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 August 25th, 2017, 06:28 PM #1 Senior Member   Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2 Neighbourhood of a point Neighbourhood : A subset N of R is a neighbourhood of a point p if $\displaystyle \exists$ $\displaystyle \epsilon > 0$ such that $\displaystyle (x- \epsilon, x+\epsilon) \subset N$ Q : A set which is a neighbourhood of all its points except $n$ points $\displaystyle (n \geq 1)$. Can this set be an Interval ? A : The set $\displaystyle (0,1) \cup$ {$1,2,...,n$} is a neighbourhood of all its points except $n$ points $\displaystyle 1,2,3...n$ If $\displaystyle n=1$, the set can be an interval. Ex : $\displaystyle (2,3]$ If $\displaystyle n=2$, the set can be an interval. Ex : $\displaystyle [2,3]$ If $\displaystyle n>2$, the set cannot be an interval. Can someone explain me the set ($\displaystyle (0,1) \cup$ {$1,2,...,n$}) defined here ?
August 25th, 2017, 08:45 PM   #2
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 Originally Posted by Lalitha183 Can someone explain me the set ($\displaystyle (0,1) \cup$ {$1,2,...,n$}) defined here ?
You mean just explain what that set is? Well, it's the open unit interval (0,1) unioned with the finite set of points {1, 2, 3, ..., n}. That's pretty self explanatory. It's the set of points that are either in the open unit interval or else they're one of the positive integers between 1 and n inclusive.

August 25th, 2017, 08:59 PM   #3
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 Originally Posted by Maschke You mean just explain what that set is? Well, it's the open unit interval (0,1) unioned with the finite set of points {1, 2, 3, ..., n}. That's pretty self explanatory. It's the set of points that are either in the open unit interval or else they're one of the positive integers between 1 and n inclusive.
Why did they choose this set as an example and can you explain me the answer given in detail ?

Why it is not an interval if $\displaystyle n \geq 2$

August 25th, 2017, 09:58 PM   #4
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 Originally Posted by Lalitha183 Why did they choose this set as an example and can you explain me the answer given in detail ? Why it is not an interval if $\displaystyle n \geq 2$
First, can you see that N is not a neighborhood of 2?

August 25th, 2017, 10:20 PM   #5
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 Originally Posted by Maschke First, can you see that N is not a neighborhood of 2?
Yes. Since $\displaystyle (x- \epsilon, x+\epsilon) \subset N$ does contain infinitely many points which are not in$N$. $N$ is not a neighborhood of any of it's points.

August 26th, 2017, 08:04 AM   #6
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 Originally Posted by Lalitha183 Yes. Since $\displaystyle (x- \epsilon, x+\epsilon) \subset N$ does contain infinitely many points which are not in$N$. $N$ is not a neighborhood of any of it's points.
Can you say that with more precision, relating it to the precise definition of neighborhood?

In other words I need to see a formal proof that 2 is not a neighborhood of N.

What does containing infinitely many points have to do with it? Which values of $\epsilon$? etc.

"$N$ is not a neighborhood of any of it's points." -- Isn't N a neighborhood of, say, 1/2?

Last edited by Maschke; August 26th, 2017 at 08:52 AM.

August 26th, 2017, 06:48 PM   #7
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 Originally Posted by Maschke Can you say that with more precision, relating it to the precise definition of neighborhood? In other words I need to see a formal proof that 2 is not a neighborhood of N. What does containing infinitely many points have to do with it? Which values of $\epsilon$? etc. "$N$ is not a neighborhood of any of it's points." -- Isn't N a neighborhood of, say, 1/2?
For $N$ to be a neighbourhood of $x$, $\displaystyle \exists \epsilon > 0$ such that $\displaystyle (x- \epsilon, x+ \epsilon)$ is a subset of $N$.

Let us take $\displaystyle \epsilon = 1/2=0.5 >0$ and the open Interval $\displaystyle (2-0.5,2+0.5) = (1.5,2.5) \not\subset N$

Hence, $N$ is not a neighbourhood of its points.

The Infinitely many points refer to the Rational and Irrational numbers, together the Reals.

I'm not much proficient in writing the proofs, please let me know if any corrections to be made.

August 26th, 2017, 07:08 PM   #8
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 Originally Posted by Lalitha183 For $N$ to be a neighbourhood of $x$, $\displaystyle \exists \epsilon > 0$ such that $\displaystyle (x- \epsilon, x+ \epsilon)$ is a subset of $N$. Let us take $\displaystyle \epsilon = 1/2=0.5 >0$ and the open Interval $\displaystyle (2-0.5,2+0.5) = (1.5,2.5) \not\subset N$ Hence, $N$ is not a neighbourhood of its points.
The definition says that THERE EXISTS some $\epsilon$ that works. You showed that one particular epsilon fails to work. But what if you chose a smaller $\epsilon$? Could you find SOME epsilon that works, thereby satisfying the definition of a neighborhood?

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 The Infinitely many points refer to the Rational and Irrational numbers, together the Reals.
This does not make sense in the context of this problem.

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 I'm not much proficient in writing the proofs, please let me know if any corrections to be made.
Your conclusion is wrong. N is in fact a neighborhood of $\frac{1}{2}$. We need to fully understand this. Can you carefully review the definition? We only need to find ONE epsilon that works. It doesn't matter how many fail.

I'm only talking about the unit interval. Let's just deal with that case first. There's some confusion. I'm taking N to be the unit interval plus some isolated points, and I'm asking if it's a neighborhood of $\frac{1}{2}$.

In your proof that N is not a neighborhood of 2, wouldn't you have to prove that EVERY epsilon fails?

Do you need to review your quantifier negation? The negation of $\exists x P(x)$ is $\forall x \neg P(x)$.

Last edited by Maschke; August 26th, 2017 at 07:15 PM.

 August 26th, 2017, 07:14 PM #9 Senior Member   Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2 If we take the natural number $1$, in case of $x=2$, the definition will work with the open interval $\displaystyle (1,3) \subset N$, since you are telling that we do only consider Natural numbers in this case.
August 26th, 2017, 07:16 PM   #10
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 Originally Posted by Lalitha183 If we take the natural number $1$, in case of $x=2$, the definition will work with the open interval $\displaystyle (1,3) \subset N$, since you are telling that we do only consider Natural numbers in this case.
We are talking past each other. I would like to go back to first principles.

Do you see why the open unit interval $(0,1)$ is a neighborhood of $\frac{1}{2}$? We need to understand this before looking at the rest of it.

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