# neighborhood proof!

• Jan 26th 2011, 09:37 PM
mremwo
neighborhood proof!
Here's the exercise:

$\mbox{Prove that if a,b}\in\Re\ \mbox{and a does not equal b then there exists}\ \varepsilon\mbox{-neighborhoods U of a and V of b such that U}\cap\mbox{V=}\emptyset$

Here's what I did, but I don't know if it's right and I also have some questions:

$\mbox{Suppose that a,b}\in\Re \ \mbox{and a does not equal b}$
$\\ \mbox{\\Let} \ \varepsilon \ \mbox{be greater than 0.}$
$\\ \mbox{Set U to be the} \ \varepsilon \ \mbox{-neighborhood or a and V to be the} \ \varepsilon \ \mbox{-neighborhood of b}\\$

$\mbox{So U}_{\varepsilon}(a)=\{x\in\Re \ \mbox{:} \mid x-a \mid \mbox{is less than} \ \varepsilon\}$
$\\ \mbox{And V}_{\varepsilon}(b)=\{y\in\Re \ \mbox{:} \mid y-b \mid \mbox{is less than} \ \varepsilon\}$

-One question I have here is do I have to change the variable to y for the set V? If so, why? I think that may be why I am getting stuck later on, but I'm not sure....

$\mbox{Now let}\ x_0 \in \mbox{U}_{\varepsilon}(a) \ \mbox{and let} \ y_0 \in \mbox{V}_{\varepsilon}(b)$

Or if both sets are sets of x such that... then would I pick an arbitrary element of V to be maybe $x_1$?

$\mbox{By the theorm, if}\ x_0 \in \mbox{U}_{\varepsilon}(a)\ \mbox{then}\ x_0\ \mbox{equals a and if}\ y_0 \in \mbox{V}_{\varepsilon}(b)\ \mbox{then}\ y_0\ \mbox{equals b}$

$\mbox{Then by the hypothesis,}\ x_0 \ \mbox{does not equal y_0}$

Then I was thinking something along the lines of that since when you pick two arbitrary elements of each set, they will never equal each other says that there is no element in both sets... But I'm stuck and I don't even know if I am on the right track? How would I show that the intersection of these two sets is empty?

Thank you!

PS- this was my first time using LATEX...it took forever!
• Jan 26th 2011, 10:58 PM
roninpro
I think that you have misunderstood what you need to do. The problem is asking you to find $\varepsilon>0$ so that the neighbourhoods about $a$ and $b$ are disjoint. The intuitive choice would be to pick $\varepsilon=(b-a)/3$. (Draw a picture if it isn't clear!) All you have to do now is to show that $U_\varepsilon(a)$ and $U_\varepsilon(b)$ are disjoint.

Give it a try, and let us know if you get stuck.
• Jan 27th 2011, 05:30 AM
mremwo
ahh how did I not even realize this is an existence proof. always trying to make things harder than they are. thanks!
• Jan 27th 2011, 05:33 AM
Ackbeet
Quote:

PS- this was my first time using LATEX...it took forever!
It's totally worth it, though. One of the best pieces of advice I ever got in undergraduate years was to start doing all my homework in LaTeX. By the time I got to where I needed to write my Ph.D. dissertation, I didn't need to learn LaTeX as I went along, and my writing took probably half the time it might have otherwise.
• Jan 27th 2011, 05:40 AM
mremwo
one question though....why is (b-a)/3 the intuitive choice?
• Jan 27th 2011, 05:53 AM
roninpro
Here is a picture. Note that the circles / neighbourhoods have radius $(b-a)/3$.
• Jan 27th 2011, 10:27 AM
Using the estimate $\epsilon = \frac {b-a}{3}$, which we pick because it will guarantee that the distance will be small enough, we can construct the two neighborhoods:

$U = \{ x \in \mathbb R : |a-x| < \epsilon \}$

$V = \{ y \in \mathbb R : |b-y| < \epsilon \}$

Our claim is that U and V are disjoint.

Now, pick an element, say $x_0 \in U$, our job is to show that this element cannot be in V. Well, what is the definition of being in V? Namely, the distance between y and $x_0$ must be smaller than $\epsilon$

That is, $|y-x_0| < \epsilon$ must hold true. But we can find a way to contradict that, i.e. show that $|y-x_0| \geq \epsilon$

Consider $|x-x_0| < \epsilon = \frac {b-a}{3}$, you can derive from this and get $x_0 > x - \frac {y-x}{4}$

Now, plug this in the expression between y and $x_0$ and see if you can get it to be bigger than the estimate.
• Jan 27th 2011, 10:41 AM
Plato
Here is another way. I prefer to have $\epsilon=\frac{b-a}{2}$.

Let $U=\{y:|a-y|<\epsilon\}~\&~ V=\{y:|b-y|<\epsilon\}$.

Suppose that $z\in U\cap V$.

Then $b-a\le |b-z|+|z-a|<\epsilon +\epsilon =b-a$

That is a contradiction so $U\cap V=\emptyset.$