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I am trying to prove that given a metric d using only the properties that it $d(a,b)=0 iff a=b$ and $d(a,c)\le d(a,b)+d(b,c)$ that $d(a,b)=d(b,a)$ and $d(a,b) \gt 0$ I understand that it is part of the definition in most texts but it is left as an exercise in mine and I can not figure it out.

I edited my original question. I stated wrong we do not have that $d(a,a)=0$ I have that $d(a,b)=0 iff a=b$

userm92
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  • If you have symmetry, nonnegativity follows from $0=d(x,x) \leq d(x,y) + d(y,x) = 2 d(x,y)$. I don't think that symmetry follows automatically from the remaining axioms. But I don't have a counterexample right now. – PhoemueX Jan 22 '16 at 16:23
  • @PhoemueX That doesn't prove that $d(x,y)\neq 0$? – user160738 Jan 22 '16 at 16:26
  • @user160738: That surely does not follow from the stated axioms. Consider $d \equiv 0$ as an example. So I assumed (without mentioning it) that the OP just meant $d(a,b) \geq 0$. – PhoemueX Jan 22 '16 at 16:27
  • @PhoemueX Yes, that's true. This is a bit weird because had it been the case other axioms followed from the two stated then we would not require them as axioms. I'm not sure what OP actually wants... – user160738 Jan 22 '16 at 16:28
  • I edited the original question I misstated one of the axioms and @phoemuex I have the nonnegativity part if I can prove the symmetry part but I have had absolutely no luck with that – userm92 Jan 22 '16 at 16:43
  • This could be helpful: http://math.stackexchange.com/questions/23390/examples-of-non-symmetric-distances – PhoemueX Jan 22 '16 at 16:54
  • Maybe the OP wants to state the triangle inequality like $d(a,c) \leq d(b,a) + d(b,c)$ or something like this? Then (taking $b=c$) we get $d(a,c) \leq d(c,a)+d(c,c) = d(c,a)$, which easily yields symmetry. – PhoemueX Jan 22 '16 at 16:57
  • @PhoemueX : this definition would lead to the solution, but I feel as if it forgot the intuitive part of the inequality : the shortest path is straight ahead ! ^^ – Balloon Jan 22 '16 at 17:04
  • @phoemuex yes you are correct I apparently can not read. Thank you for your help. – userm92 Jan 22 '16 at 17:10

1 Answers1

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Note that $d:\mathbb{R}\times\mathbb{R}\to \mathbb{R}$ defined as $d(a,b)=0$ is such that :

  • $\forall a\in\mathbb{R}, d(a,a)=0,$

  • $\forall a,b,c\in\mathbb{R},d(a,c)=0\leq d(a,b)+d(a,c)=0$

but you have not that $a\neq b\implies d(a,b)>0,$ so this point should be an axiom.

For the symmetry part, consider $d':\{0,1\}\to\mathbb{R}$ defined as : $$d'(0,0)=0,d'(1,1)=0,d'(1,0)=1 \text{ and } d'(0,1)=2.$$
Now note that :

  • $\forall a\in\{0,1\},d'(a,a)=0,$

  • $\forall a,b,c\in\{0,1\},d'(a,c)\leq d'(a,b)+d'(b,c)$

but $d'(0,1)=2\neq 1=d'(1,0),$ so that point should be an axiom too.

Balloon
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