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Property of a metric space

  1. Aug 30, 2009 #1
    I want to show the triangle inequality, d(x,x)=0, d(x,y)[tex]\neq[/tex]0 for x[tex]\neq[/tex]y
    implies that d(x,y)=d(y,x).

    Note that I do not have d(x,y)>0. But I know how to show this if I can get the transitive property.

    I have been trying to use the triangle ineq. to establish
    d(x,y)>=d(y,x) and d(x,y)<=d(y,x) but I cant get it to fall out.

    I don't have to many choices of things to do here but I still can not make the inequality work. Any hints?
     
  2. jcsd
  3. Aug 30, 2009 #2
    d(x,y) < d(x,z) + d(z,y) < d(x,y) + d(y,z) + d(z,y)
     
  4. Aug 30, 2009 #3
    To the OP: Are you sure you're not supposed to deduce all the properties of a metric space from these two axioms, 1. d(x,y) = 0 iff x = y and 2. [tex]d(x,y) \leq d(z,x) + d(z,y) ?[/tex]

    The only difference between your assumptions and the assumptions I listed is that my inequality is not exactly the triangle inequality, though it would be if we could presuppose the symmetry of the argument (which is what you are trying to prove); that or I am completely missing something in VeeEight's post.
     
  5. Aug 31, 2009 #4
    snipez90 you're right! I was using the wrong assumption... Unfortunately I am still having trouble figuring this out.

    The assumption I should have been using was:
    d(x, x) = 0, d(x, z) , 0 for x , z, and
    d(x, z) [tex]\leq[/tex] d(z,w) + d(w, x)
     
  6. Aug 31, 2009 #5
    I also have trouble seeing how that inequality will allow you to reach the desired conclusion. Note that in the inequality I wrote down, non-negativity is almost immediate since you know which side you want to be 0 and replacing the appropriate variable(s) establishes this.
     
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