A non-empty set is bounded from above if there exists such that
The number is called an upper bound of .
If a set is bounded from above, then it has infinitely many upper bounds, because every number greater then the upper bound is also an upper bound. Among all the upper bounds, we are interested in the smallest.
Let be bounded from above. A real number is called the supremum of the set if the following is valid:
(i) is an upper bound of :
(ii) is the least upper bound:
The supremum of we denote as
If , then we say that is a maximum of and we write
If the set it is not bounded from above, then we write .
Proposition 1. If the number is an upper bound for a set , then .
The question is, does every non- empty set bounded from above has a supremum? Consider the following example.
Example 1. Determine a supremum of the following set
The set is a subset of the set of rational numbers. According to the definition of a supremum, is the supremum of the given set. However, a set does not have a supremum, because is not a rational number. The example shows that in the set there are sets bounded from above that do not have a supremum, which is not the case in the set .
In a set of real numbers the completeness axiom is valid:
Every non-empty set of real numbers which is bounded from above has a supremum.
It is an axiom that distinguishes a set of real numbers from a set of rational numbers.
In a similar way we define terms related to sets which are bounded from below.
A non-empty set is bounded from below if there exists such that
The number is called a lower bound of .
Let be bounded from below. A real number is called the infimum of the set if the following is valid:
(i) is a lower bound:
(ii) is the greatest lower bound:
The infimum of we denote as
If , then we say that is the minimum and we write
If the set it is not bounded from below, then we write .
The existence of a infimum is given as a theorem.
Theorem. Every non-empty set of real numbers which is bounded from below has a infimum.
Proposition 2. Let such that . Then
Example 2. Determine , , and if
Solution. Firstly, we have to check what are the -s:
The inequality above will be less then zero if the numerator and denominator are both positive or both negative. We distinguish two cases:
1.) and , that is, and . It follows .
2.) and , that is, and . It follows .
From the proposition 2. follows that and .
The minimum and maximum do not exist ( because we have no limits of the interval).
Example 3. Determine , , and if
Firstly, we will write first few terms of :
We can assume that the smallest term is and there is no largest term, however, we can see that all terms do not exceed . That is, we assume , and do dot exists. Let’s prove it!
To prove that is the supremum of , we must first show that is an upper bound:
which is always valid. Therefore, is an upper bound. Now we must show that is the least upper bound. Let’s take some and show that then exists such that
and such surely exists. Therefore, .
However, is not the maximum. Namely, if , then such that
which is the contradiction. It follows that the maximum of does not exists.
Now we will prove that .
Since , it is enough to show that is a lower bound of . According to this, we have
which is valid for all . Therefore, .