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Homework Help: Distance Between Closed sets in a metric space

  1. Sep 26, 2006 #1
    Hey guys, thanks for looking at this.

    Ok, so we're given the distance, d(x,C) between a point, x, and a closed set C in a metric space to be: inf{d(x,y): for all y in C}. Then we have to generalize this to define the distance between two sets I'm fairly certain you can define it as:

    the distance between closed sets D and C in a metric space, d(C,D) = inf{d(y,D): for all y contained in C}. Which should be equivalent to inf{d(x,y): for all x,y contained in C,D respectively}.

    My question is this: How to construct an example of two closed, disjoint sets whose distance is zero under this definition? I feel like I need to find two sets containing points that can be made arbitrarily close, but am unsure how to do this without some point being a limit point of both sets, contradicting C,D disjoint.

    If you guys have an idea that would be great, I'd much prefer a hint or nudge in the right direction if possible.

    Thanks again.
     
    Last edited: Sep 26, 2006
  2. jcsd
  3. Sep 26, 2006 #2

    Hurkyl

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    I presume you get to pick your metric space too...
     
  4. Sep 26, 2006 #3
    yes, sorry, forgot to mention that we're talking about a generalized metric space.

    I must say that I have a hard time imagining a scenario where this (the aforementioned problem) is possible, I formerly operated under the assumption(read:intuition) that closed and disjoint in a metric space implied some distance between sets.
     
    Last edited: Sep 26, 2006
  5. Sep 26, 2006 #4

    Hurkyl

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    Here's my approach -- figure out your sets first, then decide upon the metric space. :smile:
     
  6. Sep 26, 2006 #5
    thanks for the help!
     
  7. Mar 30, 2010 #6
    Consider unbounded sets. The distance of two unbounded sets in Euclidean spaces (with the usual metric) can be 0.

    Example: Let A = {(t,0): t>=0}, B={(t,1/t): t>=0}. Both are closed, unbounded and their distance is 0.

    If one of the sets compact, then the distance can never be zero.

    Proof: Let A be compact, B be closed. Since d(A,B) = inf{d(x,y): x in A, y in B}, there exists two sequences (x_n) in A and (y_n) in B, s.t. lim d(x_n,y_n) = d(A,B). Since A is compact, (x_n) has a convergent subsequence (x_n_k), say lim (x_n_k) = x0 in A. Since (d(x_n,y_n)) is convergent, so is its subsequence (d(x_n_k,y_n_k)). In sum,
    d(A,B) = lim d(x_n,y_n) = lim (d(x_n_k,y_n_k)) = lim (d(x0,y_n_k)).

    Now, if d(A,B) = 0, then x0 is an accumulation point of B. Since B is closed, x0 must be in B. But, x0 cannot be both in A and B: the two sets are disjoint. Thus, we must have d(A,B)>0.
     
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