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How to prove a set is a bounded set?

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  1. Jan 30, 2017 #1

    Cyn

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    1. I have to show that
    S1 = {x ∈ R2 : x1 ≥ 0,x2 ≥ 0,x1 + x2 = 2}
    is a bounded set.



    2. So I have to show that sqrt(x1^2+x2^2)<M for all (x1,x2) in S1.


    3. I have said that M>0 and we have 0<=x1<=2 and 0<=x2<=2.
    And x2 = 2-x1
    We can fill in sqrt(x1^2 + (2-x1)^2) = sqrt (0^2 + (2-0)^2) = 2 < M = 3.
    And we can fill in sqrt (x1^2 + (2-x1)^2) = sqrt (2^2 + (2-2)^2) = 2 < M = 3.
    Every value between the 0 and the 2 that satisfy x1+x2 = 2 is smaller than this M. So the set is bounded.
    Is this correct?
     
  2. jcsd
  3. Jan 30, 2017 #2

    mfb

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    2016 Award

    Staff: Mentor

    Can you prove this?

    The approach works, although I would make it easier: Just use 0<=x1<=2 and 0<=x2<=2 to find a maximal value the square root can get, and use that as limit. You don't have to find the smallest possible M.
     
  4. Jan 30, 2017 #3

    Ray Vickson

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    Bounded usually means that ##L_1 \leq x_1 \leq U_1## and ##l_2 \leq x_2 \leq U_2## for some finite ##L_1,L_2,U_1,U_2## with ##U_i \geq L_i## for all ##i##. Looking at ##\sqrt{x_1^2 + x_2^2}## is not necessary (but is harmless); one version of "bounded" is true if and only if the other is true as well. That is, the two versions are equivalent.

    In your case you have ##0 \leq x_i \leq 2## for ##i = 1,2##, and that is all you need. You can always enclose the square with vertices (0,0), (),2), (2,0), (2,2) inside a larger circle if you want to, and that will bound the norm ##\sqrt{x_1^2 + x_2^2}##.
     
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