Answer to Question #108076 in Real Analysis for Pratibha

Question

Is V=(1,2) a neighbourhood of the point 1.6 ? what epsilon neighbourhood would work ? Five value of epsilon

Answer 1

For a given $\epsilon >0 ,$ A ϵ -neighborhood of a point p is a set $N_{\epsilon}(p)$ consisting of all the point q such that ∣p−q∣<ϵ. The number ϵ is called the radius of $N_{\epsilon}(p)$

$\therefore \ N_{\epsilon}(p)=\{ q\in\R:|p-q|<\epsilon\}$

Again. $\mid p-q\mid<\epsilon \ \iff \mid q-p\mid<\epsilon \ \iff \ -\epsilon<q-p<\epsilon$

$\iff \ p-\epsilon<q<p+\epsilon$ .

$\therefore \ N_{\epsilon}(p)=\{ q\in\R: \ p-\epsilon<q<p+\epsilon\}$

$=(p-\epsilon ,p+\epsilon)$ .

If possible let $(1,2)$ is a $\epsilon$ -neighborhood of 1.6 then,

$1.6-\epsilon=1 \ and \ 1.6+\epsilon=2$

$\implies \ \epsilon=0.6 \ and \ \epsilon=0.4$

Which is a contradiction.

Hence, V=(1,2) is not a ϵ -neighborhood of 1.6 .

Since $\epsilon>0$ be an arbitrary real number, so for five value of $\epsilon$

take any five positive real number.

For example:

$\ If \ \epsilon=0.4 \ then \ V_1=(1.2,2)$ is a neighborhood of 1.6.

If $\ \epsilon=0.6 \ then \ V_2=(1,2.2)$ is a neighborhood of 1.6

If $\epsilon=1 \ then \ V_3=(0.6,2.6)$ is a neighborhood of 1.6

If $\epsilon=2 \ then \ V_4=(-0.4,3.6)$ is a neighborhood of 1.6

$If \ \epsilon=1.5 \ then \ V_5=(0.1,3.1)$ is a neighborhood of 1.6.