Showing $d$ is a metric

Question

The metric $d: \mathbb{R}^2 \times \mathbb{R}^2 \rightarrow \mathbb{R}$ defined by

$$d(x, y) = \left\{ \begin{array}{lr} ||x||+||y|| & \mbox{if } x \neq y \\ 0 & \mbox{if } x = y \\ \end{array} \right.$$ is a variation of the SNCF metric. Let $x = (a_1, b_1)$ and $y = (a_2, b_2)$. Prove that $d$ is a metric.

My attempt:

Axiom 1.

Suppose $d(x, y) = 0$. Then $||x|| + ||y|| = \sqrt{a_1^2 + b_1^2} + \sqrt{a_2^2 + b_2^2} = 0$.

This is a sum of roots so $a_1^2 + b_1^2 = 0$ and $a_2^2 + b_2^2 = 0$. The sum of squares can only be $0$ if the terms are $0$. So $a_1 = 0$, $b_1 = 0$ and $a_2 = 0$, $b_2 = 0$. Therefore $(a_1, b_1) = (a_2, b_2)$.

Going the other way, suppose $x = y$. From the very definition of the metric $d(x, y) = 0$.

Axiom 2.

Notice that

$$d(y, x) = \left\{ \begin{array}{lr} ||y||+||x|| & \mbox{if } y \neq x \\ 0 & \mbox{if } y = x \\ \end{array} \right.$$

You can see that in the top piece of the proposed metric, $||x|| + ||y|| = ||y|| + ||x||$ by commutativity of addition. On the other hand the bottom piece is $0$ which is present in both cases. Hence $d(x, y) = d(y, x)$.

Axiom 3.

Case 1. $x \neq y \neq z$ and $x \neq z$.

$||x|| \leq ||x|| + ||z||$ and $||y|| \leq ||y|| + ||z||$

Adding these: $||x|| + ||y|| \leq ||x|| + ||z|| + ||z|| + ||y|| \implies d(x, y) \leq d(x, z) + d(z, y)$

Case 2. $x \neq y$ and $y = z$.

Since $y = z$ we know $d(z, y) = 0$ and $||y|| = ||z||$.

$||x|| + ||y|| = ||x|| + ||z|| + 0 \implies d(x, y) = d(x, z) + d(z, y) \implies d(x, y) \leq d(x, z) + d(z, y)$

Case 3. $x \neq y$, $y \neq z$ and $x = z$.

Since $x = z$ we know $d(x, z) = 0$ and $||x|| = ||z||$.

$||x|| + ||y|| = 0 + ||z|| + ||y|| \implies d(x, y) = d(x, z) + d(z, y) \implies d(x, y) \leq d(x, z) + d(z, y)$

Case 4. $x = y$.

Since $x = y$ we know $d(x, y) = 0$ and $||x|| = ||y||$.

If $y = z$ then $x = y = z$ and so $d(x, y) = d(x, z) = d(y, z) = 0$ $\implies d(x, y) \leq d(x, z) + d(z, y)$.

If $y \neq z$ and $x = z$ then $d(x, y) = d(x, z) = 0$. But $d(x, y) = 0 \leq 0 + ||y|| + ||z|| = d(x, z) + d(y, z)$

If $y \neq z$ and $x \neq z$ then $||x|| + ||z|| + ||z|| + ||y|| > 0 \implies d(x, z) + d(z, y) \geq d(x, y)$.

So in any case the triangle inequality holds.

QED.

Is this proof correct? Did I go through all the cases for the triangle inequality or did I miss something there? Also is there a simpler way to tackle the triangle inequality axiom or is it necessary to do it in cases as I have done. Any assistance is much appreciated.

Answer 1

Your argument for Axiom $1$ isn’t right, because $d(x,y)=0$ does not imply that $\|x\|+\|y\|=0$. What is true is that either $x=y$, which is fine, or $\|x\|+\|y\|=0$. And since $\|z\|\ge 0$ for all $z\in\Bbb R^2$, $\|x\|+\|y\|=0$ if and only if $\|x\|=\|y\|=0$, which in turn is true if and only if $x=\langle 0,0\rangle=y$. Thus, in all cases we find that $x=y$.

It’s actually a little easier to prove the contrapositive: if $x\ne y$, then $d(x,y)\ne 0$. If $x\ne y$, then without loss of generality we may assume that $x\ne\langle 0,0\rangle$, in which case $\|x\|+\|y\|\ge\|x\|>0$.

For Axiom $3$ you’ve covered the cases, but the argument could safely be shortened considerably:

If $z=x$, then $d(x,y)=d(z,y)=d(x,z)+d(z,y)$. Similarly, if $z=y$, then $d(x,y)=d(x,z)=d(x,z)+d(z,y)$. And if $x\ne z\ne y$, then

$$\begin{align*}d(x,z)+d(z,y)&=\|x\|+\|z\|+\|z\|+\|y\|\\ &\ge\|x\|+\|y\|\\ &\color{red}\ge d(x,y) \end{align*}$$

since $\|z\|\ge 0$. (Note that the red inequality is true whether or not $x=y$, since $\|x\|+\|y\|$ is always non-negative.)

It would be a good exercise to try to figure out what topology on $\Bbb R^2$ is generated by this metric; I’ve left a brief description in the spoiler-protected box.

If $x\ne\langle 0,0\rangle$, then $B(x,\|x\|)=\{x\}$, so $\{x\}$ is open for every $x\in\Bbb R^2\setminus\{\langle 0,0\rangle\}$. On the other hand, for each $r>0$ we have $$B(\langle 0,0\rangle,r)=\left\{x\in\Bbb R^2:\|x\|<r\right\}\,,$$ the usual open ball of radius $r$ centred at the origin. Thus, the topology is discrete everywhere except at the origin, which has its usual nbhds.