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Is there a metric like space $(X,d)$ where $d$ only satisfies triangle inequality, and not the first two conditions of a metric? That is, $d:X\times X\longrightarrow [0,\infty)$ only satisfies: $$d(x,z)\leq d(x,y)+d(y,z)$$

Amer Bhat
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    Welcome to MSE. Please read this text about how to ask a good question. – José Carlos Santos Sep 05 '22 at 13:19
  • Related: there is a minimalistic formulation of being a metric, which is that a function $d:X\times X\to\Bbb R$ is a metric if and only if it satisfies: 1) $\forall x,\forall y, (d(x,y)=0\leftrightarrow x=y)$; 2) $\forall x,\forall y,\forall z, d(x,z)\le d(y,x)+d(y,z)$. https://math.stackexchange.com/questions/2000046/metric-space-axioms – Sassatelli Giulio Sep 05 '22 at 13:32

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Consider $X=\mathbb R$ and $$d(x,y)=|x|+1$$

This definition satisfies the triangle inequality but is not a metric. It is neither symmetric nor does it satisfy $d(x,x)=0$.

Andreas Tsevas
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Presumably the first two conditions here are $d(x,y)=d(y,x)$ and $d(x,y)=0$ if and only if $x=y$. If so, consider $$d:\mathbb{N}\times\mathbb{N}\to\mathbb{R}^{+},\quad (x,y)\mapsto \begin{cases}2&\text{if $x\leq y$}\\ 0&\text{if $x>y$}\end{cases}.$$

Andijvie
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