Page 62 - DMTH401_REAL ANALYSIS

Real Analysis

Notes may even be a NBD of each of its points as in the case of the interval ]a, b[. Such a set is called an
open set.
Definition: X a set S is said to be open if it is a neighbourhood of each of its points.
Thus, a set S is open if for each x in S, there exists an open interval ]x – 6, x + [,  > 0 such that

x] x – , x + [ S.
It follows at once that a set S is not open if it is not a NBD of even one of its points.

Example: An open interval is an open set
Solution: Let ]a, b[ be an open interval. Then a < b. Let c] a, b[. Then a < c < b and therefore

c – a > 0 and b – c > 0
Choose
 = Minimum of {b – c, c – a)
= Min (b – c, c – a).
Note that b – c > 0, c – a > 0. Therefore > 0.
Now  c – a  a  c – 
and  b – c c + < b.
i.e.
Therefore, ]c – 6, c + [  ]a, b[ and hence ]a, b[ is a NBD of c.

Example: (i) The sat R of real numbers is an open set
(ii) The null set f is an open set
(iii) A finite set is not an open set

(iv) The interval ]s, b] is not an open set.

Example: Prove that the intersection of any two open sets is an open set.
Solution: Let A and B be any two open sets. Then we have to show that A  B is also an open set.
If A  B = , then obviously A  B is an open set. Suppose A  B .
Let x be an arbitrary element of A  B. Then xA B  xA and xB.
Since A and B are open sets, therefore A and B are both NBDS of x. Hence A  B is a NBD of x. But
xA  B is chosen arbitrarily. Therefore, A B is a NBD of each of its points and hence A  B is
an open set. This proves the result. In fact, you can prove that the intersection of a finite number
of open sets is an open set. However, the intersection of an infinite number of open sets may not
be an open set.

Task
1. Give an example to show that intersection of an infinite number of open sets need
not be an open set.
2. Prove that the union of any two open sets is an open set. In fact, you can show that
the union of an arbitrary family of open sets is an open set.

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