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Is it a metric space?

  1. Sep 16, 2010 #1
    1. The problem statement, all variables and given/known data

    prove that the given function is a metric, or give a counterexample to show how it fails to be a metric: d(x,y)=|x3-y3|

    2. Relevant equations

    ok, out of the 3 requirements to be a metric, 2 are trivial. The third is to prove the triangle inequality holds: d(x,y)[tex]\leq[/tex]d(x,z)+d(z,y)

    should i just go through the different cases of x<y<z , x<z<y , z<x<y?? and if the inequality holds true for all of them I am done? I feel like there must be a better way to take care of this... Oh, and please don't spoil the question for me. Thanks
     
  2. jcsd
  3. Sep 16, 2010 #2
    As you have written the inequality to prove, you may assume [itex]x \leq y[/itex] since otherwise you can just switch the positions of the two variables. You still have to consider further cases but this should simplify the problem slightly.
     
  4. Sep 16, 2010 #3
    since i already have that |x-y|[tex]\leq[/tex]|x-z|+|z-y| is a metric space, there must be a real direct way to show this holds true also right???
     
  5. Sep 16, 2010 #4
    Yes, well precisely that is the triangle inequality for the usual metric on the reals. You're still working with the usual metric on R, so try to use the same triangle inequality you just wrote, except with each variable replaced by its cube.
     
  6. Sep 16, 2010 #5
    I don't see another case other than the 3 in my first post assuming that x[tex]\leq[/tex]y
     
  7. Sep 16, 2010 #6

    Dick

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    There's no cases necessary. snipez90 already told you how to do it directly. This is a LOT easier than you think it is.
     
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