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August 27th, 2017, 08:39 PM   #21
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Originally Posted by Maschke View Post
So now we have a better sense of what a neighborhood is. By the way, is [0,1] a neighborhood of 0? Why or why not?
$\displaystyle [0,1] $ is not a neighbourhood of 0.
Since the epsilon interval $\displaystyle (0-1/2,0+1/2) \not\subset [0,1]$.

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Ok onward. Now what is the next step in your question? I assume we're working the problem in the original post.
We were talking about the set $\displaystyle (0,1)\cup ${$\displaystyle 1,2,...,n$}, why it is not an interval when $\displaystyle n \geq 2?$
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August 27th, 2017, 08:43 PM   #22
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$\displaystyle [0,1] $ is not a neighbourhood of 0.
Since the epsilon interval $\displaystyle (0-1/2,0+1/2) \not\subset [0,1]$.
But what if some other epsilon does work? Do you understand that to prove it's a neighborhood we'd only have to find ONE epsilon that works? Finding that some particular epsilon fails to work is not enough. Do you understand this?


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Originally Posted by Lalitha183 View Post
We were talking about the set $\displaystyle (0,1)\cup ${$\displaystyle 1,2,...,n$}, why it is not an interval when $\displaystyle n \geq 2?$
Ok now I'm oriented to where we are. Let's return to this once you have a proof that [0,1] is not a neighborhood of 0.

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 08:47 PM.
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August 27th, 2017, 08:55 PM   #23
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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 $\displaystyle (-0.1,0.1)\not\subset [0,1]$.
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August 27th, 2017, 09:06 PM   #24
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Originally Posted by Maschke View Post
But what if some other epsilon does work? Do you understand that to prove it's a neighborhood we'd only have to find ONE epsilon that works? Finding that some particular epsilon fails to work is not enough. Do you understand this?
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 ?
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August 27th, 2017, 09:12 PM   #25
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I will workout for the proof of this.
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?

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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 ?
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?
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August 27th, 2017, 09:24 PM   #26
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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?
Well, $\displaystyle (0,1) \subset [0,1]. $ and $\displaystyle (0,1)$ is not a neighbourhood of $\displaystyle 0$.

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 ?
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August 27th, 2017, 09:33 PM   #27
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Originally Posted by Maschke View Post
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?
We have $\displaystyle [0,1] \subset R $
let $\displaystyle \epsilon = 0.2 > 0 $ and $\displaystyle x = 0 $.
we need to show that $\displaystyle (x- \epsilon, x+ \epsilon) \not\subset [0,1] $
$\displaystyle (0-0.2, 0+0.2)$
$\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 ?
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August 28th, 2017, 08:23 AM   #28
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Originally Posted by Lalitha183 View Post
We have $\displaystyle [0,1] \subset R $
let $\displaystyle \epsilon = 0.2 > 0 $ and $\displaystyle x = 0 $.
we need to show that $\displaystyle (x- \epsilon, x+ \epsilon) \not\subset [0,1] $
$\displaystyle (0-0.2, 0+0.2)$
$\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 ?
$(0,1)$ is not a neighborhood of $\frac{1}{2}$. If I take $\epsilon = 1000$ then the interval $(\frac{1}{2} - \epsilon, \frac{1}{2} + \epsilon$ is not a subset of $(0,1)$.

What's wrong with that proof?

You are still confused about quantifiers.
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August 28th, 2017, 08:30 AM   #29
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Quote:
Originally Posted by Maschke View Post
$(0,1)$ is not a neighborhood of $\frac{1}{2}$. If I take $\epsilon = 1000$ then the interval $(\frac{1}{2} - \epsilon, \frac{1}{2} + \epsilon$ is not a subset of $(0,1)$.

What's wrong with that proof?

You are still confused about quantifiers.
Well, I have asked you if a subset is not a neighborhood of a point then it's superset is also not a neighborhood of that point?

You have asked to prove for closed set $\displaystyle [0,1]$ right ?
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August 28th, 2017, 08:41 AM   #30
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Originally Posted by Lalitha183 View Post
Well, I have asked you if a subset is not a neighborhood of a point then it's superset is also not a neighborhood of that point?
You asked me, and I gave a counterexample at the end of post #25. {1/2} is not a neighborhood of 1/2, but (0,1) is a superset of {1/2} and it IS a neighborhood of 1/2.

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Originally Posted by Lalitha183 View Post
You have asked to prove for closed set $\displaystyle [0,1]$ right ?
Yes, I asked you to prove that [0,1] is not a neighborhood of 0 using the definition you originally gave.

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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