Download MATH 337 The Real Number System

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 ∈ 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
€ of two rational numbers are also rational numbers.
to show that the sum and product
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
1
1
1
1
× ( x + y) = z . Moreover, x = (x + x ) < (x + y) < (y + y) = y ; thus, x < z < y . QED
2
2
2
2
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
2
2
2
hypotenuse of a right triangle with two other sides of length 1, then c = 1 + 1 = 2 .
2
Thus, c is a positive number such that c = 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 , 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
lub E or sup E , 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 = sup E 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.
sup E = 10 .
{
Let E = 9 +
1
2
3
n
, 9 + , 9 + , . . .,9 +
, . . . ⊆ ℜ.
2
3
4
n+1
}
We claim that
Dr. Neal, WKU
n
n+1
<9+
= 10 for all integers n ≥ 1; thus, 10 is an upper bound of E .
n +1
n+1
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 .
n
= 1 , we can choose an integer n large enough such that
Because lim
n →∞ n + 1
n
n
λ −9<
< 1 . But then λ < 9 +
< 10 , which contradicts the fact that λ is an
n+1
n+1
upper bound of E . So we must have 10 < λ which makes 10 = sup E .
First, 9 +
Proposition 1.3. Let β be an upper bound of a subset E of an ordered set S . If β ∈ E ,
then β = sup E .
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
glb E or inf E , 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 sup E = 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 sup E exists and is an element in S .
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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 = sup E 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 sup E 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 sup E 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 a N ≥ 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 a N ≥ 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, . . .}
⎧
a
a
= ⎨1 + 11 + . . .+ kk k ≥ 1, where ai is the ith decimal place value in
⎩ 10
10
⎫
2⎬.
⎭
Then E is a non-empty subset of the rational numbers and E is bounded above by 2 .
In fact 2 = sup E , 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 = sup L exists and is an
element of S . We claim that α = inf E .
sup L = α = inf E
L
(All lower bounds of E)
E
S
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 = sup L .
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,
⎛
a
a ⎞
x = ± ⎜ n + 11 + . . . + kk ⎟ is the sum of rational numbers which is rational. (If all ai are
⎝
10
10 ⎠
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 . . .a j a j+1 . . .a j +k a j +1 . . . a j+ k a j+1 . . . a j +k . . .
⎛
aj
a j +k ⎞
a j +k ⎞
⎞
a
1 ⎛ a j+1
1 ⎛ a j +1
= ± ⎜ n + 11 + . . .+ j + j ⎜
+ . . . + k ⎟ + j ⎜ k +1 + . . .+ 2k ⎟ + . . .⎟
⎝
⎠
10
10
10 ⎝ 10
10 ⎠ 10 ⎝ 10
10 ⎠
⎛
a
a
⎛ 1
⎞
a
⎛ 1
⎞⎞
a
1
1
1
1
= ± ⎜ n + 11 + . . + jj + j +1
⎜
+ k+1 + 2 k+1 +. .⎟ + . . + j +kj ⎜ k + 2k + 3k +. .⎟ ⎟
j
⎠
⎠⎠
⎝
10
10
10 ⎝ 10 10
10
10 ⎝ 10
10
10
i
i
⎛
aj
a j +1 ∞ ⎛ 1 ⎞
a j +k ∞ ⎛ 1 ⎞ ⎞⎟
a1
⎜
= ± ⎜ n + 1 + . . .+ j + j +1 ∑ ⎜ k ⎟ + . . .+ j +k ∑ ⎜ k ⎟ ⎟
⎝
⎠
⎝
⎠ ⎠
10
10
10
10
⎝
i = 0 10
i = 0 10
⎛
a j +k ⎞ ∞ ⎛ 1 ⎞ i ⎞
a j ⎛ a j+1
a1
⎜
= ± ⎜ n + 1 + . . .+ j + ⎜ j +1 + . . . + j+ k ⎟ ∑ ⎜ k ⎟ ⎟⎟
⎝ 10
⎠ i = 0 ⎝ 10 ⎠ ⎠
10
10
10
⎝
⎛
aj
a j +k ⎞
a1
1 ⎛ a j+1
10 k ⎞⎟
⎜
= ± ⎜ n + 1 + . . .+ j + j ⎜ 1 + . . . +
⎟×
⎟ , which is rational .
⎝
10
10
10 ⎝ 10
10 k ⎠ 10 k − 1 ⎠
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 = a j +1 b + r j +1 where 0 ≤ r j+1 < b and 0 ≤ a j +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
a a1 1 ⎛ r1 ⎞
=
+ ⎜ ⎟.
b 10 10 ⎝ b ⎠
r
a
1 ⎛r ⎞
But then 10r1 = a2 b + r2 , which gives 1 = 2 + ⎜ 2 ⎟ ; hence,
b 10 10 ⎝ b ⎠
a a1 1 ⎛ a2 1 ⎛ r2 ⎞ ⎞ a1 a2
1 ⎛r ⎞
=
+ ⎜⎝
+ ⎜⎝ ⎟⎠ ⎟⎠ =
+ 2 + 2 ⎜⎝ 2 ⎟⎠ .
b 10 10 10 10 b
10 10
b
10
Continuing, we obtain
a a1
a
a
a
a +1
1 ⎛r
⎞
=
+ 22 + . . .+ ii + ii+1
+. . . + kk+1
+ k +1 ⎜⎝ k +1 ⎟⎠ ,
+1
b 10 10
b
10
10
10
10
where rk +1 = ri . But then we have
a a1
a
a
a
a +1
a
1 ⎛r ⎞
=
+ 22 + . . .+ ii + ii+1
+. . . + kk+1
+ ik+1
+ k +2 ⎜ i+1 ⎟
+1
+2
⎝ b ⎠
b 10 10
10 10
10
10
10
a
a
a
a
a +1
a
a 2
1 ⎛r ⎞
= 1 + 22 + . . .+ ii + ii+1
+. . . + kk+1
+ ik+1
+ i+
+ k +3 ⎜ i +3 ⎟
+1
+2
k
+3
⎝ b ⎠
10 10
10 10
10
10
10
10
a
a
a
a
ak +1
ai +1
ai+ 2
ak +1
= 1 + 22 + . . .+ ii + ii+1
+.
.
.
+
+
+
+.
.
.
+
10 10
10 10 +1
10 k+1 10k +2 10k +3
10 2(k+1)−i
⎛r ⎞
1
+ 2(k+1)−i ⎜⎝ k+1 ⎟⎠
b
10
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 . . .
expansion of 3/8. (c) Find the decimal expansion of 3/14.
(b) Find the decimal
Solution. (a) By Theorem 1.2, we have
8
6
1
103
+
+
× 0.342 × 3
10 100 100
10 − 1
8
6
1
342
=5+
+
+
×
10 100 100 999
8
6
342
16271
=5+
+
+
=
.
10 100 99900
2775
5.86342 = 5 +
(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 + 6
60 = 7 × 8 + 4
40 = 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 + 2
20 = 1 × 14 + 6
60 = 4 × 14 + 4
40 = 2 × 14 + 12
120 = 8 × 14 + 8
80 = 5 × 14 + 10
100 = 7 × 14 + 2 ← repeated remainder
20 = 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
∞ ⎛ 1 ⎞i
(1 / 10)k+1
(1 / 10)k+1 ⎛ 1 ⎞ k
9 ∑ ⎜⎝ ⎟⎠ = 9 ×
= 9×
= ⎜⎝ ⎟⎠ .
1 − 1 / 10
9 / 10
10
i = k +1 10
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.
∞ a
IV. Every element in Q+ can be written uniquely in the form n + ∑ ii , where n ∈ W ,
i = 1 10
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
∞ a
n + ∑ ii , where n ∈ W , 0 ≤ ai ≤ 9 for all i , and the sequence {ai } neither terminates
i = 1 10
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 + .
∞ a
i
VII. Negative numbers can be written in the form − n ± ∑
i
i =1 10
.
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.
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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. a1a2 a3 . . . and y = n. b1b2 b3 . . . 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. a1a2 a3 . . .(ai + 1) , which is rational, and let w = m. a1a2 a3 . . .(ai + 1)c1c2 c3 . . . ,
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 a j = b j 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. a1a2 a3 . . .ai . . .(ak + 1) . Then z is rational
and x < z < y . Finally, let w = m. a1a2 a3 . . .ai . . . (ak + 1)c1c2 c3 . . . 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:
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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 / 102 is a lower bound of E .
Continuing, we thereby obtain a real number α = n + ∑i∞=1 ai / 10i with α ≤ n + 1.
We claim α = inf E . So let x ∈ E . By construction, n + ∑ik=1 ai / 10i ≤ 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 + ∑i∞=1 bi / 10i 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 + ∑ki =1 ai / 10i , 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
i
write λ = m + ∑∞
i =1 ci / 10 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 < c k for some k and we choose the smallest k where this occurs, then we would
k
have n + ∑ ci / 10i ≤ λ ≤ x for all x ∈ E . But then this ck contradicts the choice of ak .
i=1
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 ] . In particular,
i=1
∞
∩ [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 ] . In particular,
i=1
∞
∩ [ai, bi ] contains a and b .
QED
i=1
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 ≺ sup E .
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) = −glb S .
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 sup E = b .
(ii) For the closed interval F = [a, b], prove that inf F = a and sup F = b .