August 26th, 2017, 07:18 PM  #11 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2  Are you referring $N$ as a subset of Reals or $N$ as the set of Natural numbers ?

August 26th, 2017, 07:19 PM  #12  
Senior Member Joined: Aug 2012 Posts: 1,960 Thanks: 547  Quote:
You introduced the notation N and now you are asking me what it means. You said it's a subset of R but you did not mark up R as $\mathbb R$. I assumed that N is an arbitrary subset of the reals, and I had to assume R is the reals. Isn't that what you meant? In other words in your first post when you wrote: "A subset N of R" I had to assume that: 1) N is not mathbb'd so it must be some arbitrary set, N meaning "neighborhood"; and 2) R is not marked up because it IS the reals and you just didn't mark it up for some reason. So that's where some confusion exists right from the start. Can you clarify your notation? Last edited by Maschke; August 26th, 2017 at 07:36 PM.  
August 26th, 2017, 07:35 PM  #13 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2 
Sorry about that. Since you have asked whether $N$ is a neighbourhood of $2$, I thought you might be asking about Natural Numbers. The open Interval $\displaystyle (0,1)$ is a neighbourhood of the point $1/2$ , since the Real numbers $\displaystyle 1/2 \epsilon $ and $\displaystyle 1/2+ \epsilon $ where $\displaystyle \epsilon > 0 $ lies in the open interval. 
August 26th, 2017, 07:38 PM  #14  
Senior Member Joined: Aug 2012 Posts: 1,960 Thanks: 547  Quote:
Quote:
Please refer back to the exact definition of neighborhood and convince me that $(0,1)$ either is or isn't a neighborhood of $\frac{1}{2}$. The art of proofs is to refer back to the EXACT definition, and then to see if some particular example does or doesn't fit the definition. Last edited by Maschke; August 26th, 2017 at 07:43 PM.  
August 26th, 2017, 07:49 PM  #15  
Senior Member Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2  Quote:
 
August 26th, 2017, 07:56 PM  #16  
Senior Member Joined: Aug 2012 Posts: 1,960 Thanks: 547  Quote:
I understand that you're not (yet) good at proofs. You've mentioned your educational objective. It's with that in mind that I'm being insistent. Let's see if we can construct a proof. Last edited by Maschke; August 26th, 2017 at 08:02 PM.  
August 26th, 2017, 08:49 PM  #17 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2 
We have $\displaystyle (0,1) = ${$\displaystyle x0<x<1$} Since $\displaystyle 0<1/2<1$, it is enough to show that $\displaystyle (0,1)$ is a neighbourhood of $1/2$. 1. Let $x$ be any point in the open interval $\displaystyle (0,1)$. $\displaystyle x \epsilon (0,1) \subset (0,1)$ (since every set is a subset of itself). Since x is any point of $\displaystyle (0,1)$ $\displaystyle (0,1)$ is a neighbourhood of each of its points. Hence $\displaystyle (0,1)$ is a neighbourhood of $1/2$. (or) 2. As $\displaystyle (0,1)$ is an open interval which implies that $\displaystyle (0,1)$ is an open set (Since Every Open interval is an open set) because it is a neighbourhood of each of its points. Hence $\displaystyle (0,1)$ is a neighbourhood of $1/2$. Is it necessary to prove that an open interval is the neighbourhood of each of its points? Or can we just name it like (Every set is a subset of itself)? Last edited by skipjack; August 26th, 2017 at 09:18 PM. 
August 27th, 2017, 12:13 PM  #18  
Senior Member Joined: Aug 2012 Posts: 1,960 Thanks: 547  Quote:
Secondly, what is wrong with the following FALSE argument, which is identical to yours: (FALSE) Claim: $[0,1]$ is a neighborhood of $0$. (FALSE) Proof: Let $x \in [0,1] \subset [0,1]$. Since $x$ is any point of $[0,1]$, then $[0,1]$ is a neighborhood of any of its points, hence it's a neighborhood of $0$. Now this is false and nonsensical. But it's identical to your proof. Do you understand that you need to prove the proposition directly from the definition of neighborhood you gave in your first post? Do you see that you have not done that? Quote:
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You initially wrote the following definition: Given that definition, you need to show that $(0,1)$ is a neighborhood of $\frac{1}{2}$. Last edited by Maschke; August 27th, 2017 at 12:18 PM.  
August 27th, 2017, 07:48 PM  #19 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2 
We have $\displaystyle (0,1) \subset R $ let $\displaystyle \epsilon = 0.5 > 0 $ and $\displaystyle x = 1/2 $. we need to show that $\displaystyle (x \epsilon, x+ \epsilon) \subset (0,1) $ $\displaystyle (1/20.5, 1/2+0.5)$ $\displaystyle (0,1) \subset (0,1) $ ( Since every set is a subset of itself) Hence $\displaystyle (0,1) $ is a neighborhood of $\displaystyle 1/2$. Please let me know more errors!ðŸ˜… 
August 27th, 2017, 08:03 PM  #20  
Senior Member Joined: Aug 2012 Posts: 1,960 Thanks: 547  Quote:
Also it's confusing that you say every set is a subset of itself. That's not relevant here. What you need is that the epsiloninterval of 1/2 is a subset of (0,1). That's clear, right? So now we have a better sense of what a neighborhood is. By the way, is [0,1] a neighborhood of 0? Why or why not? Ok onward. Now what is the next step in your question? I assume we're working the problem in the original post. Last edited by Maschke; August 27th, 2017 at 08:20 PM.  

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