September 18th, 2017, 02:51 PM  #81  
Senior Member Joined: Nov 2015 From: hyderabad Posts: 211 Thanks: 2  Quote:
$\displaystyle \forall$ $\displaystyle x\in (0,1) $$\displaystyle \exists $ some $\displaystyle \epsilon >0$ such that $\displaystyle (x\epsilon,x+\epsilon) \subset (0,1)$ Since $x$ is any point, $\displaystyle (0,1)$ is a neighborhood of each of its points. (ii) $\displaystyle \forall$ $\displaystyle x\in $ {$1,2,...,n$} $\displaystyle \exists $ some $\displaystyle \epsilon >0$ such that $\displaystyle (x\epsilon,x+\epsilon) \not\subset (0,1)\cup${$1,2,...,n$} Since $x$ is any point, $\displaystyle (0,1)\cup${$1,2,...,n$} is not a neighborhood of $n$ points.  
September 18th, 2017, 02:56 PM  #82  
Senior Member Joined: Aug 2012 Posts: 1,681 Thanks: 437  Quote:
Quote:
Last edited by Maschke; September 18th, 2017 at 03:06 PM.  
September 18th, 2017, 03:17 PM  #83 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 211 Thanks: 2 
(i) Let us take $\displaystyle \epsilon =x/2$. Clearly, $\displaystyle 0<x/2<x<1$ $\displaystyle \forall x$, $\displaystyle (xx/2,x+x/2)=(x/2,3x/2)\subset(0,1)$. (ii) $\displaystyle \forall$ $\displaystyle x\in $ {$1,2,...,n$} $\displaystyle \exists $ $\displaystyle \epsilon >0$ such that $\displaystyle (x\epsilon,x+\epsilon) \not\subset (0,1)\cup${$1,2,...,n$} Since $x$ is any point, $\displaystyle (0,1)\cup${$1,2,...,n$} is not a neighborhood of $n$ points. So I need to take the $\displaystyle \epsilon$ that works $\displaystyle \forall x$? 
September 18th, 2017, 03:53 PM  #84  
Senior Member Joined: Aug 2012 Posts: 1,681 Thanks: 437  Quote:
You have got to draw yourself some pictures and try to understand what's happening. Quote:
Do you understand what a neighborhood is? You need to go back to your book and draw pictures and work through some examples. Last edited by Maschke; September 18th, 2017 at 04:00 PM.  
September 18th, 2017, 04:50 PM  #85 
Senior Member Joined: Aug 2012 Posts: 1,681 Thanks: 437 
Is this helpful? 
September 19th, 2017, 10:30 AM  #86 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 211 Thanks: 2  Yes, it is. Let $\displaystyle \epsilon=min${$\displaystyle x,1x$} We know that $\displaystyle 0<1x<x$ => $\displaystyle 0<1x$ or $\displaystyle 1x<x$ (i) $\displaystyle 0<1x$ => $\displaystyle x<1$ Multiply by $2$ on both sides $\displaystyle 2x<2$ subtract $1$ on both sides $\displaystyle 2x1<1$ ($1$) (ii) $\displaystyle 1x<x$ => $\displaystyle 12x<0$ => $\displaystyle 2x1>0$ ($2$) By ($1$) & ($2$), $\displaystyle 0<2x1<1$ ($3$) $\displaystyle \forall x, \exists \epsilon=min${$\displaystyle x,1x$}$\displaystyle >0$ such that $\displaystyle (x1x,x+1x) = (2x1, 1)\subset(0,1)$ (Since by ($3$)) Therefore $\displaystyle (0,1)$ is a neighborhood of each of its points. 
September 19th, 2017, 10:33 AM  #87 
Senior Member Joined: Aug 2012 Posts: 1,681 Thanks: 437  
September 19th, 2017, 11:19 AM  #88 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 211 Thanks: 2  Let $\displaystyle \epsilon = min${$\displaystyle x,1x$} If $\displaystyle \epsilon \leq 0.5$, $\displaystyle min${$\displaystyle x,1x$}$\displaystyle =x$. for $\displaystyle \epsilon =x, (xx,x+x)=(0,2x).$ ($1$) If $\displaystyle \epsilon > 0.5$, $\displaystyle min${$\displaystyle x,1x$}$\displaystyle =1x$. for $\displaystyle \epsilon =1x, (x1+x,x+1x)=(2x1,1)$ ($2$) Since $\displaystyle 2x1<2x,$ By ($1$) & ($2$), $\displaystyle 0<2x1<2x<1$ which implies that for any $\displaystyle x \in (0,1) \exists \epsilon>0$, such that $\displaystyle (x\epsilon, x+\epsilon) \subset (0,1)$ 

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neighbourhood, point 
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