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Consider a metric on V such that $$i)\quad d(x,y)=0\iff x=y,$$ $$ii)\quad d(x,y)=d(y,x)\forall x,y\in V,$$ $$iii)\quad d(x,y)+d(y,z)\geq d(x,z).$$ So we do not take $d(x,y)\geq 0$ as an axiom. With this we will try to show that this set of axioms is inconsistent i.e. there exists a statement for which both itself and its denial can be derived from the axioms.

Consider $d(a,b)<0 $. Then by axiom $i$ we have that $$d(a,a)=0.$$ Now axiom $iii$ implies that $$d(a,b)+d(b,a)\geq 0\iff d(a,b)\geq -d(b,a)\implies d(a,b)>0.$$ This contradicts our initial assumption that $d(a,b)<0.$ Now is it correct to be stated that without the positivity axiom these statements are inconsistent or can the four axioms be considered redundand i.e. that positivity can be implied by the other 3 axioms?

I. Violaris
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    The correct conclusion is that including the positivity axiom is redundant and that the other axioms imply it. Not having an axiom is not the same as assuming that the axiom is false. – Dylan Nov 19 '15 at 22:23
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    First of all, axiom i does not imply $d(a,a) = 0$, the implication only goes in the other direction. Second, your axioms can not be inconsistent because they are a subset of the usual axioms for a metric, hence any metric is a model for your axioms (and thus the axioms are consistent). Third, if you add the other direction in axiom i, positivity is indeed implied and hence redundant. – Eike Schulte Nov 19 '15 at 22:25
  • Yes thanks about the answers. The one way implication is a slip I will correct it. What confused me mainly is the fact that all the sources I happened to look at have all 4 axioms (or positivity axiom included as a part of axiom i.) This confused me a little exactly because as you say it is implied. Do you by any chance know of a reliable source that does not include the positivity as an axiom? – I. Violaris Nov 19 '15 at 22:35
  • The following is a tautology: $( p \to \neg p)\to \neg p$. So if you assume $p$ (in this case $d(a,b) < 0$) and can prove $\neg p$, then $\neg p$ is true. So you've shown that not $d(a,b)<0$, which by the way is $d(a,b)\ge 0$ not $>$. As $a,b$ were arbitrary, you have deduced the nonnegativity axiom. – BrianO Nov 19 '15 at 22:37
  • No according to the initial assumption it should be d(a,b)>0 since the assumption was d(a,b)<0. What you say would be true if I assumed $d(a,b)\leq 0.$ It can be proven very easily though that $2d(a,b)\geq0$ without following this approach and yes $a,b$ are arbitrary obviously. – I. Violaris Nov 19 '15 at 22:40
  • And yes $\neg p$ is $d(a,b)\geq 0$ it is just not stated but this is what we want to show. – I. Violaris Nov 19 '15 at 22:49
  • I've never thought about this. In my experience there have been only 3 axioms and d(a,b) >= 0 is part of the definition rather than an axiom. I suppose it is redundant but there really isn't any point in hindering instruction showing that as d(a,b) >=0 is really something to be defined; not devined. – fleablood Nov 19 '15 at 23:01
  • Yes what you say makes perfect sense. That is most probably the reason. – I. Violaris Nov 19 '15 at 23:05

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