Page 64 - DMTH401_REAL ANALYSIS

Real Analysis

Notes The question, therefore, arises: "How to know whether or not a set has a limit point?" One
obvious fact is that a finite set can not have a limit point. Can you give a reason for it? Try it. But
then there are examples where even an infinite set may not have a limit point e.g. the sets N and
Z do not have a limit point even though they are infinite 'sets. However, it is certainly clear that
a set which has a limit point, must necessarily be an infinite set. Thus our question takes the
following form:

"What are the conditions for a set to have a limit point?"
This question was first studied by a Czechoslovakian Mathematician, Bemhard Bulzano [1781-
1848] in 1817 and he gave some ideas.

Unfortunately, his ideas were so far ahead of their time that the world could not appreciate the
full significance of his work. It was only much later that Bulzario's work was extended by Karl
Weierstrass [1815-1897], a great German Mathematician, who is known as the "father of analysis".
It was in the year 1860 that Weierstrass proved a fundamental result, now known as Bulzano-
Weierstrass Theorem for the existence of the limit points of a set. We state and prove this
theorem as follows.

3.5.1. Bulzano Weierstrass Theorem

Theorem 1: Every infinite bounded subset of set R has a limit point (in K).
Proof: Let S be an infinite and bounded subset of R. Since A is bounded, therefore A has both
a- lower bound as well as an upper bound.

Let m be a lower bound and M be an upper bound of A. Then obviously
m  x  M, " xA.

Construct a set S in the following way:
S = (xR: x exceeds at most finite number of the elements of A}. Now, let us examine the
following two questions:
(i) Is S a non-empty set?
(ii) Is S also a bounded set?

Indeed, S is non-empty because m  x,  M, " xA, implies that mS. Also M is an upper bound
of S because no number greater than or equal to M can belong to S. Note that M cannot belong
to S because it exceeds an infinite number of elements of A.
Since the set S is non-empty and bounded above, therefore, by the axiom of completeness, S has
its supremum in R. Let p be the supremum of S. We claim that p is a limit point of the set A.
In order to show that p is a limit point of A, we must establish that every NBD of p has at least
one point of the set A other than p. In other words, we have to show that every NBP of p has an
infinite number of elements of A. For this, it is enough to show that any open interval ]p – ,
p +  [, for  > 0, contains an infinite number of members of set A. For this, we proceed as follows.
Since p is the supremum of S, therefore, by the definition of the Supremum of a set, there is at
least one element y in S such that y > p – , for  > 0. Also y is a member of S, therefore, y exceeds
at the most a finite number of the elements of A. In other words, if you visualise it on the line as
shown in the Figure below, the number of elements of A lying on the left of p –  is finite at the
most. But certainly, the number of elements of A lying on the right side of the point p –  is
infinite.

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