60
2.2
CHAPTER 2. IMPORTANT PROPERTIES OF R
Some Consequences of the Completeness Axiom
In this section, we use the fact that R is complete to establish some important
results. First, we will prove that Z is unbounded and establish the Archimedean
principle. Second, we will prove that the rational numbers are dense in R.
Finally, we will prove that Q is not complete.
2.2.1
Archimedean Property
We …rst establish that Z is unbounded. While this result seems obvious, it turns
out that it is not as easy to prove. It depends upon the completeness property
of R. We establish this result by proving several lemmas and then use these
lemmas to establish the main result.
Lemma 2.2.1 Every non-empty subset S of the integers which is bounded above
has a largest element.
Proof. Since S is a non-empty subset of Z (hence of R) which is bounded above,
by the supremum property (completeness axiom), sup S exists. Let w = sup S.
It is enough to that w 2 S. In a problem from the previous section, it was
established that if a set contains its supremum, then the supremum is also the
largest element. We do a proof by contradiction. Suppose that w 2
= S. Since
w 1 < w, by theorem 2.1.33, there exists m 2 S such that
w
1<m
w
Since we assumed w 2
= S, we can even write
w
1<m<w
We repeat the same argument using the fact that m < w to …nd n 2 S such that
m < n < w. So, we have
w 1<m<n<w
The inequality n < w is equivalent to w < n. If we add up the two inequalities
w 1 < m and w < n, we obtain 1 < m n which is equivalent to n m < 1.
But since m < n, we have n m > 0 and therefore 0 < n m < 1. Since both
m and n are in S, they are integers. Therefore, n m is also an integer. Since
there are no other integers between 0 and 1, the inequality
0<n
m<1
is not possible. It means that our assumption w 2
= S cannot happen. Hence,
w 2 S.
Remark 2.2.2 This lemma looks very similar to the least upper bound property.
But its conclusion is quite di¤ erent. The result of this lemma is obviously not
true for subsets of R. An example is (0; 5) which is not empty, bounded above but
2.2. SOME CONSEQUENCES OF THE COMPLETENESS AXIOM
61
has no largest element. For this example, we would like to say that the largest
element is the number "just before 5". The problem is that such a number does
not exist. However, in the case of a subset of Z, such a number exists. Every
integer has a successor and a predecessor.
Lemma 2.2.3 Every non-empty subset S of the integers which is bounded below
has a smallest element.
Proof. Similar to the previous lemma and left as an exercise.
Theorem 2.2.4 Z is unbounded both above and below.
Proof. If Z were bounded above, by lemma 2.2.1 it would have a largest element,
which we know it does not. A similar arguments is used to show that Z is not
bounded below.
We are now ready to state the Archimedean principle.
Theorem 2.2.5 (Archimedean Property) For each strictly positive real num1
ber x, there exists a positive integer n such that < x.
n
1
cannot be an upper bound of Z
Proof. Let x be as in the theorem. Then,
x
since Z is not bounded above. Therefore, we can …nd n 2 Z and n > 0 such that
1
<n
x
It follows that
1
< x.
n
This theorem tells us that by choosing n large enough, we can make
1
as
n
1
: n 2 Z+ = 0 where Z+ denotes
n
the set of positive integers. We will prove this in the exercises.
The Archimedean property is sometimes stated di¤erently. We give an equivalent statement which proof is left as an exercise.
close to 0 as we want. In other words, inf
Theorem 2.2.6 (Archimedean Property 2) If x and y are real numbers,
x > 0, then there exists a positive integer n such that nx > y.
Proof. See exercises.
2.2.2
Denseness of Q in R
We have already mentioned the fact that if we represented the rational numbers
on the real line, there would be many holes. These holes would correspond
to the irrational numbers. If we think of the rational numbers as dots on the
real line and the irrational numbers as holes, one might ask how the holes are
distributed with respect to the dots. The next concept gives us a partial answer.
62
CHAPTER 2. IMPORTANT PROPERTIES OF R
De…nition 2.2.7 (Dense) A subset S of R is said to be dense in R if between
any two real numbers there exists an element of S.
Another way to think of this is that S is dense in R if for any real numbers
a and b such that a < b, we have S \ (a; b) 6= ?.
Theorem 2.2.8 Q is dense in R. That is, between any two real numbers, there
exists a rational number.
Proof. Let a and b be two arbitrary real numbers such that a < b. Then,
b a > 0. By the Archimedean principle, we can choose n 2 Z such that
1
0< <b a
n
This is equivalent to saying that
1
a<a+ <b
(2.1)
n
We will …nish the proof by showing there exists an integer m such that
m
a<
<b
n
We de…ne
o
n
x
S =
x2Z: >a
n
= fx 2 Z : x > nag since n > 0
This set is bounded below by na. It is also non-empty since Z is not bounded
above. Hence, by lemma 2.2.3, S has a smallest element. Call it m. We will
show m is the integer we are looking for. We have
m
>a
(2.2)
n
but
m 1
a
(2.3)
n
m
since m is the smallest integer for which
> a. If we write
n
m
m 1
1
=
+
(2.4)
n
n
n
then, combining equations 2.2 and 2.4 gives
m
m 1
1
a<
=
+
n
n
n
If we use equation 2.3, we obtain
m
1
a<
a+
n
n
Finally, using equation 2.1 gives
m
<b
a<
n
which is what we wanted to prove.
2.2. SOME CONSEQUENCES OF THE COMPLETENESS AXIOM
2.2.3
63
Completeness
De…nition 2.2.9 A set S is said to be complete if every non-empty bounded
subset of S has both a supremum and an in…mum in S.
Example 2.2.10 R is complete, by the completeness axiom.
Example 2.2.11 [0; 1] is complete. If T
[0; 1] and T 6= ? then by the completeness axiom, inf T and sup T exist since T
R. Furthermore, 0 is a lower
bound of T and 1 is an upper bound of T hence 0
inf T
sup T
1 hence
both inf T and sup T belong to [0; 1]. Since T was arbitrary, the result follows.
Example 2.2.12 (0; 1) is not complete since sup ((0; 1)) = 1 2
= (0; 1).
Corollary 2.2.13 Q is not complete.
Proof. De…ne
n
p o
S =
x2Q:0<x< 2
p
=
0; 2 \ Q
Clearly, S is a subset of Q. We will show that p
Q is not complete by showing
that S does not
have
a
supremum
in
Q.
Because
2 is anpupper bound of S, we
p
2. We want to show that
in
fact,
sup
S = 2. We do a proof by
have sup S
p
contradiction. Suppose that sup S < 2.pSince Q is dense in R, we can …nd a
rational number q such that sup S < q p
< 2. Thus q 2 S. Therefore, we
p should
have q
sup S and not sup S < q < 2. So, we must have sup S = 2. We
see that S, a subset of Q has a supremum which is not in Q.
2.2.4
Exercises
1. Prove lemma 2.2.3
2. Prove that inf
1
: n 2 Z+
n
= 0 where Z+ denotes the set of positive
integers.
3. Find the supremum and in…mum of the sets below.
(a) A =
1 1 1
1; ; ; ; :::
2 4 8
(b) E =
2+n
:n2N .
n
=
1
2n
1
:n2N .
4. Prove theorem 2.2.6 using the results of this section.
5. Prove that between any two real numbers there exists an irrational number. This proves that the set of irrational numbers, R n Q is dense in
R.
64
CHAPTER 2. IMPORTANT PROPERTIES OF R
6. Prove that every interval which contains more than one element must
contain an in…nite number of rational as well as irrational numbers (hint:
do it by contradiction).
7. Given x 2 R, prove there is a unique (9!) n 2 Z such that n
1
x < n.
8. The goal of this problem is to give a "real analysis" proof of the division
algorithm which states that if a and b are positive integers, b 6= 0, then
there exists integers r and q where 0 r < b such that a = bq + r. We do
the proof in several steps. Let a and b be positive integers and de…ne
S = fn 2 Z : nb
ag
(a) Prove that S 6= ?.
(b) Prove that S is bounded above.
(c) Explain why S has a largest member that we will call q.
(d) Prove that if we de…ne r = a
the same as a = bq + r):
bq then 0
9. Are the integers complete? Justify your answer.
r < b. (Note that this is
2.2. SOME CONSEQUENCES OF THE COMPLETENESS AXIOM
2.2.5
65
Hints for the Exercises
For all the problems, remember to …rst clearly state what you have to prove or
do.
1. Prove lemma 2.2.3.
Hint: Proof is similar to that of lemma 2.2.1
2. Prove that inf
1
: n 2 Z+
n
= 0 where Z+ denotes the set of positive
integers.
Hint: Use the de…nition of the in…mum and the Archimedean property.
3. Find the supremum and in…mum of the sets below.
1
1 1 1
:n2N .
1; ; ; ; ::: =
2 4 8
2n 1
Hint: Use the de…nition of supremum and in…mum.
2+n
(b) E =
:n2N .
n
Hint: Use the de…nition of supremum and in…mum. For in…mum,
also use the Archimedean property.
(a) A =
4. Prove theorem 2.2.6 using the results of this section.
Hint: Consider di¤erent cases on y and use 2.2.5.
5. Prove that between any two real numbers there exists an irrational number. This proves that the set of irrational numbers, R n Q is dense in R.
Hint: Use the de…nition of denseness, but p
instead ofpapplying it to two
arbitrary real numbers a and b, apply it to 2a and 2b.
6. Prove that every interval which contains more than one element must
contain an in…nite number of rational as well as irrational numbers (hint:
do it by contradiction).
Hint already given.
7. Given x 2 R, prove there is a unique (9!) n 2 Z such that n 1 x < n.
Hint: You need to prove existence and uniqueness. For existence, de…ne
S = fk 2 Z : k > xg. Explain why S is bounded below and not empty.
What does this imply? For uniqueness, assume that there are two such
numbers, say m and n. Using a proof similar to that of lemma 2.2.1, prove
that m = n by showing that 1 < m n < 1.
8. The goal of this problem is to give a "real analysis" proof of the division
algorithm which states that if a and b are positive integers, b 6= 0, then
there exists integers r and q where 0 r < b such that a = bq + r. We do
the proof in several steps. Let a and b be positive integers and de…ne
S = fn 2 Z : nb
ag
66
CHAPTER 2. IMPORTANT PROPERTIES OF R
No hints since you are already guided through the proof.
(a) Prove that S 6= ?.
(b) Prove that S is bounded above.
(c) Explain why S has a largest member that we will call q.
(d) Prove that if we de…ne r = a
the same as a = bq + r):
bq then 0
r < b. (Note that this is
9. Are the integers complete? Justify your answer.
Hint: Use the de…nition and the results of this section.