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September 18th, 2017, 01:51 PM   #81
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Quote:
 Originally Posted by Maschke Yes that is exactly what needs to be proved. Yes! Now the proof ...
(i)
$\displaystyle \forall$ $\displaystyle x\in (0,1) $$\displaystyle \exists some \displaystyle \epsilon >0 such that \displaystyle (x-\epsilon,x+\epsilon) \subset (0,1) Since x is any point, \displaystyle (0,1) is a neighborhood of each of its points. (ii) \displaystyle \forall \displaystyle x\in {1,2,...,n} \displaystyle \exists some \displaystyle \epsilon >0 such that \displaystyle (x-\epsilon,x+\epsilon) \not\subset (0,1)\cup{1,2,...,n} Since x is any point, \displaystyle (0,1)\cup{1,2,...,n} is not a neighborhood of n points. September 18th, 2017, 01:56 PM #82 Senior Member Joined: Aug 2012 Posts: 2,311 Thanks: 706 Quote:  Originally Posted by Lalitha183 (i) \displaystyle \forall \displaystyle x\in (0,1)$$\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.
This is what you claim but it is not a proof. Suppose I say I don't believe you? You say there's some epsilon. I say prove it!

Quote:
 Originally Posted by Lalitha183 (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.
No that's not right. So what if there exists some epsilon that doesn't work? What if some other epsilon does? Suppose I claim (0,1) is not a neighborhood of 1/2? And I say look, take epsilon = 1000. See? $(1/2 -1000, 1/2 + 1000)$ is not a subset of $(0,1)$. What's wrong with my proof?

Last edited by Maschke; September 18th, 2017 at 02:06 PM.

 September 18th, 2017, 02:17 PM #83 Senior Member   Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2 (i) Let us take $\displaystyle \epsilon =x/2$. Clearly, $\displaystyle 00$ 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, 02:53 PM   #84
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Quote:
 Originally Posted by Lalitha183 (i) Let us take $\displaystyle \epsilon =x/2$. Clearly, $\displaystyle 0 Take$x = \frac{9}{10}$. What is$\frac{x}{2}$? And what is$x + \frac{x}{2}$? Is the sum still within the unit interval? You have got to draw yourself some pictures and try to understand what's happening. Quote:  Originally Posted by Lalitha183 (ii)$\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$\displaystyle \forall x$? Let's do the first one first and get to ii later. Did you understand my counterexample "proving" that (0,1) is not a neighborhood of 1/2? 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 03:00 PM.  September 18th, 2017, 03:50 PM #85 Senior Member Joined: Aug 2012 Posts: 2,311 Thanks: 706 Is this helpful? September 19th, 2017, 09:30 AM #86 Senior Member Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2 Quote:  Originally Posted by Maschke Is this helpful? Yes, it is. Let$\displaystyle \epsilon=min${$\displaystyle |x|,|1-x|$} We know that$\displaystyle 0<|1-x|<|x|$=>$\displaystyle 0<1-x$or$\displaystyle 1-x<x$(i)$\displaystyle 0<1-x$=>$\displaystyle x<1$Multiply by$2$on both sides$\displaystyle 2x<2$subtract$1$on both sides$\displaystyle 2x-1<1$-----($1$) (ii)$\displaystyle 1-x<x$=>$\displaystyle 1-2x<0$=>$\displaystyle 2x-1>0$-----($2$) By ($1$) & ($2$),$\displaystyle 0<2x-1<1$----($3$)$\displaystyle \forall x, \exists \epsilon=min${$\displaystyle |x|,|1-x|$}$\displaystyle >0$such that$\displaystyle (x-|1-x|,x+|1-x|) = (2x-1, 1)\subset(0,1)$(Since by ($3$)) Therefore$\displaystyle (0,1)$is a neighborhood of each of its points. September 19th, 2017, 09:33 AM #87 Senior Member Joined: Aug 2012 Posts: 2,311 Thanks: 706 Quote:  Originally Posted by Lalitha183 Yes, it is. Let$\displaystyle \epsilon=min${$\displaystyle |x|,|1-x|$} We know that$\displaystyle 0<|1-x|<|x|$Where did that come from? What if$x = \frac{1}{10}$? September 19th, 2017, 10:19 AM #88 Senior Member Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2 Quote:  Originally Posted by Maschke Where did that come from? What if$x = \frac{1}{10}$? Let$\displaystyle \epsilon = min${$\displaystyle |x|,|1-x|$} If$\displaystyle \epsilon \leq 0.5$,$\displaystyle min${$\displaystyle |x|,|1-x|$}$\displaystyle =|x|$. for$\displaystyle \epsilon =|x|, (x-x,x+x)=(0,2x).$----($1$) If$\displaystyle \epsilon > 0.5$,$\displaystyle min${$\displaystyle |x|,|1-x|$}$\displaystyle =|1-x|$. for$\displaystyle \epsilon =|1-x|, (x-1+x,x+1-x)=(2x-1,1)$---($2$) Since$\displaystyle 2x-1<2x,$By ($1$) & ($2$),$\displaystyle 0<2x-1<2x<1$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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