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Closed and bounded

  1. May 1, 2010 #1
    1. The problem statement, all variables and given/known data
    [tex] A = \left\{(x,y): 0\leq xy \leq 1\right\}, A \in R^{2} [/tex]

    I'm trying to determine if this set is bounded and/or closed.


    2. Relevant equations

    if X = (x,y)

    euclidean metric: [tex] ||X|| = \sqrt{x^{2}+y^{2}} [/tex]

    3. The attempt at a solution



    I know a bounded set => ||X|| [tex] \leq [/tex] k

    so I need to show somehow

    [tex] ||X|| = \sqrt{x^{2}+y^{2}} \leq k [/tex] (somehow)

    and closed => every limit point belongs to the set. So take an arbitrary X'= (x',y') [tex] \in [/tex] A'. Then there exists Xn = (x,y) [tex] \in [/tex] A such that Xn -> X' and Xn [tex]\neq[/tex] X'.

    Xn [tex] \in [/tex] A => [tex] 0 \leq xy \leq 1 [/tex]

    Need to show X' is such that [tex] 0 \leq x'y' \leq 1 [/tex] (somehow)
     
  2. jcsd
  3. May 1, 2010 #2
    If x=0 for what values of y would you have 0<=xy<=1?
     
  4. May 1, 2010 #3
    I see, all values of y, so this set is not finite and therefore not bounded?
     
  5. May 1, 2010 #4
    Correct.

    Have you sketched it? The sketch wouldn't prove anything, but it can be helpful to suggest a proof for the next bit.
     
  6. May 1, 2010 #5

    Mark44

    Staff: Mentor

    Martin's advice of sketching the set seems to me a good one. For each nonzero value of k, with 0 < k <= 1, you have xy = k, or y = k/x, a hyperbola.
     
  7. May 1, 2010 #6
    In general though, "not finite" does not imply "not bounded", correct?
     
  8. May 1, 2010 #7
    Correct. Just take [0,1] in the reals
     
  9. May 2, 2010 #8
    Yes, sorry, I shouldn't have written, "Correct". But I knew what you meant.
     
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