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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

    Mark44

    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
     
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