# Metric on the unit cube

Let $X$ be $\mathbb{R}^3$ with the sup norm $\|\cdot\|_{\infty}$. Let $Y=\{x\in X: \|x\|_{\infty}=1\}$. For $x,y\in Y,y\neq -x$ define $d(x,y)$ to be the arc length of the path $$Y\cap \{\lambda x+\mu y: \lambda\ge 0, \mu\ge 0\}.$$ Define $d(x,-x)=4$. Note that the arc lengths are computed using the sup norm. My question is: Does $d$ define a metric on $Y$? A related question was answered in Shortest path on unit sphere under $\|\cdot\|_\infty$

-
In fact the answer can be found in the given link. Let $A=(1,3/4,1/4), B=(3/4,1,3/4), C=(1,1,1/2)$ and $M=(1,1,4/7)$. Then $d(A,C)=1/4, d(C,B)=1/4$ and $d(A,B)=d(A,M)+d(M,B)=4/7$. So $d(A,C)+d(C,B)<d(A,B)$, proving that $d$ is not a metric. – TCL Feb 11 '12 at 5:00
TCL: Why not post an answer? – Jonas Meyer Feb 11 '12 at 5:58
@TCL I'm not sure what the relevance is more than a year later, but it's better to ask a new question instead of editing in this case -- your edit made this question wholly different from its original. – Lord_Farin Nov 11 '13 at 12:54

Let: $$A=(1,3/4,1/4),\ B=(3/4,1,3/4),\ C=(1,1,1/2)\text{ and }M=(1,1,4/7).$$ Then: $$d(A,C)=1/4,\ d(C,B)=1/4\text{ and }d(A,B)=d(A,M)+d(M,B)=4/7.$$ So $d(A,C)+d(C,B)<d(A,B)$, proving that $d$ is not a metric.