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Using the generalized triangle inequality

  1. Nov 3, 2012 #1
    1. The problem statement, all variables and given/known data
    Using the generalized triangle inequality, prove |d(x,y) - d(z,w)| ≤ d(x,z) + d(y,w)



    2. Relevant equations
    d(x,y) is a metric
    triangle inequality: d(x,y) ≤ d(x,z) + d(z,y)

    3. The attempt at a solution
    I know that this needs to be proved with cases: a) d(x,y) - d(z,w) ≥ 0, and b) d(x,y) - d(z,w) < 0. I know that for case a), we have |d(x,y) - d(z,w)| = d(x,y) - d(z,w). The part I am stuck at is that by applying the generalized triangle inequality to this expression, we are supposed to get d(x,y) - d(z,w) ≤ d(x,z) + d(w,y). I just do not understand why. Can anyone help please?
     
  2. jcsd
  3. Nov 3, 2012 #2

    micromass

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    Why would this prove anything?? Why do you even think (a) and (b) are true?
     
  4. Nov 3, 2012 #3
    OK, I used the two cases a) d(x,y) - d(z,w) ≥ 0 and b) d(x,y) - d(z,w) < 0 because of the absolute value in the problem. So for case a) i need to prove d(x,y) - d(z,w) ≤ d(x,z) + d(y,w) and for case b) i need to prove d(z,w) - d(x,y) ≤ d(x,z) + d(y,w)
     
  5. Nov 3, 2012 #4

    micromass

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    OK, that makes sense.

    So you need to prove d(x,y)-d(z,w) ≤ d(x,z) + d(y,w). What happens if you add d(z,w) to both sides?
     
  6. Nov 3, 2012 #5
    Ok, so then adding d(z,w) to both sides gives d(x,y) ≤ d(x,z) + d(y,w) + d(z,w), so d(x,z) + d(z,w) + d(y,w) ≥ d(x,w) + d(w,y) by the triangle inequality, and d(x,w) + d(w,y) ≥ d(x,y) by the triangle inequality, so d(x,y) ≤ d(x,z) + d(y,w) + d(z,w). Thanks so much, now I understand where this is coming from!
     
  7. Nov 3, 2012 #6

    micromass

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    Note: if you actually write up the proof then you have to reverse it. You can't write: from [itex]d(x,y)-d(z,w) \leq d(x,z) + d(y,w)[/itex] follows [itex]d(x,y)\leq d(x,z)+d(y,w)+d(w,z)[/itex]. This is true, but it isn't helpful since you started with what you wanted to proof.

    What you have to do is write: from [itex]d(x,y)\leq d(x,z)+d(y,w)+d(w,z)[/itex] follows [itex]d(x,y)-d(z,w) \leq d(x,z) + d(y,w)[/itex].
     
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