Neighbourhood of a point

Neighbourhood of a point

Def. A subset N of R is said to be a neighbourhood of a point p ∈ R If there exists a number ε > 0 such that

( p – ε , p + ε ) ⊂ N.

Since p ∈ ( p – ε , p + ε ) ⊂ N,

Therefore if N is a neighbourhood of p, we must have p ∈ N.

Thus N ⊂ R is nbd of a point p ∈ R if there exists an open interval contained in N whose centre is the point p.

Exp- let S be an open interval (1,3), then S is a neighbourhood of the point 2 ∈ (1,3) because ε > 0 , ε = 0.04 is a positive real number such that

(2 – 0.004 , 2+ 0.004) ⊂ (1,3)

Illustrative Examples

1. Any open interval (a,b) is a nbd of each of its points.

Sol. Let x be an arbitrary point of the given open interval ( a,b) . Since every set is a subset of itself, we have

x ∈ (a,b) ⊂ (a,b)

Showing that (a,b) is an nbd of x. But x is a arbitrary point of (a,b) is a nbd of its points.

2. A closed set [a.b] is nbd of each of its points except the two end points a and b.

Sol. Let x be a an arbitrary point of the point of open interval ( a,b) . Then

x ∈ (a,b) ⊂ [a,b].

showing that ( a,b) is a nbd of each point of (a,b).

Now [a,b] cannot be a nbd of a because to be a nbd of a. it must contain an open interval [a – ε, a + ε] , i.e. [a,b] has to contain points less than a, which it does not. Similarly, [a,b] cannot be a nbd of b because to be a nbd of b. it must contain an open interval ( b -ε , b + ε) i.e. [a,b] has to contain points greater than b. which it does not.

Thus [a,b] is a nbd of each points except the two points a and b.

3. The set of rational numbers Q is not a nbd of any of its points.

Sol. Let p ∈ Q . For any positive real number , p – εand p + ε are two distinct real numbers and we know that between any two distinct real numbers there lie infinite irrational numbers which are not umbers of Q.

∴ ( p- ε, p + ε) ⊄ Q ∀ ε> 0.

∴ Q is not a nbd of p.