August 27th, 2017, 09:39 PM  #21  
Senior Member Joined: Nov 2015 From: hyderabad Posts: 211 Thanks: 2  Quote:
Since the epsilon interval $\displaystyle (01/2,0+1/2) \not\subset [0,1]$. We were talking about the set $\displaystyle (0,1)\cup ${$\displaystyle 1,2,...,n$}, why it is not an interval when $\displaystyle n \geq 2?$  
August 27th, 2017, 09:43 PM  #22  
Senior Member Joined: Aug 2012 Posts: 1,704 Thanks: 454  Quote:
Quote:
Do you need practice or explanation in negating quantifiers? In other words if I am required to prove that there exists an epsilon, trying one epsilon and finding it's not the one I wanted isn't enough to prove that NO epsilon will work. Last edited by Maschke; August 27th, 2017 at 09:47 PM.  
August 27th, 2017, 09:55 PM  #23 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 211 Thanks: 2 
But you have asked to prove that $\displaystyle [0,1]$ is not a neighbourhood of $\displaystyle 0$ only right ? $\displaystyle [0,1]$ is a neighbourhood of each of its points, excluding the two extreme ends 0,1. That is why I have taken an $\displaystyle \epsilon>0$ for which the epsilon interval is not a subset of $\displaystyle [0,1]$. Even if we take $\displaystyle \epsilon=0.1$, it is still $\displaystyle (0.1,0.1)\not\subset [0,1]$. 
August 27th, 2017, 10:06 PM  #24 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 211 Thanks: 2  I will workout for the proof of this. And let me know if the subset of a set is not a neighbourhood of that point then it's not always true that the superset is also not a neighbourhood of that point right ?

August 27th, 2017, 10:12 PM  #25 
Senior Member Joined: Aug 2012 Posts: 1,704 Thanks: 454  Are you unsure about the quantifiers? That's important to nail down. You understand that to satisfy an $\exists$ you only have to find one that works. So if you try one and it doesn't work ... that doesn't prove anything. Is that clear? I'm afraid I don't understand that question. The set {1/2} is not a neighborhood of 1/2 but the superset (0,1) is a neighborhood of 1/2. Yes? 
August 27th, 2017, 10:24 PM  #26  
Senior Member Joined: Nov 2015 From: hyderabad Posts: 211 Thanks: 2  Quote:
I'm asking if $\displaystyle (0,1)$ not being a neighbourhood of $\displaystyle 0$ effect the relation between $\displaystyle 0$ and $\displaystyle [0,1]$ ? Since we know that Every super set of the set which is a neighbourhood of x, is also a neighbourhood of that point. I'm asking if the converse true ?  
August 27th, 2017, 10:33 PM  #27  
Senior Member Joined: Nov 2015 From: hyderabad Posts: 211 Thanks: 2  Quote:
let $\displaystyle \epsilon = 0.2 > 0 $ and $\displaystyle x = 0 $. we need to show that $\displaystyle (x \epsilon, x+ \epsilon) \not\subset [0,1] $ $\displaystyle (00.2, 0+0.2)$ $\displaystyle (0.2,0.2) \not\subset [0,1] $ Hence $\displaystyle [0,1] $ is not a neighborhood of $\displaystyle 0$. Is this enough ? or Is it to be more generalized ?  
August 28th, 2017, 09:23 AM  #28  
Senior Member Joined: Aug 2012 Posts: 1,704 Thanks: 454  Quote:
What's wrong with that proof? You are still confused about quantifiers.  
August 28th, 2017, 09:30 AM  #29  
Senior Member Joined: Nov 2015 From: hyderabad Posts: 211 Thanks: 2  Quote:
You have asked to prove for closed set $\displaystyle [0,1]$ right ?  
August 28th, 2017, 09:41 AM  #30  
Senior Member Joined: Aug 2012 Posts: 1,704 Thanks: 454  Quote:
Quote:
Are you drawing pictures for yourself to try to understand what a neighborhood is? And working with the quantifiers to understand what you need to show?  

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