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Supremum and Infimum
Mika Seppälä
Distance in the Set of Real Numbers
Definition
The absolute value x of a number x 
is defined by setting
 x if x  0
x 
 x if x  0
Triangle Inequality
x  y  xy  x  y
Triangle inequality for the absolute value is almost obvious. We have equality
on the right hand side if x and y are either both positive or both negative (or
one of them is 0).
We have equality on the left hand side if the signs of x and y are opposite
(or if one of them is 0).
Definition
The distance between two real numbers x and y is |x-y|.
Mika Seppälä: Sup and Inf
Properties of the Absolute Value
3
6
a  b  a  b
a  a
1
a 0
2
4
ab  a b
5
6
Let b  0. a  b  a  b or a  b
7
ab  a  b
Example
Proof
a a a
2
a2  a
Triangle Inequality
Let x, y,w  . Show that x  y  x  w  w  y .
x  y  x w w  y  x w  w  y
by the triangle inequality.
Problem
When do we have equality in the above estimate?
Mika Seppälä: Sup and Inf
Solving Absolute Value Equations
Example
Solution
2x  1  5
For those values of x for which 2x  1  0 we have
2x  1  5  2x  1  5  2x  4  x  2.
If x  2, then 2x  1  5  0. Hence x  2 is a solution.
For those values of x for which 2x  1  0 we have
2x  1  5    2x  1  5  2x  6  x  3.
If x  3, then 2x  1  5  0. Hence x  3
is also a solution.
Conclusion
The equation has two solutions: x = 2 and x = -3.
Mika Seppälä: Sup and Inf
Solving Absolute Value Inequalities
Example
Solution
2x  3  5
By the property 6 of absolute values,
2x+3  5  2x  3  5 or 2x  3  5.
2x  3  5  2x  2  x  1.
2x  3  5  2x  8  x  4
Conclusion
Solution is x  1 or x  4.
Mika Seppälä: Sup and Inf
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.
Mika Seppälä: Sup and Inf
Supremum
Let 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 

Mika Seppälä: Sup and Inf

Infimum
Let 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 

Mika Seppälä: Sup and Inf

Characterization of the Supremum (1)
Let 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.
Mika Seppälä: Sup and Inf
Characterization of the Supremum (2)
Theorem
 1) a  A : s  a and
s  sup  A   
 2)   0 : a  A such that s  a  
Proof
To prove the converse, assume that the number s
satisfies the conditions 1) and 2). We have to show
Cont’d

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   .
By the definition of the number  this means that a  sup  A  .
This is impossible, hence s  sup  A  .
Mika Seppälä: Sup and Inf
Characterization of the Infimum
Theorem
 1) a  A : a  m and
m  inf  A   
 2)   0 : a  A such that a  m  
The proof of this result is a repetition of the argument the previous proof
for the supremum.
Mika Seppälä: Sup and Inf
Usage of the Characterizations
Example
Assume that sup  A  , and let 2A  2a | a  A.
Claim
sup  2A  2sup  A.
Proof of
the Claim
2
1
Let m  sup  A  . Then a  A : m  a. Hence
a  A : 2m  2a  sup  2 A   2m  2sup  A  .
Let   0. Apply the above characterization for sup  A  replacing
 by

to conclude: a  A such that m  a 
2
implying sup  2A   2m  2sup  A  .
1
and
2
 sup  2A  2sup  A.
Mika Seppälä: Sup and Inf

2
. Then 2m  2a   .