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The 3 conditions for a metric space $(X,d)$ are that for all $x,y,z\in X, $ $$d(x,y)=d(y,x)$$ $$d(x,y)\geq0,d(x,y)=0 \iff x=y$$ $$d(x,y)+d(y,z)\geq d(x,z)$$ Are there any interesting results in altering the third condition so that we get$$d(x,y)+d(y,z)= d(x,z)$$ where the elements $x,y,z$ act in the way that vectors do, in so much as we get $xy→+yz→=xz→ $, we get in this "metric" that $d(x,y)+d(y,z)=d(x,z)$. ?

Thanks to @Arthur for the edit suggestion

W M Seath
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    What does that give us in the case $x=z$? – Arthur May 07 '19 at 18:19
  • What distinguishes vectors from other objects is that we can add them and multiply them by scalars. Are you trying to introduce operations in your metric space? If not, then I don't understand the question. The best you can do is define a closed interval $$I(x,z)={y\in X \mid d(x,y)+d(y,z)=d(x,z)}$$and study the geometry of this. But I don't see how this relates to vectors. – Ivo Terek May 07 '19 at 18:25
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    @IvoTerek I think what he is referring to is that the same way that vectors have $\vec{xy}+\vec{yz}=\vec{xz}$, we get in this "metric" that $d(x,y)+d(y,z)=d(x,z)$. – Arthur May 07 '19 at 18:29
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    Mathematicians make up sets of axioms in order to formalize intuitive concents (like distance) so that they can prove theorems. They don't usually arbitrarily change one axiom just to see what happens. The motivation for change comes from new situations they want to understand, not the other way around. There is one strengthening change to that metric space axiom that does cover some interesting cases: $d(x,y) = \max(d(x,z), d(z,y))$. – Ethan Bolker May 07 '19 at 18:29
  • @EthanBolker That should be a $\leqslant$ and not an equality. – Math1000 May 07 '19 at 22:49

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