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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)\$ 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

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