CONSEQUENCES OF THE COMPLETENESS AXIOM
1. Archimedean Property
The first consequence we will discuss today is the so-called Archimedean Property. This is the property responsible for the fact that given any real number we
can find an integer which exceeds it.
Archimedean Property. For all ε > 0 and all b > 0 there exists n ∈ N such that
nε > b.
Proof. We assume the opposite: that for all n ∈ N we have nε ≤ b. Under this
assumption the set
! "
#
S := nε " n ∈ N
is bounded from above by b. It follows from the Completeness Axiom that the set
S has the supremum
M = sup S.
Since M is the smallest upper bound for S we see that M − ε cannot be an upper
bound for S. This means that there is some element of S, say n0 ε, such that
M − ε < n0 ε ≤ M.
The last inequality implies M < n0 ε + ε i.e. M < (n0 + 1)ε. Since (n0 + 1)ε ∈ S
we have a contradiction with our assumption that M = sup S. This contradiction
proves the Archimedean Property.
!
An immediate consequence of the Archimedean Property is that
$
%$
%
(1.1)
∀ε < 0 ∀b < 0 (∃n ∈ N) nε < b.
Indeed, one only needs to set δ = −ε, c = −b and apply the Archimedean Property
to δ > 0 and c > 0. Thus, there exists n ∈ N such that nδ > c i.e. nε < b.
Corollary. We have the following;
(1) For all x ∈ R there is n ∈ N with n > x;
(2) For all x ∈ R there is z ∈ Z such that z < x;
(3) For all x ∈ R there is m ∈ Z such that
m ≤ x < m + 1.
Proof. The claim listed under (1) is immediate if x ≤ 0. So, assume x > 0. We may
use the Archimedean Property with ε = 1 and b = x. Consequently, there is some
n ∈ N with n > x. The proof of (2) is analogous: the case of x ≥ 0 is immediate
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2
CONSEQUENCES OF THE COMPLETENESS AXIOM
while in the case of x < 0 we use the version of the Archimedean Property stated
in (1.1). More specifically, if ε = −1 and x < 0 then there is some n ∈ N with
−n = nε < x.
In particular, z = −n satisfies z < x. We now focus on showing the property listed
under (3). First note that there is nothing to show if x ∈ Z. So, we may assume
that x '∈ Z.
CASE I: Let us first assume that x ≥ 0. Consider the set
"
!
#
"
S = k ∈ N ∪ {0}" k ≤ x .
According to the property listed under (1) there is some n ∈ N with n > x and so
{0} ⊆ S ⊆ {0, 1, 2, ..., n − 1}.
The set S is a non-empty finite set and therefore it has its maximum:
m = max S.
Note that m ∈ S and m + 1 '∈ S so that m ≤ x < m + 1, as desired. In fact, since
x '∈ Z we really have m < x < m + 1.
CASE II: Let us now assume that x < 0. Since −x > 0 we may use the conclusion
of the previous case. Thus, there exists some m̃ ∈ Z such that
m̃ < −x < m̃ + 1.
It follows that −m̃−1 < x < m̃. If m = −m̃−1 then we exactly have m < x < m+1.
This completes the proof of our Corollary.
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Remark. The value of m ∈ N whose existence is guaranteed by part (3) of the
previous corollary gives rise to the so called floor function of x ∈ R:
m = *x+ ⇐⇒
m ∈ Z, m ≤ x < m + 1.
Remark. It is also true that for all x ∈ R there is m ∈ Z with m < x ≤ m + 1. The
proof is very analogous to those above, and is left for the readers.
2. Denseness of Q
The following is an extremely important property of the set of rational numbers
Q. Without this property there would be no such thing as, for example, the function
x .→ 2x on the set of real numbers. The property states that rational numbers are
“dense” among real numbers, that between any two real numbers there is a rational
number.
Denseness of Q. Let a, b ∈ R with a < b. Then there exists q ∈ Q such that
a < q < b.
Reasoning behind the proof. Rational numbers take the form of m
n with m ∈ Z
and n ∈ N. With this in mind the inequality we need to show reads
m
(2.1)
a<
< b i.e. an < m < bn.
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CONSEQUENCES OF THE COMPLETENESS AXIOM
3
By the consequences of the Archimedean Property we may as well assume that the
integer m is the smallest satisfying (2.1) for given a, b and n , i.e. that
m − 1 ≤ an < m < bn.
Note that if m − 1 ≤ an then also m ≤ an + 1. Thus, if we can arrange that
an + 1 < bn, i.e. n(b − a) > 1, then we can also arrange that m < bn.
Proof. By Archimedean Property we know that there exists n ∈ N with n(b−a) > 1.
It follows from the Archimedean Property and its consequences that there is m ∈ N
such that
m − 1 ≤ an < m.
The inequalities m − 1 ≤ an and 1 < n(b − a) imply that m < an + n(b − a), i.e.
that m < bn. Overall, we have proved that for m and n chosen above we have:
an < m < bn.
It now follows that a <
m
n
!
< b.
Here is one simple example where we use the denseness of Q.
Example. Show that
"
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#
"
inf x ∈ Q" x > 0 and x2 > 4 = 2.
Reasoning behind the solution. As usual, we should start by proving that 2 is
a lower bound for the set, i.e. that if x > 0
(2.2)
x2 > 4 =⇒ x ≥ 2.
√
This would be very easy if we had established (the properties of) , but since we
√
have not yet discussed the properties of
in class (much less its existence), we
√
should try to prove the above without using
. So we should re-write (2.2) in a
form which uses operations we can rely on at this point, such as taking the square.
In other words, we are better off trying to prove the contrapositive: that for x > 0
x < 2 =⇒ x2 ≤ 4.
This comes out easily as a consequence of the transitivity: we can argue that
0 < x < 2 implies x2 < 2x and 2x < 4. Next, we need to prove that 2 is the
greatest lower bound. Consider l > 2; we need to find q ∈ Q such that q > 0,
q 2 > 4 and
l > q > 2.
“Finding” such rational number is easy using the denseness of Q. Indeed, we can
always find q ∈ Q such that l > q > 2; the same argument from the above will show
that q 2 > 4.
Solution. To show that 2 is a lower bound we assume the opposite: that there is
some x ∈ Q such that 0 < x < 2 and x2 > 4. Since x > 0 the assumption x < 2
implies x2 < 2x. In addition, we also have 2x < 4. Using transitivity we see that
2
x2 < 4, which contradicts
"our assumption that
!
# x > 4. Thus, 2 must be a lower
"
2
bound for the set x ∈ Q" x > 0 and x > 4 . To show that it is the greatest
lower bound consider l ∈ R with l > 2. By denseness of Q we know that there is
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CONSEQUENCES OF THE COMPLETENESS AXIOM
some q ∈ Q such that l > q > 2. It now follows that q > 0, q 2 > 2q, 2q > 4 and
q 2 > 4. This means that we found an element of our set which is smaller than l.
Thus, l is not a lower bound for our set. We now see that 2 is the greatest lower
bound of the set, i.e. that
"
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#
"
inf x ∈ Q" x > 0 and x2 > 4 = 2.
Homework.
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(1) Show that the set n2 " n ∈ N is unbounded from above. Please do not
√
but argue as in the proof of the Archimedean Property.
use
(2) Use denseness of Q to show that
"
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"
sup x ∈ Q" x2 < 4 = 2.