Page 35 - DMTH401_REAL ANALYSIS

Unit 2: Algebraic Structure and Countability

This number is called an upper bound of the corresponding set and such a set is said to be bounded Notes
above. Thus, we have the following definition:
Definition 4: Upper Bound of a Set

Let S  R. If there is a number u  R such that x  u, for every x  S, then S is said to be bounded
above. The number u is called an upper bound of S.

Example: Verify whether the following sets are bounded above. Find an upper bound of
the set, if it exists.

(i) The set of negative integers
{–1, –2, –3, ......}.
(ii) The set N of natural numbers.
(iii) The sets Z, Q and R.

Solution:
(i) The set is bounded above with –1 as an upper bound,
(ii) The set N is not bounded above.
(iii) All these sets are not bounded above.

Now consider a set S = {2, 3, 4, 5, 6, 7). You can easily see that this set is bounded above because
7 is an upper bound of S. Again this set is also bounded below because 2 is a lower bound of S.
Thus S is both bounded above as well as bounded below. Such a set is called a bounded set.
Consider the following sets:
S = {... –3, –2, –1, 0, 1, 2, ......},
1
S = {0, 1, 2, ......},
2
S = (0, –1, –2, ......}.
3
You can easily see that S, is neither bounded above nor bounded below. The set S is not bounded
4
above while S, is not bounded below. Such sets are known as Unbounded Sets.
Thus, we can have the following definition.
Definition 5: Bounded Sets
A set S is bounded if it is both bounded above and bounded below.

In other words, S has an upper bound as well as a lower bound. Thus, if S is bounded, then there
exist numbers u (an upper bound) and v (a lower bound) such that v  x  u, for every x  S.
If a set S is not bounded then S is called an unbounded set. Thus S is unbounded if either it is not
bounded above or it is not bounded below.

Example:
(i) Any finite set is bounded.
(ii) The set Q of rational numbers is unbounded.
(iii) The set R of real numbers is unbounded.

(iv) The set P = {sin x, sin 2x, sin 3x,......, sin nx, ......} is bounded because –1  sin nx  1, for every
n and x.

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