August 25th, 2017, 06:28 PM  #1 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 206 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 
Senior Member Joined: Aug 2012 Posts: 1,514 Thanks: 364  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  
Senior Member Joined: Nov 2015 From: hyderabad Posts: 206 Thanks: 2  Quote:
Why it is not an interval if $\displaystyle n \geq 2$  
August 25th, 2017, 09:58 PM  #4 
Senior Member Joined: Aug 2012 Posts: 1,514 Thanks: 364  
August 25th, 2017, 10:20 PM  #5 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 206 Thanks: 2  
August 26th, 2017, 08:04 AM  #6  
Senior Member Joined: Aug 2012 Posts: 1,514 Thanks: 364  Quote:
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  
Senior Member Joined: Nov 2015 From: hyderabad Posts: 206 Thanks: 2  Quote:
Let us take $\displaystyle \epsilon = 1/2=0.5 >0 $ and the open Interval $\displaystyle (20.5,2+0.5) = (1.5,2.5) \not\subset N$ 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  
Senior Member Joined: Aug 2012 Posts: 1,514 Thanks: 364  Quote:
Quote:
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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: 206 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  
Senior Member Joined: Aug 2012 Posts: 1,514 Thanks: 364  Quote:
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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