Page 37 - DMTH401_REAL ANALYSIS

Unit 2: Algebraic Structure and Countability

Very often in our discussion, we have used the expressions ‘the supremum’, rather than a supremum. Notes
What does it mean? It simply means that the supremum of a set, if it exists, is unique i.e. a set can
not have more than one supremum. Let us prove it in the form of the following theorem:
Theorem 1: Prove that the supremum of a set, if it exists, is unique.
Proof: If possible, let there be two supremums (Suprema) say m and m’ of a set $.
Since m is the least upper bound of S, therefore by definition, we have

m  m
Similarly, since m the least upper bound of S, therefore, we must have
m  m.
This shows that m = m which proves the theorem.

You can now similarly prove the following result:

Task Prove that the infimum of a set, if it exists, is unique.

In example 3, you have seen that supremum or the infimum of a set may or may not belong to
the set. If the supremum of a set belongs to the set, then it is called the greatest member of the set.
Similarly, if the infimum of a set belongs to it, then it is called the least member of the set.

Example:
(i) Every finite set has the greatest as well as the least member.
(ii) The set N has the least member but not the greatest. Determine that number.
(iii) The set of negative integers has the greatest member but not the least member. What is
that number?
You have seen that whenever a set S is bounded above, then S has the supremum. In fact this is
true in general. Thus, we have the following property of R without proof:

Property 4: Completeness Property
Every non-empty subset S of R which is bounded above, has the supremum.
Similarly, we have
Every non-empty subset S of R that is bounded below, has the infimum.

In fact, it can be easily shown that the above two statements are equivalent.
Now, if you consider a non-empty subset S of Q, then S considered as a subset of R must have, by
property, a supremum. However, this supremum may not be in Q. This fact is expressed by
saying that Q considered as a field in its down right is not Order-Complete. We illustrate this
observation as follows:

Construct a subset S of Q consisting of all those positive rational numbers whose squares are less
than 2 i.e.
2
S = {x  Q: x > 0, x < 2}.
Since the number 1 ES, therefore S is non-empty. Also, 2 is an upper bound of S because every
element of S is less than 2. Thus the set S is non-empty and bounded, above. According to the

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