APPLICATIONS OF THE SUPREMUM

PROPERTY

Rajendra Pant
CONTENTS

1 Applications of the Supremum Property

Bounded Functions
The Archimedean Property
√
The Existence of 2
Density of Rational Numbers in R
Applications of the Supremum Property: It is an important fact that
taking supremum and infimum of sets is compatible with the algebraic
properties of R. As an example, we present here the compatibility of taking
supremum and addition.

Example 2.4.1
(a) Let S be a nonempty subset of R that is bounded above, and let a ∈ R.
Define the set a + S := {a + s : s ∈ S}. Then

sup(a + S) = a + sup S.

Proof. Let u := sup S. Then x ≤ u for all x ∈ S, so that a + x ≤ a + u.

Therefore, a + u is an upper bound for the set a + S. Consequently, we have

sup(a + S) ≤ a + u.
Now if v is a an upper bound for the set (a + S) then a + x ≤ v for all x ∈ S.
Consequently x ≤ v – a for all x ∈ S, so that v – a is an upper bound of S.
Therefore, u = sup S ≤ v – a, which gives us a + u ≤ v. Since v is an upper
bound of a + S, we can replace v by sup(a + S) to get

a + u ≤ sup(a + S).

Combining these inequalities, we conclude that

sup(a + S) = a + u = a + supS.

If the supremum or infimum of two sets are involved, it is often necessary

to establish results in two stages, working with one set at a time. Here is an
example.
Example
(b) Suppose that A and B are nonempty subsets of R that satisfy the
property:
a ≤ b for all a ∈ A and all b ∈ B.

Then sup A ≤ inf B

Proof. For, given b ∈ B, we have a ≤ b for all a ∈ A. This means that b is an

upper bound of A, so that
sup A ≤ b.

Next, since the last inequality holds for all b ∈ B, we see that the number
sup A is a lower bound for the set B. Therefore, we conclude that

sup A ≤ inf B.
Bounded Functions
Let f : D → R be a function. Then
1. f is said to be bounded above if the set f (D) = {f (x) : x ∈ D} is bounded
above in R i.e. there exists a b ∈ R such that f (x) ≤ b for all x ∈ D.
2. Similarly, f is said to be bounded below if the set f (D) is bounded
below i.e. there exists a c ∈ R such that c ≤ f (x) for all x ∈ D.
3. f is bounded if it is bounded above and below i.e. there exists a b ∈ R
such that |f (x)| ≤ b for all x ∈ D.
Example
1. The function f : [0, 2] → R defined by f (x) = x 2 is bounded.
1
2. The function f : [–1, 1] → R defined by f (x) = is not bounded, since for
x
all n ∈ N there exists n1 ∈ [–1, 1] such that f ( n1 ) = n, and we know that
the set of natural numbers is not bounded, that is there does not exist a
b ∈ R, b > 0 such that |f ( n1 )| = n < b for all n ∈ N.
Example 2.4.2
Let f , g : D → R be bounded functions. Then
(a) If f (x) ≤ g(x) for all x ∈ D, then sup f (D) ≤ sup g(D), that is,

sup f (x) ≤ sup g(x) .

x∈D x∈D

(b) Note that (a) doesn’t imply a relationship between sup f (D) and
inf g(D).
For example, if f (x) = x 2 and g(x) = x with D = {x : 0 ≤ x ≤ 1}, then
f (x) ≤ g(x) for all x ∈ D. However, we see that sup f (D) = 1 and
inf g(x) = 0. Since sup g(D) = 1, the conclusion of (a) holds.
(c) If f (x) ≤ g(y) for all x, y ∈ D, then sup f (D) ≤ inf g(D), that is,

sup f (x) ≤ inf g(y) .

x∈D y∈D
Theorem 2.4.3 (The Archimedean Property)
If x ∈ R, then there exists nx ∈ N such that x < nx .

Proof. Suppose to the contrary that n ≤ x for all n ∈ N. Hence x is an upper

bound of N, and so by the Completeness Property of R, N has a supremum,
say u ∈ R. Then u – 1 < u and since u = sup N, u – 1 is not an upper bound
for the set. Hence there is a natural number m such that u – 1 < m. It follows
then that u < m + 1. However, m + 1 ∈ N and so we have a contradiction
with the assumption that N is bounded above. The result now follows.

The Archimedean Property implies that the naturals are not bounded.
Together with the following corollaries, it provides a useful tool for analysis.
Corollary 2.4.4
n o
If S = 1 : n ∈ N , then inf S = 0.
n

Proof. Clearly, 0 ≤ n1 for all n ∈ N. So, S is bounded below. By the

Completeness Property of R, it follows that S has an infimum, say w. Clearly
w ≥ 0. Let ε > 0. From the Archimedean Property, there exists an n ∈ N
such that 1ε < n. Thus n1 < ε and so

1
0≤w≤ < ε.
n

Since ε > 0 was chosen arbitrarily, it follows from Theorem 2.1.9 that
w = 0.
Corollary 2.4.5

If t > 0, then there exists nt ∈ N such that 0 < n1t < t.

Proof. This follows directly from Corollary 2.4.4. Since the set S in
Corollary 2.4.4 is such that inf S = 0, it follows that for every t > 0, t is not a
lower bound for S and hence there exists an nt ∈ N such that 0 < n1t < t.

Corollary 2.4.6

If y > 0, then there exists an n y ∈ N such that n y – 1 ≤ y < n y .

Proof. Let E y = {m ∈ N : y < m}. By the Archimedean property the set E y is

nonempty. The Well-Ordering Property of the naturals says that E y must
have a smallest element which we denote by n y . Therefore n y – 1 ∉ E y and
we have n y – 1 ≤ y < n y .
 √
The Existence of 2: The importance of the Supremum Property lies in
the fact that it guarantees the existence of real numbers under certain
hypotheses. We shall make use of it in this way many times. At the moment,
we shall illustrate this use by proving the existence of a positive real number
x such that x 2 = 2; that is, the positive square root of 2. It was shown earlier
(see Theorem 2.1.4) that such an x cannot be a rational number; thus, we
will be deriving the existence of at least one irrational number.

Theorem 2.4.7
There exists a positive real x such that x 2 = 2.

Proof. Let S := {s ∈ R : 0 ≤ s, s2 < 2}. We first note that S is nonempty,

since 1 ∈ S, that is, 0 ≤ 1 and 12 = 1 < 2. Furthermore, we note that this set
is also bounded above by 2 since for any t ∈ R, if t > 2 then t2 > 4 so t ∉ S.
Therefore this set has a supremum in R. Let x = sup S. Note that x > 1. We
√
show that our supremum x = 2 by ruling out the other two possibilities:
x 2 < 2 and x 2 > 2.

First assume taht x 2 < 2. We will show that this assumption contradicts the
fact that x = sup S by finding a natural number n ∈ N such that x + n1 ∈ S
which would imply that x is not an upper bound for S. To see how to choose
n, note that n12 ≤ n1 so that

1 2
 
2x 1
x+ = x2 + +
n n n2
2x 1
≤ x2 + +
n n
1
= x 2 + (2x + 1).
n
Hence if we can choose n so that

1 1
x 2 + (2x + 1) < 2 or rather (2x + 1) < 2 – x 2 ,
n n

then we get
1 2
 
x+ < x 2 + (2 – x 2 ) = 2.
n
2–x 2 > 0 since the numerator is positive because x 2 < 2 so
Note that 2x+1
0 < 2 – x 2 and the denominator is positive since x ≥ 0. Therefore by The
Archimedean Property there exists some natural number n such that

1 2 – x2
≤ .
n 2x + 1
 
Therefore we have obtained such a number n where x + n1 ∈ S which
contradicts the fact that x = sup S since x < x + n1 .
Now assume that x 2 > 2. Now we show that it is possible to find a natural
1 is also an upper bound of S contradicting
number m ∈ N such that x – m
the fact that x = sup S. We note that:

1 2
 
2x 1
x– = x2 – + 2
m m m
2x
> x2 – .
m

Hence if we can choose m such that

2x 2x
x2 – > 2 or equivalently < x 2 – 2,
m m

then
1 2
 
x– > x 2 – (x 2 – 2) = 2.
m
 2
By assumption x 2 – 2 > 0and 2x > 0 so that x2x–2 > 0. Once again by the
Archimedean property there exists m ∈ N such that

1 x2 – 2
< .
m 2x

Thus there exists m ∈ N such that

1 2
 
x– > 2.
m

1 which implies that x – 1 is an

Now suppose that s ∈ S. Then s2 < 2 < x – m m
upper bound for S which contradicts the fact that x = sup S, since x – m1 < x.

√
Therefore our only possibility is that x 2 = 2and so x = 2 = sup S. Since the
√
supremum of a set is a real number, we conclude 2 ∈ R.
Theorem 2.4.8 (The Density Theorem)
If x, y ∈ R such that x < y, then there exists an r ∈ Q such that x < r < y.

Proof. Assume that x > 0. Since y – x > 0, as a consequence of the

Archimedean Property (Corollary 2.4.5) there exists an n ∈ N such that
0 < n1 < y – x. Therefore nx + 1 < ny (after multiplying both sides by n and
simplifying). From Corollary 2.4.6 applied to 0 < nx, there exists an m ∈ N
such that
m – 1 ≤ nx < m.

From the left inequality above it follows that m ≤ nx + 1 and so m < ny. This,
together with the right inequality, gives nx < m < ny. Dividing by
n, x < m m
n < y. Noting that n ∈ Q gives the desired result.

Now, in the case x ≤ 0, there exists k ∈ N such that k > |x|. Since
k – |x| = k + x is positive and k + x < k + y, the above argument proves that
there is a rational number r such that k + x < r < k + y. Then, letting r ′ = r – k,
r ′ is a rational number such that x < r ′ < y.
Corollary 2.4.9
If x, y ∈ R with x < y, then there exists an irrational number z such that
x < z < y.

Proof. Consider the real numbers √x and √y . By the density theorem,

2 2
there exists a rational number r such that

x y
√ <r< √
2 2
√
We may assume that r ̸= 0 without loss of generality. Hence z = r 2 is
√ √
irrational (since if it were rational then 1r r 2 = 2 would be rational, being
the product of rational numbers, which is clearly false). Hence x < z < y,
where z is irrational.