Here's the exercise:

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

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

$\displaystyle \mbox{Suppose that a,b}\in\Re \ \mbox{and a does not equal b}$

$\displaystyle \\ \mbox{\\Let} \ \varepsilon \ \mbox{be greater than 0.}$

$\displaystyle \\ \mbox{Set U to be the} \ \varepsilon \ \mbox{-neighborhood or a and V to be the} \ \varepsilon \ \mbox{-neighborhood of b}\\$

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

$\displaystyle \\ \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....

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

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

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

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

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!