Prove: If c is rational, c ≠ 0 , and x is irrational, then c x and c + x are irrational. 2. Prove that 12 is irrational. 3. Let E be a non-empty subset of an ordered set. Suppose α is a lower bound of E and β is an upper bound of E . Prove that α≺β . 4. Let E be a non-empty subset of an ordered set S that has the least upper bound property. Suppose E is bounded above and below. Prove that inf E≺ supE . 5. Let S be a non-empty subset of the real numbers that is bounded below. Define −S by −S = {− x x ∈S }. Prove: (i) −S is bounded above. (ii) lub(−S) = −glbS . 6. Prove: Let α be a lower bound of a subset E of an ordered set S . If α ∈E , then α = inf E . 7. Let S be an ordered set with the greatest lower bound property. Prove that S has the least upper bound property. 8. Find a rational number and an irrational number between x = 8.9949926 and y = 8.995 . 9. Let a ,b ∈ℜ with a < b , (i) For the open interval E = (a, b) , prove that inf E = a and supE = b . (ii) For the closed interval F = [a , b], prove that inf F = a and supF = b .