Neighbourhood of a point
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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
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( p – ε , p + ε ) ⊂ N.
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Since p ∈ ( p – ε , p + ε ) ⊂ N,
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Therefore if N is a neighbourhood of p, we must have p ∈ N.
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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.
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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
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1. Any open interval (a,b) is a nbd of each of its points.
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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)
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Showing that (a,b) is an nbd of x. But x is a arbitrary point of (a,b) is a nbd of its points.
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2. A closed set [a.b] is nbd of each of its points except the two end points a and b.
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Sol. Let x be a an arbitrary point of the point of open interval ( a,b) . Then
x ∈ (a,b) ⊂ [a,b].
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showing that ( a,b) is a nbd of each point of (a,b).
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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.
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Thus [a,b] is a nbd of each points except the two points a and b.
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3.The set of rational numbers Q is not a nbd of any of its points.
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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.
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∴ ( p- ε, p + ε) ⊄ Q ∀ ε> 0.
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∴ Q is not a nbd of p.
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