Page 60 - DMTH401_REAL ANALYSIS

Real Analysis

Notes Neighbourhood of a Point

A set P is said to be a Neighbourhood (NBD) of a point V if there exists an open interval which
contains c and is contained in P.
This is equivalent to saying that there exists an open interval of the form ]c – , c + [, for some
6 > 0, such that
]c – 6, c + [  P.

Example: (i) Every open interval ]a, b[ is a NBD of each of its points.
(ii) A closed interval [a, b] is a NBD of each of its points except the end point i.e. [a, b] is not a
NBD of the points a and b, because it is not possible to find an open interval containing a
or b which is contained in [a, b]. For instance, consider the closed interval [0,1]. It is a NBD
of every point in ]0, l[. But, it is not a NBD of 0 because for every  > 0, ]-, [[0, 1].
Similarly [0, 1] is not a NBD of 1.
(iii) The null set 0 is a NBD of each of its point in the sense there is no point in 0 of which it is
not a NBD.
(iv) The set R of real numbers is a NBD of each real number x because for every 5 > 0, the open
interval ]x - 6, x + [ is contained in R.
(v) The set Q of rational numbers is not a NBD of any of its points x because any open interval
containing x will also contains an infinite number of irrational numbers and hence the
open interval can not be a subset of Q.

1 1 1 1
Now consider any two neighbourhoods of the point 0 say ] – , [ and ] – , [ as shown in
10 10 5 5
the Figure below.

The intersection, of these two neighbourhood is

1 1 1 1 1 1
] – , [] – , [=] – , [
10 10 5 5 10 10

1 1
which is again a NBD of 0. The union of these two neighbourhoods is ] – , [, which is also a
5 5
NBD of 0. Let us now examine these results in general.

Example: The intersection of any two neighbourhoods of a point is a neighbourhood of
the point.
Solution: Let A and B be any two NBDS of a point c in R. Then there exist open intervals ]c –  ,
1
c +  ] and ]c – , c +  [such that]c –  , c +  ] A, for some  > 0, and ]c –  , c +  [B, for some
1 2 2 1 1 1 2 2
 > 0.
2
Let 6 = Min. { ,  ) = minimum of  ,  .
1 2 1 2
This implies that ]c – 6, c + [C A B which shows that A  B is a NBD of c.

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