I feel rather embarrassed and ashamed asking this question , since it is so obvious.
Anyway here I go...
I am trying to proof rigorously that the Discrete Metric , which is defined as: $$ d_D(x,y) = \begin{cases} 0 & x =y \\ 1 & x\neq y \end{cases}$$ is a metric on any set $X$
In particular show that it satisfies the following condition:
$$d(x ,y) = 0 \iff x =y $$
My Attempt:
If $x =y$ then by definition of $d_D(x,y) =0$
Now suppose that $d_D(x,y) =0$ but $x\neq y$. Then by definition $d_D(x,y) =1$.
We have a contradiction so there if $d_D(x ,y) =0$ then $x=y$
Is this rigorous enough?