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Is the following correct? (concerns sets and convergence)

  1. Dec 1, 2011 #1
    Let A = {(x,y) in R^2 | x^2 + y^2 <= 81}
    Let B = {((x,y) in R^2 | (x-10)^2 + (y-10)^2 <= 1}

    then here "A intersection B" is the empty set.
    Then let x_n be the sequence (0,10-(2/n)) which is a sequence in A and y_n be the sequence (10/n,10) which is a sequence in B.
    would |x_n - y_n| tend to 0 in this case?
     
  2. jcsd
  3. Dec 1, 2011 #2

    jgens

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    Why do you think xn is a sequence in A? It is not.
     
  4. Dec 1, 2011 #3
    Is it not? Isn't a sequence in A that converges to a point in A compliment?
     
  5. Dec 1, 2011 #4

    jgens

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    A sequence in A is a sequence all of whose values are in A. Almost all of the values of xn are not in A.
     
  6. Dec 1, 2011 #5

    micromass

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    A is a disk with radius 9. The sequence [itex]x_n[/itex] converges to (0,10). So the sequence is not in A.
     
  7. Dec 1, 2011 #6
    But can't a sequence x_n [itex]\subseteq[/itex] S converge to a point in S compliment?
     
  8. Dec 1, 2011 #7

    micromass

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    Not if S is closed, no.
     
  9. Dec 1, 2011 #8
    Oh right I see. Thank you.

    Can there exist a pair of non-empty closed sets A,B [itex]\subseteq[/itex] R^2 with A [itex]\cap[/itex] B equals the empty set with a pair of sequences x_n [itex]\subseteq[/itex] A and y_n [itex]\subseteq[/itex] B such that ||x_n - y_n|| -> 0.

    I was trying to find a specific example and hence my original thought was my first attempt.
     
  10. Dec 1, 2011 #9

    micromass

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    Yes, such a thing exists. Take

    [tex]A=\{(x,0)~\vert~x\in \mathbb{R}\}[/tex]

    and

    [tex]B=\{(x,1/x)~\vert~x\in \mathbb{R}\}[/tex]

    (Note the distance between these two sets is 0). Can you find suitable sequences in these sets?
     
  11. Dec 1, 2011 #10
    Could you explain why the distance between the two sets is 0 please. I thought it was because d|x-y| = (x,0) - (x,1/x) = (0, -1/x) but wasn't sure how this means the distance is 0.

    Also I'm finding it hard to grasp the concept of sequences in 2 dimension in the sense of how to represent them but my guess would be,
    x_n = (3n, 0) and y_n = (3n, 1/3n) ?
     
  12. Dec 1, 2011 #11
    The distance between two sets is usually defined as the greater lower bound of the set of distances between 2 points in the sets. In symbols:
    [tex]
    d(A,B) = \inf \{d(a,b) : a \in A, b \in B\} \, .
    [/tex]
    Take a look at this: http://en.wikipedia.org/wiki/Distance#Distances_between_sets_and_between_a_point_and_a_set

    So you were right that [itex] d(a,b) = |a-b| = |(x,0) - (x, 1/x)| = |(0,-1/x)| = 1/|x| [/itex]. But now you need to find the greatest lower bound of the set [itex] \{d(a,b) : a \in A, b \in B\} [/itex]. Do you see why it will be zero?
     
  13. Dec 2, 2011 #12
    Is it because as x gets smaller and tends to -infinity but you take the modulus and so u get limn-> infinity 1/x = 0?

    Also were my choices of sequences okay? That is the main issue I am stuck with? I need x_n in A and y_n in B such that ||x_n - y_n|| tends to 0.
    I think x_n=(3n,0) and y_n=(3n/1/3n) would work right?
     
  14. Dec 2, 2011 #13

    micromass

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    Yes, that choice of sequence is a good one!
     
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