Let $X$ be a set and $d: X \times X \to X$ be a function such that $d(a,b)=0$ if and only if $a=b$.
Suppose further that $d(a,b) ≤ d(z,a)+d(z,b)$ for all $a,b,z \in X$.
Show that $d$ is a metric on $X$.
Let $X$ be a set and $d: X \times X \to X$ be a function such that $d(a,b)=0$ if and only if $a=b$.
Suppose further that $d(a,b) ≤ d(z,a)+d(z,b)$ for all $a,b,z \in X$.
Show that $d$ is a metric on $X$.
This must be the discrete metric. 1) The first condition follows by definition that d(a,b)=0 iff a=b; 2) Symmetry: this is trivial, because if a=b you have d(a,b)=0 and b=a gives you d(b,a)=0; 3) The triangle inequality: from the symmetry you can write d(z,a)=d(a,z). Consider two cases: - if a=b, clear - if a is not equal to b, then either a not equal to z or z not equal to b. There you have $1\leqslant 1$ or $1\leqslant 2$.
Let $X$ be a set and $d: X \times X \to X$ be a function such that $d(a,b)=0\text{ if and only if}\;\; a=b,\text{ and}\tag{1}$ $d(a,b) ≤ d(z,a)+d(z,b)\forall a,b,z \in X.\tag{2}$
There's additional criterion that needs to be met for a function $d$ to be a metric on $X$:
You must have that $d(a, b) = d(b,a)$ for all $a, b \in X$ (symmetry).
You can use the two properties you have been given to prove this.
$d(a,b)\leq d(b,a)+d(b,b)= d(b, a) + 0 = d(b,a)$ and vice versa, hence we get equality.
Having proven symmetry, you will then have that
$d(a,b) \leq d(z,a) + d(z, b) \iff d(a, b) \leq d(a, z) + d(z, b)$.
Finally, using the property immediately above, along with the $(1)$, you can establish that for all $a, b\in X$ such that $a\neq b$, we must have $d(a, b) > 0$.
Then you are done.
The first condition of a metric is $d(a,b)\geq 0$ with equality if and only if $a=b$. Obviously that latter portion is satisfied by hypothesis. To show it is greater than zero otherwise, just observe $0=d(b,b)
Next, we want to show $d(a,b)=d(b,a)$. This is clear, though, since $d(a,b)\leq d(b,a)+d(b,b)=d(b,a)$ and vice versa, hence we get equality.
Finally, your last hypothesis is precisely the triangle inequality. Hence, $d$ is a metric.