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August 27th, 2017, 09:39 PM   #21
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Quote:
 Originally Posted by Maschke 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 $\displaystyle (0-1/2,0+1/2) \not\subset [0,1]$.

Quote:
 Originally Posted by Maschke 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 $\displaystyle n \geq 2?$

August 27th, 2017, 09:43 PM   #22
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Quote:
 Originally Posted by Lalitha183 $\displaystyle [0,1]$ 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?

Quote:
 Originally Posted by Lalitha183 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 09:47 PM.

 August 27th, 2017, 09:55 PM #23 Senior Member   Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2 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 $\displaystyle (-0.1,0.1)\not\subset [0,1]$.
August 27th, 2017, 10:06 PM   #24
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Quote:
 Originally Posted by Maschke 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 ?

August 27th, 2017, 10:12 PM   #25
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Quote:
 Originally Posted by Lalitha183 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?

Quote:
 Originally Posted by Lalitha183 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?

August 27th, 2017, 10:24 PM   #26
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Quote:
 Originally Posted by Maschke 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 ?

August 27th, 2017, 10:33 PM   #27
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Quote:
 Originally Posted by Maschke 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 ?

August 28th, 2017, 09:23 AM   #28
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Quote:
 Originally Posted by Lalitha183 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.

August 28th, 2017, 09:30 AM   #29
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Quote:
 Originally Posted by Maschke $(0,1)$ 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 ?

August 28th, 2017, 09:41 AM   #30
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Quote:
 Originally Posted by Lalitha183 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.

Quote:
 Originally Posted by Lalitha183 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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