1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Showing two epsilon balls intersection is empty

  1. Dec 1, 2012 #1
    1. The problem statement, all variables and given/known data
    Suppose [itex]x,y \in X[/itex] which is a normed linear space and [itex]x\neq y[/itex]
    . Prove that [itex]\exists r>0[/itex] such that [itex]B(x,r) \cap B(y,r)=∅[/itex]

    2. Relevant equations
    Epsilon Ball

    [itex]B(x,r)={z \in X:||x-z||<r}[/itex]

    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. jcsd
  3. Dec 1, 2012 #2
    So you've assumed, toward a contradiction, that for all [itex]r>0[/itex], [itex]B(x,r)\cap B(y,r)\neq\emptyset[/itex] and shown that this implies that for all [itex]r>0[/itex], [itex]2r>\|x-y\|[/itex].

    What happens as [itex]r\rightarrow 0[/itex], what does this say about [itex]x[/itex] and [itex] y[/itex], 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.
  4. Dec 1, 2012 #3


    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).
  5. Dec 1, 2012 #4
    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook