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1.2 Supremum and Infimum
Integers, Rational Numbers and Real Numbers
Completeness of Real Numbers
Supremum and Infimum
Characterizations of Sup and of Inf
Calculus and Elementary Analysis
1.2.1 Integers, Rational and Real Numbers
 0,1,2,3,
Notations

Natural Numbers
 0, 1, 2, 3,
p
  p
q
Theorem
Notation
,q 

Integers
This condition means
that p and q have no
common factors.

, q  0, gcd  p, q   1Rational Numbers

There are no rational numbers r such that r2=2.
Real Numbers
Real numbers consist of rational and of irrational numbers.
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Calculus and Elementary Analysis
Integers, Rational and Real Numbers
Proposition
Proof
There are not rational numbers r such that r2=2.
Assume the contrary.
Then there are positive integers p and q such that
p2/q2=2 and p and q do not have common factors.
This means that p2=2q2.
Hence the area of a square with side length p is twice the area of
the square with side length q. Furthermore p and q are
smallest such numbers since they do have common factors.
Graphically:
p
G
q
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B
Now p and q are integers
so that
B=p2=2q2=2G.
I.e. the area B of the large
brown square is twice the
area G of the green square.
Calculus and Elementary Analysis
Integers, Rational and Real Numbers
There are not rational numbers r such that r2=2.
Theorem
Proof
(cont’d)
p
Furthermore, p and q are smallest
integers such that the area B of the
large brown square is twice the area G
of the smaller green square, B=2G.
q
Now move a copy of the green square to
p
the upper right hand corner of the larger
square.
The intersection I of the two copies of the
green square is the square I with the area I.
A
I
A
p-q
q
The two squares marked by A in the picture have the same area A.
Furthermore, by the assumptions, I=2A. This is impossible, since
B and G were the smallest squares with integer side lengths such
that B=2G.
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Calculus and Elementary Analysis
1.2.2 Upper and Lower Bounds
Definition
Let A be a non-empty set of real numbers.
We say that a number M is an upper bound for the set A if
a  A : a  M.
A number m is an lower bound for the set A if
a  A : a  m.
A set A need not have neither upper nor lower bounds.
The set A is bounded from above if A has a finite upper bound.
The set A is bounded from below if A has a finite lower bound.
The set A is bounded if it has finite upper and lower bounds.
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Calculus and Elementary Analysis
1.2.3 Bounds of Integers, Rationals and Reals
Observations
1
2
Any non-empty bounded set of integers has the
largest (and the smallest) element which is also an
integer.
Consider the set of rational numbers r such that r2 ≤ 2.
This set is clearly bounded both from the above and
from below. Observe that the set of rational upper
bounds for the elements of this set does not have a
smallest element. This follows from the previous
considerations showing that 2 is not rational.
This means that the set of rational numbers is not
complete in the sense that the set of rational upper
bounds of a bounded set does not necessarily have the
smallest rational upper bound.
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Calculus and Elementary Analysis
1.2.4 Supremum
Let A  , A  , be bounded from above.
Completeness of Real Numbers
The set A has finite upper bounds. An important
completeness property of the set of real numbers is that
the set A has a unique smallest upper bound.
Definition
The smallest upper bound of the set A is
called the supremum of the set A.
Notation
sup(A) = the supremum of the set A.
Example
1 3 7 
Let A   , , ,   1  2 n n  , n  0 . Then sup  A   1.
2 4 8 
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
Calculus and Elementary Analysis

Infimum
Let A  , A  , be bounded from below.
The set A has finite lower bounds. As in the case of
upper bounds, the set of real numbers is complete in the
sense that the set A has a unique largest lower bound.
Definition
The largest lower bound of the set A is called
the infimum of the set A.
Notation
inf(A) = the infimum of the set A.
Example
3 5 9 
Let A   , , ,   1  2 n n  , n  0 . Then inf  A   1.
2 4 8 
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
Calculus and Elementary Analysis

1.2.5 Characterization of the Supremum (1)
Let A  , A  , be bounded from above. This is equavalent to
saying that sup  A   .
Theorem
Proof

 1) a  A : s  a and
s  sup  A   
 2)   0 : a  A such that s  a  
Assume that s  sup  A  . Then since s is an upper bound of A,
the condition 1) is trivially satisfied.
To prove the condition 2) assume the contrary, i.e.
assume that there is an   0 such that there are no elements a  A
such that s  a   .
If this is the case, then also s 

is an upper bound for the set A.
2
But this is impossible, since s is the smallest upper bound for A.
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Calculus and Elementary Analysis
Characterization of the Supremum (2)
Theorem
For a non-empty set A:
 1) a  A : s  a, and
s  sup  A   
 2)   0 : a  A such that s-  a  s.
Proof
Cont’d

To prove the converse, assume that the number s
satisfies the conditions 1) and 2). We have to show
that s  sup  A  .
By the condition 1), s is an upper bound for the set A.
If s  sup  A  , then s  sup  A  , and   s  sup  A   0.
By the condition 2), a  A such that s-  a  s.
By the definition of the number  this means that a  sup  A  .
This is impossible, hence s  sup  A  .
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Calculus and Elementary Analysis
1.2.6 Characterization of the Infimum
Theorem
For a non-empty set A:
 1) a  A : a  m, and
m  inf  A   
 2)   0 : a  A such that m<a<m+ .
The proof of this result is a repetition of the argument the
previous proof for the supremum.
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Calculus and Elementary Analysis
1.2.7 Usage of the Characterizations
Example
Assume that sup  A  , and let 2A  2a | a  A.
Claim
sup  2A  2sup  A.
Proof of
the Claim
1
Let m  sup  A  . Then a  A : m  a. Hence
a  A : 2m  2a  sup  2 A   2m  2sup  A  .
2 Let   0. Apply the above characterization for sup  A  replacing
 by

to conclude: a  A such that a-
2
implying sup  2 A   2m  2sup  A  .
1
and
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2

2
 a  m. Then m-  2a  m.
 sup  2A  2sup  A.
Calculus and Elementary Analysis