User Name Remember Me? Password

 Topology Topology Math Forum

August 26th, 2017, 07:18 PM   #11
Senior Member

Joined: Nov 2015

Posts: 232
Thanks: 2

Quote:
 Originally Posted by Maschke 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 $\frac{1}{2}$.
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: 2,193
Thanks: 645

Quote:
 Originally Posted by Lalitha183 Are you referring $N$ as a subset of Reals or $N$ as the set of Natural numbers ?
IT"S YOUR N!!!!!!!! You first used the notation. What do YOU mean by it? Sorry for expressing a little frustration 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: 2,193
Thanks: 645

Quote:
 Originally Posted by Lalitha183 Sorry about that. Since you have asked whether $N$ is a neighbourhood of $2$, I thought you might be asking about Natural Numbers.
I can see that. I was using N as in your notation. If I mean $\mathbb N$ I'd write that.

Quote:
 Originally Posted by Lalitha183 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.
No this is not true. What if I take $\epsilon = 100000$? It's still greater than zero, right? But 1/2 plus/minus epsilon is no longer a subset of the unit interval.

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

Posts: 232
Thanks: 2

Quote:
 Originally Posted by Maschke I can see that. I was using N as in your notation. If I mean $\mathbb N$ I'd write that. No this is not true. What if I take $\epsilon = 100000$? It's still greater than zero, right? [It's true that the unit interval is a neighborhood of 1/2, but not for the reason you give].
This works only for some $\displaystyle \epsilon >0$ and also it depends on the x value, because x should lies in between a and b (a<x<b) to have $N$ as as the neighbourhood. August 26th, 2017, 07:56 PM   #16
Senior Member

Joined: Aug 2012

Posts: 2,193
Thanks: 645

Quote:
 Originally Posted by Lalitha183 This works only for some $\displaystyle \epsilon >0$ and also it depends on the x value, because x should lies in between a and b (a
I want to see a formal proof that (0,1) is a neighborhood of 1/2.

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 $\displaystyle (0,1) =${$\displaystyle x|0 August 27th, 2017, 12:13 PM #18 Senior Member Joined: Aug 2012 Posts: 2,193 Thanks: 645 Quote:  Originally Posted by Lalitha183 We have$\displaystyle (0,1) = ${$\displaystyle x|0
I see no reference to the definition you gave of a neighborhood. This is not a proof from that definition.

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:
 Originally Posted by Lalitha183 (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 $1/2$.
I see no reference to the definition of neighborhood you gave in your first post. You are simply invoking well-known facts that are more powerful than the proposition you are trying to prove.

Quote:
 Originally Posted by Lalitha183 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)?
It's necessary to prove the proposition directly from the definition you have given, not from more general well-known facts that are not givens in this problem.

You initially wrote the following definition:

Quote:
 Originally Posted by Lalitha183 Neighbourhood : A subset N of R is a neighbourhood of a point p if $\displaystyle \exists$ $\displaystyle \epsilon > 0$ such that $\displaystyle (x- \epsilon, x+\epsilon) \subset N$
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 $\displaystyle (x- \epsilon, x+ \epsilon) \subset (0,1)$ $\displaystyle (1/2-0.5, 1/2+0.5)$ $\displaystyle (0,1) \subset (0,1)$ ( 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!😅 August 27th, 2017, 08:03 PM   #20
Senior Member

Joined: Aug 2012

Posts: 2,193
Thanks: 645

Quote:
 Originally Posted by Lalitha183 We have $\displaystyle (0,1) \subset R$ let $\displaystyle \epsilon = 0.5 > 0$ and $\displaystyle x = 1/2$. we need to show that $\displaystyle (x- \epsilon, x+ \epsilon) \subset (0,1)$ $\displaystyle (1/2-0.5, 1/2+0.5)$ $\displaystyle (0,1) \subset (0,1)$ ( 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!������
Yes that's the proof, good. I would have added that we know it's a neighborhood since we have found the epsilon that we are required to find.

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 epsilon-interval 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. Tags neighbourhood, point Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post abcdefgh123 Algebra 2 January 4th, 2014 03:08 PM medvedev_ag Real Analysis 0 June 5th, 2013 01:38 AM Vibonacci Algebra 5 September 17th, 2012 12:28 AM mathsiseverything Algebra 1 March 4th, 2008 06:41 AM medvedev_ag Calculus 0 December 31st, 1969 04:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top      