Page 61 - DMTH401_REAL ANALYSIS

Unit 3: Matric Spaces

Notes
Example: Show that the superset of a NBD of a point is also a NBD of the point.
Solution: Let A be a NBD of a point c. Then there exists an open interval ]c – 6, – c + 6[, for some
6 > 0 such that
]c – 6, c + [ C A.
Now let S be a super set which contains A. Then obviously

A  S  [c – , c + [  S
which shows that S is also a NBD of c.

1 1
For instance, if ] , [ is a NBD of the point 0.
10 10
1 1
Then, ] – [ is also a NBD of 0 as can be seen from Figure below.
5 5

Is a subset of a NBD of a point also a NBD of the point? Justify your answer.
Now you can try the following exercise.

Task Prove that the Union of any two NBDS of a point is a NBD of the point.

The conclusion of the Exercise, in fact, can be extended to a finite or an infinite or an arbitrary
number of the NBDS of a point.
However, the situation is not the same in the case of intersection of the NBDS. It is true that the
intersection of a finite number of NBDS of a point is a NBD of the point. But the intersection of
an infinity collection of NBDS of a point may not be a NBD of the point. For example, consider
the class of NBDS given by a family of open intervals of the form
1 1 1 1
I = ] –1, 1 [, I = ] – , [ I = ] – , [,.
1 2 3
2 2 3 3
1 1
I = ] – , [...
n
n n
which are NBDS of the point 0. Then you can easily verify that
I  I  I I  I 
1 2 3 4 n

or  I = {0}
=
n 1 n
3.4 Open Sets

You have seen from the previous examples and exercises that a given set may or may not be a
NBD of a point. Also, a set may be a NBD of some of its points and not of its other points. A set

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