1. ## Advanced Calculus - Epsilon-Neighborhoods

This is from a real analysis textbook so I'm not 100% sure if it should go in here or in analysis.

The problem:
Show that if a,b belong in R, and a=/=b, then there exist epsilon-neighborhoods U of a and V of b such that the intersection of U and V equals to an empty set.

This is what I have so far:
Assume that there does not exist epsilon-neighborhoods U of a and V of b such that the intersection of U and V equals an empty set. Let, 0< epsilon < absolute value of (1/2(b-a)). Therefore, since the intersection of U and V does not equal an empty set, there must exist a y with which y-a < epsilon and y-b < epsilon.

Then I calculated the bounds of the union of all epsilon neighborhoods within the range of 0< epsilon < absolute value of (1/2(b-a)) giving U the bounds 1/2(3a - b) and 1/2(a+b); V the bounds 1/2(3b-a) and 1/2(a+b).

Am I on the right track? I know I have to somehow prove that if y-a < epsilon and y-b < epsilon then it cannot be within both of those bounds, but I don't how to do that...

2. Just take $\varepsilon=\frac{|b-a|}{4}$.
That single value works for both.

3. Originally Posted by Plato
Just take $\varepsilon=\frac{|b-a|}{4}$.
That single value works for both.
Just out of curiosity, maybe I am reading the problem wrong but why does $\delta=\frac{d(a,b)}{2}$ work? I am only asking because usually one picks the minimum radius for the open balls to be disjoint.

4. Originally Posted by Drexel28
Just out of curiosity, maybe I am reading the problem wrong but why does $\delta=\frac{d(a,b)}{2}$ work? I am only asking because usually one picks the minimum radius for the open balls to be disjoint.
Yes it works. But there are times I want their closures to be disjoint.

5. So taking this new approach where, $\varepsilon=\frac{1}{2}|b-a|$. I find that the two sets are:

$U:=\{x \epsilon R : |x-a| < \frac{1}{4}|b-a|\}$
$V:=\{y \epsilon R: |y-b| < \frac{1}{4}|b-a|\}$

To prove they are disjoint, I would have to find that no element of U can be found in V.

So I let $x_1 \epsilon U$.

Let us assume that $x_1 \epsilon V$, also.
Therefore, I have to prove that $|x-b| > \frac{1}{4} |b-a|$ to find the 'contradiction'.

Or am I thinking about it too complicated again?

6. Suppose that $z \in U \cap V$ then
$\left| {a - b} \right| \leqslant \left| {a - z} \right| + \left| {z - b} \right| <\frac{{\left| {a - b} \right|}}{4} + \frac{{\left| {a - b} \right|}}{4} = \frac{{\left| {a - b} \right|}}{2}$

7. Yeah, I was just about to post that I got it. Triangle Inequality, right?

Thanks!

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