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September 18th, 2017, 01:51 PM   #81
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Originally Posted by Maschke View Post
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.
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September 18th, 2017, 01:56 PM   #82
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Originally Posted by Lalitha183 View Post
(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!

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Originally Posted by Lalitha183 View Post
(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.
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September 18th, 2017, 02:17 PM   #83
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(i)
Let us take $\displaystyle \epsilon =x/2$. Clearly, $\displaystyle 0<x/2<x<1$
$\displaystyle \forall x$, $\displaystyle (x-x/2,x+x/2)=(x/2,3x/2)\subset(0,1)$.

(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 $\displaystyle \forall x$?
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September 18th, 2017, 02:53 PM   #84
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Originally Posted by Lalitha183 View Post
(i)
Let us take $\displaystyle \epsilon =x/2$. Clearly, $\displaystyle 0<x/2<x<1$
$\displaystyle \forall x$, $\displaystyle (x-x/2,x+x/2)=(x/2,3x/2)\subset(0,1)$.
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.

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(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.
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September 18th, 2017, 03:50 PM   #85
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Is this helpful?

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September 19th, 2017, 09:30 AM   #86
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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.
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September 19th, 2017, 09:33 AM   #87
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Originally Posted by Lalitha183 View Post
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 $x = \frac{1}{10}$?
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September 19th, 2017, 10:19 AM   #88
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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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