Showing two epsilon balls intersection is empty

1. Dec 1, 2012

Visceral

1. The problem statement, all variables and given/known data
Suppose $x,y \in X$ which is a normed linear space and $x\neq y$
. Prove that $\exists r>0$ such that $B(x,r) \cap B(y,r)=∅$

2. Relevant equations
Epsilon Ball

$B(x,r)={z \in X:||x-z||<r}$

3. The attempt at a solution

So my attempt here is via contradiction and its not workout out. I assume the intersection of the balls around x and y are not empty, and try to use the fact that x is not equal to y and use some typical triangle inequality tricks to attempt to get a contradiction. I always end up getting what I assumed to be true.

For instance, if the intersection is not empty then there exists an element in both balls. This would imply that the 2*r>ε=||x-y||. However I cant contradict this. What am I doing wrong? Please give any suggestions if you can. I missed this problem on a test and I still can't figure it out.

2. Dec 1, 2012

gopher_p

So you've assumed, toward a contradiction, that for all $r>0$, $B(x,r)\cap B(y,r)\neq\emptyset$ and shown that this implies that for all $r>0$, $2r>\|x-y\|$.

What happens as $r\rightarrow 0$, what does this say about $x$ and $y$, and how does this contradict the assumptions of the problem?

For the record, you don't need to do a proof by contradiction here. There is a fairly simple constructive proof (i.e. you can actually find a specific r that works) using the triangle inequality. Try to figure out how that works.

3. Dec 1, 2012

Staff: Mentor

If x ≠ y, then ||x - y|| is a positive number, say d.

If you take r smaller than d/2, then no point in B(x, r) can also be in B(y, r).

4. Dec 1, 2012

Visceral

gopher_p- I see what you mean, and thank you for the help

Mark44-This is exactly what I just realized! Thanks

I think I got it. I let ||x-y||= ε and let the balls have radius of ε/3 and there is no intersection. That way was really incredibly easy... Can anyone give me a hint on how you can do this by contradiction? I got 7/10 points for this proof on a test and got stuck at the end. Apparently by the professors grading I was "on the right track" but missed some final trick.

Thanks again