1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Proving the triangle inequality property of the distance between sets

  1. Mar 18, 2012 #1
    Proving the "triangle inequality" property of the distance between sets

    Here's the problem and how far I've gotten on it:

    Untitled-3.png

    If you are unfamiliar with that notation, S(A, B) = (A \ B) U (B \ A), which is the symmetric difference.

    And D(A, B) = m^*(S(A, B)), which is the outer measure of the symmetric difference.
     
  2. jcsd
  3. Mar 18, 2012 #2
    Re: Proving the "triangle inequality" property of the distance between sets

    Hm, can you show that [itex]m^*(A\backslash B)+m^*(B\backslash C)\ge m^*(A \backslash C)[/itex], same for [itex]m^*(B\backslash A)+[/itex]...

    One question, is [itex]m^*[/itex] outer Jordan measure?
     
  4. Mar 18, 2012 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Re: Proving the "triangle inequality" property of the distance between sets

    Note that A and B cannot be just "arbitrary subsets of an arbitrary set X". They must be measurable subsets of the measurable set X.
     
  5. Mar 18, 2012 #4
    Re: Proving the "triangle inequality" property of the distance between sets

    m^* is Lebesgue Outer Measure.

    I can see your point, but I'm copying the problem verbatim out of my textbook, and it uses "arbitrary subsets of an arbitrary set X." Would you say that is a mistake on the author's part?
     
  6. Mar 19, 2012 #5
    Re: Proving the "triangle inequality" property of the distance between sets

    Ok.

    Well, if [itex]A[/itex] and [itex]B[/itex] are disjoint sets, then [itex]m^*(A\cup B)\le m^*(A)+m^*(B)[/itex].

    You can prove this by definition Lebesgue Outer Measure.
     
  7. Mar 19, 2012 #6
    Re: Proving the "triangle inequality" property of the distance between sets

    Well, I got really close. Using the fact that S(A, C) subset S(A, B) U S(B, C), I obtained the following:

    close.png
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Proving the triangle inequality property of the distance between sets
Loading...