Determine whether each of the following functions defines a metric on R.

Question

$d(x,y)=|x-y|^2$ this is what I have right now:

i) Since for $x, y \in R$

$|x-y|^2 \geq 0 \rightarrow d(x,y) \geq 0$ Since squaring is always non negative.

ii) $d(x,y) = |x-y|^2$

If $x=y$ then $d(x,x)=|x-x|^2=0^2 \rightarrow d(x,x)=0.$

If $d(x,y)=0 \rightarrow |x-y|^2=0 \rightarrow |x-y|=0 \rightarrow x-y=0 \rightarrow x=y$

So $d(x,y)=0 \leftrightarrow x=y$

iii)

$d(x,y)=|x-y|^2 = |y-x|^2 \rightarrow d(x,y)=d(y,x)$

Now here is where i'm stuck, I took this route, but i was hoping someone could help me do a better one.

iv) For transitivity i took this route

let $x=1, y=2, z=3$. Then:

$d(x,y)=|x-y|^2=|1-2|^2=1.$

$d(y,z)=|y-z|^2=|2-3|^2=1.$

$d(x,z)=|x-z|^2=|1-3|^2=4.$

$4>1+1 \rightarrow 4>2$. Then $d(x,z) > d(x,y)+d(y,z)$

Then the transtivity does not hold since $d(x,z) \leq d(x,y)+d(y,z).$

Thus $d(x,y)=|x-y|^2$ is not metric on R.

Answer 1

You are right: triangle inequality (not “transitivity”) doesn't work and therefore it is not a metric. There is no shorter route, other than the fact that this is enough. That is, there was no need to prove that the other properties hold (but, of course, you could not know that when you started).

Answer 2

It is sufficient to show a counterexample: $$|6-5|^2+|5-4|^2\space\text{ is not greater or equal than }|6-4|^2$$