$d(x,y) = \min\{|x_1−y_1|+|x_2−y_2|, 1−|x_1−y_1|+|x_2−(1−y_2)|\}$ defines a metric on $[0,1)\times[0,1]$? [closed]

Question

For $x,y \in [0,1)\times[0,1]$, let $d(x,y)$ be the minimum of $|x_1−y_1|+|x_2−y_2|$ and $1−|x_1−y_1|+|x_2−(1−y_2)|$. Prove or disprove that $d$ is a metric.

I was unable to find a counterexample to disprove and was able to show that

1. $d(x,y) = 0 \iff x = y$,
2. $d(x,y) = d(y,x)$,
3. $d(x,z) \leq d(x,y) + d(y,z)$ when either $(d(x,z) = |x_1−z_1|+|x_2−z_2|$ and $d(x,y) = |x_1−y_1|+|x_2−y_2|$ and $d(y,z) = |y_1−z_1|+|y_2−z_2|)$ or $(d(x,z) = 1−|x_1−z_1|+|x_2−(1−z_2)|$ and $d(x,y) = 1−|x_1−y_1|+|x_2−(1−y_2)|$ and $d(y,z) = 1−|y_1−z_1|+|y_2−(1−z_2)|)$.

How can I proceed for triangular inequality when there are mixed distances, for instance, if $d(x,z) = |x_1−z_1|+|x_2−z_2|$ and $d(x,y) = 1−|x_1−y_1|+|x_2−(1−y_2)|$ and $d(y,z) = |y_1−z_1|+|y_2−z_2|$?

I asked it on Mathematics Stack Exchange a couple of days ago, but haven't got any response yet.

Answer 1

It is a metric. Denote $a(x,y):=|x_1−y_1|+|x_2−y_2|$ (a usual $\ell^1$-type metric on the plane), $b(x,y):=1-|x_1-y_1|+|x_2+y_2-1|$. To prove that $d:=\min(a,b)$ is a metric, it suffices to check that $d$ is symmetric (obvious), non-negative (obvious), strictly positive for $x\ne y$ (holds since $b>0$ always due to $x_1,y_1\in [0,1)$), and a triangle inequality. We have:

1. $a(x,y)+a(y,z)\geqslant a(x,z)\geqslant d(x,z)$ since $a$ is a metric;

2. \begin{align*}b(x,y)+b(y,z)&=(2-|x_1-y_1|-|y_1-z_1|)+(|x_2+y_2-1|+|y_2+z_2-1|)\\&\geqslant (2\max(x_1,y_1,z_1)-2\min(x_1,y_1,z_1)-|x_1-y_1|-|y_1-z_1|))+|x_2-z_2|\\&=|x_1-z_1|+|x_2-z_2|=a(x,z)\geqslant d(x,z);\end{align*}

3. $$a(x,y)+b(y,z)=(|x_1-y_1|+1-|y_1-z_1|)+(|y_2+z_2-1|+|x_2-y_2|)\\ \geqslant 1-|x_1-z_1|+|x_2+z_1-1|=b(x,z)\geqslant d(x,z).$$

Thus, $$d(x,y)+d(y,z)\\=\min(a(x,y)+a(y,z),b(x,y)+b(y,z),a(x,y)+b(y,z),a(y,z)+b(x,y))\\\geqslant d(x,z)$$