Show that the following is a valid metric on a metric space $(M,d)$

Question

Let $M$ be any set and $d$ a real-valued, non-negative function on $M\times M$ such that for all $x,y,z \in M$:

$d(x,y)=0$ iff $x=y$ (I)

$d(x,y)\leq\max\{d(x,z),d(y,z)\}$ (II)

So I get that for this one has to use the axioms.

i) $d(x,y)\geq0$: this follows directly from definition written above and from (I)

ii) $d(x,y)\leq d(x,z)+d(y,z)$: this follows from (II) because since $d(x,y)$ is less than the maximum of that set and $d(a,b)\geq0$ for any $a,b\in M$ then it follows that if $d(x,y)$ is less than the maximum of two positive values, then it is less than their sum

iii) $d(x,y)=d(y,x)$: this is where I am stuck. I do not know how to proceed with this because it seems that there is not enough information to deduce this (I am most probably missing some key point here). The only thing I can think of is maybe justifying it with the fact that it says "for all $x,y,z \in M$" but that seems like an easy-way-out.

Answer 1

Put $z=x$ in (II) and use (I) to get

$$d(x,y)\le d(y,x).$$

Interchange $x,y$ gives the opposite inequality.

Answer 2

In order to show that $$d(x,y)=d(y,x)$$ Let $x=z$ in $$d(x,y)\leq\max\{d(x,z),d(y,z)\}$$ to get $$d(x,y)\le d(y,x)$$

Now swap $x$ and $y$ to get the other side of inequality and you are done.