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Open ball's diameter = 2r?

  1. Feb 25, 2012 #1
    1. The problem statement, all variables and given/known data
    Show that open ball [itex]B(0,r) \subset \mathbb{R}^2[/itex] has a diameter 2r.

    3. The attempt at a solution
    Intuitively i'd say right away that it can't be 2r but i presume it has to be since i'm being asked to prove it.
    To simplify the situation, i'll think about this in one dimension.

    Assume we have points x and y in the ball which means that
    [itex]|x-y| = |x-0+0-y| \leq |x-0|+|y-0|< r+r = 2r[/itex]
    And that's pretty much what i got. Based on this, the diameter should be less than 2r since the points x and y must both be less than r away from the center. What am i not understanding here?
     
    Last edited: Feb 25, 2012
  2. jcsd
  3. Feb 25, 2012 #2

    tiny-tim

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    Hi Uniquebum! :smile:

    Why can't you choose x and -x? :wink:
     
  4. Feb 25, 2012 #3
    Heh, that's obviously wiser. However, that leads to the same situation either way... or?

    Since we can't choose r to be the x as it's an open ball so [itex]x<r[/itex]
    [itex]|x-(-x)|=|2x|<2r[/itex]
    It's better to do this way :) but it leads to the same result.
     
  5. Feb 25, 2012 #4

    tiny-tim

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    Why not?

    Isn't there's still a diameter of length 2r from x to -x, even if the end-points aren't there? :wink:

    You just have to be careful about proving it! :biggrin:
     
  6. Feb 25, 2012 #5

    Fredrik

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    Based on this, I'd say that you need to have another look at the definition of "diameter".

    Consider the open interval (0,1) in ℝ, i.e. the set ##\{x\in\mathbb R|0<x<1\}##. What's the diameter of that set? OK, I'll tell you the answer. It's 1. But how does that follow from the definition of "diameter"? That's what you need to think about.
     
    Last edited: Feb 25, 2012
  7. Feb 25, 2012 #6
    I actually maybe got it! :)

    Let's call the set [itex]A[/itex] and then [itex]sup(A)=r[/itex]. If we pick points [itex]x[/itex] and [itex] -x[/itex] we have an [itex]\epsilon > 0[/itex] from which we can state that
    [itex]x+\epsilon \leq r[/itex] and [itex]|-x-\epsilon| \leq r[/itex]

    Now the distance between the points
    [itex]|x-(-x)|=|x+\epsilon - (-x)-\epsilon| \leq |x+\epsilon| + |-x-\epsilon|\leq r + r = 2r[/itex]

    That means the largest possible value of [itex]|x-(-x)|[/itex] is [itex]2r[/itex] which would be the diameter.

    I so hope that's it!
    Thanks a lot.
     
  8. Feb 25, 2012 #7

    tiny-tim

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    that's all true :smile:, but it doesn't prove …
    try again, starting "given any ε < 0 and < r, there exists an x with r - x < ε/2" :wink:
     
  9. Feb 25, 2012 #8

    Fredrik

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    I don't think so. I don't understand your proof.

    What set?

    What does this mean? I'm guessing that you mean that "for all x in B(0,r), there's an ε>0 such that...", but it's not clear from what you're saying.

    OK, now I see that you couldn't have meant "for all x in B(0,r)", because then the sum x+ε wouldn't make sense. I'm changing my guess to "for all x in (-r,r)". So you seem to be saying that for all x in (-r,r), there's an ε>0 such that |x+ε|≤r. This is true, but not very useful.

    Maybe I'm missing something obvious, but I don't see how you're using the triangle inequality here. I get ≤|x+ε|+|x-ε|. Of course, if you go back and change "for all x in (-r,r)" to "for all x in (0,r)", we have |x-ε|<|x+ε|.

    There are two major problems with this sentence:

    1. There's no x in B(0,r) such that |x-(-x)|=2r.
    2. The diameter isn't defined as the largest value of |x-(-x)|.
     
  10. Feb 25, 2012 #9
    Oh well, banged my head against the wall for a while but still can't figure it out. At least i know that i'm doing it wrong and where i'd have to start the proper proof.

    Thanks again. :)
     
  11. Feb 25, 2012 #10

    Fredrik

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    I suggest that you do a similar but simpler problem first. Can you prove that ##\sup\{x\in\mathbb R|x<0\}=0##? you should do this by proving the inequalities ##\sup\{x\in\mathbb R|x<0\}\leq 0## and ##\sup\{x\in\mathbb R|x<0\}\geq 0## separately. The first one is rather obvious if you think of the definition of "supremum". The second is trickier. I suggest that you prove that for all ε>0, it's impossible that ##\sup\{x\in\mathbb R|x<0\}<-\varepsilon##.
     
  12. Feb 25, 2012 #11

    Ray Vickson

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    Can you first prove that [itex] B(0,r)[/itex] and its closure [itex] \bar{B(0,r)}[/itex] have the same diameter? (It IS true.) Now can you convince yourself that [itex]\text{diam}(\bar{B(0,r)}) = 2r[/itex]? Computing [itex]\text{diam}(\bar{B(0,r)})[/itex] involves solving a maximization problem for a continuous function on a compact set, so has an optimal solution; you can easily solve it.

    RGV
     
  13. Feb 25, 2012 #12

    Fredrik

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    I wouldn't recommend this solution, as it requires the OP to know stuff about closures, compact sets, continuity and how to solve maximization problems. It's a longer solution, even though we're using some theorems that are much more difficult than the problem itself.

    The straightforward approach is to first prove that diam B(0,r)≤2r, and then prove that diam B(0,r)>2r-ε for all ε>0.
     
    Last edited: Feb 26, 2012
  14. Feb 26, 2012 #13

    Ray Vickson

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    Of course, but I did not want to assume anything about what the OP knows or does not know. That is why I asked him questions, and if he answered NO he could just ignore the method. Nevertheless, your direct approach is much more straightforward.

    RGV
     
  15. Feb 26, 2012 #14
    Ok one more attempt. I'm not giving up!
    So the point is to prove that [itex]sup\{d(x,y)|x,y\in B(0,r)\}=2r[/itex]?

    Thus
    1)[itex]sup(d(x,y))=2r[/itex] if and only if
    2)[itex]sup(d(x,y))>2r-\epsilon[/itex] with any given [itex]\epsilon > 0[/itex].

    1)
    Antithesis:
    There exists an [itex]\epsilon > 0[/itex] so that [itex]sup(d(x,y)) \leq 2r-\epsilon[/itex] and then [itex]2r-\epsilon[/itex] is an upper bound of the set. Now [itex]2r-\epsilon<2r[/itex] which contradicts with the presumptions since [itex]2r[/itex] had to be the supremus.

    2)
    Suppose that for any given [itex]\epsilon > 0[/itex] we find such [itex]sup(d(x,y)) > 2r-\epsilon[/itex]. Let [itex]M[/itex] be an upper bound of the set and thus [itex]2r \leq M[/itex].
    Antithesis:
    Let [itex]2r>M[/itex] and [itex]\epsilon = 2r - M > 0[/itex]. Now based on the presumptions [itex]2r - \epsilon < sup(d(x,y)) \leq M[/itex]
    Thus [itex]2r < M + \epsilon = M + 2r - M = 2r[/itex] which is a contradiction.

    This means [itex]sup\{d(x,y)|x,y\in B(0,r)\}=2r[/itex] and thus the diameter...? I think...
     
  16. Feb 26, 2012 #15

    Fredrik

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    I was hoping you would say that. :smile:

    Statement 2 is equivalent to sup(d(x,y))≥2r, not sup(d(x,y))=2r. So if you prove 2, you will have to prove sup(d(x,y))≤2r as well, before you can conclude that the equality holds.


    Here you're proving that if statement 2 is false, then 2r is not the supremum. But you can't conclude that statement 2 is true from this, because we can't assume that 2r is the supremum. That's the statement we want to prove.

    You're making two assumptions here. First you assume that sup(d(x,y))≥2r, which is one of the two inequalities we want to prove. Then (the "thus" comment) you assume that sup(d(x,y))=2r, which is to assume that both of the inequalities you want to prove are true.

    I'll give you some additional hints:

    1. I've been saying that we can prove sup(d(x,y))=2r by proving sup(d(x,y))≤2r and sup(d(x,y))≥2r separately, but this is only true if we have already proved that {d(x,y)|x,y in B(0,r)} is bounded from above. Can you prove that 2r is an upper bound of that set? What else can you conclude from this result?

    2. We want to prove that for all ε>0, sup(d(x,y))>2r-ε. You should start by saying "let ε>0 be arbitrary" and then prove that the set {d(x,y)|x,y in B(0,r)} has members that ensure that it's impossible that sup(d(x,y))≤2r-ε.
     
  17. Feb 26, 2012 #16
    Righty, here we go again.

    We want to prove that for all ε>0, sup(d(x,y))>2r-ε.
    Let [itex]\epsilon>0[/itex] be arbitrary.
    Antithesis: [itex]sup(d(x,y)) \leq 2r-\epsilon[/itex]
    This means that [itex]2r-\epsilon[/itex] is an upper bound of the set.
    Let [itex]\epsilon>\delta>0[/itex]. Now [itex]2r-\epsilon<2r-\delta<2r[/itex] which means [itex]2r-\epsilon[/itex] cannot be an upper bound and [itex]2r[/itex] is.

    Since [itex]2r[/itex] is an upper bound of the set we can state that [itex]2r\geq sup(d(x,y))[/itex]



    This is as far as i can get. What i'm wondering is the statement of [itex]2r \leq sup(d(x,y))[/itex]. As far as i understand, i'd have to prove first that 2r belongs to the set [itex]\{d(x,y)|x,y\in B(0,r)\}[/itex] and i don't know how i'd go about that. So what am i missing?
     
  18. Feb 26, 2012 #17

    tiny-tim

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    i don't understand that :confused:

    try again, starting "given any ε > 0 and < r, there exists an x with r - x < ε/2" :wink:
     
  19. Feb 26, 2012 #18

    Fredrik

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    I don't see how it follows from those inequalities that a) 2r-ε is not an upper bound, or that b) 2r is an upper bound.

    If this had been true, only sets that have a maximum element would have a supremum. It would e.g. be impossible for the set of negative real numbers to have 0 as its supremum. We have ##\sup\{x\in\mathbb R|x<0\}\geq 0## even though 0 isn't in the set.


    The issue that you're struggling with has little to do with metric spaces, open balls, etc. It's about the concept of "supremum". So I would suggest that you try the idea I've been describing on a slightly simpler problem first. Can you e.g. prove that the supremum of all negative real numbers is 0? These are the steps:

    1. Prove that 0 is an upper bound of ##\{x\in\mathbb R|x<0\}##. (OK, this step is trivial).
    2. Prove that ##\sup\{x\in\mathbb R|x<0\}\leq 0##. (Also kind of trivial, if you think of what a supremum is).
    3. Prove that ##\sup\{x\in\mathbb R|x<0\}\geq -\varepsilon## for all ε>0.
     
  20. Feb 27, 2012 #19
    1. Prove that 0 is an upper bound of [itex]S=\{x\in \mathbb{R}|x<0\}[/itex]. (OK, this step is trivial).
    2. Prove that [itex]S=sup\{x\in \mathbb{R}|x<0\}\leq 0[/itex]. (Also kind of trivial, if you think of what a supremum is).
    3. Prove that [itex]S=sup\{x\in \mathbb{R}|x<0\}\geq -\epsilon[/itex] for all [itex]\epsilon>0[/itex].

    1)
    0 is an upper bound for the set, since for all [itex]x \leq 0[/itex].

    2)
    Since 0 is an upper bound and S the least of upper bounds, [itex]S\leq 0[/itex]

    3)
    This i don't know how to go about. Since, assume S would be -2 and then it wouldn't go for all epsilon. The only thing we know is that 0 is an upper bound but the difference S-0 could be other than 0.

    And the way tiny-tim suggests leads to [itex]2x> 2r-\epsilon[/itex] which is pretty much what i've been trying to do the whole time. Only with the difference that [itex]2x=d(x,y)[/itex].
     
  21. Feb 27, 2012 #20

    tiny-tim

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    but isn't that what you want?

    try writing it out properly, and see :smile:
     
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