Trivial Metric Space Can someone show me how to prove that the trivial metric space is indeed a metric space (if $a=b$ then $d(a,b)=0$ and if $a \ne b$ then $d(a,b)=1$)? 
I'm having trouble with the triangle inequality property.
 A: Hint: You want to prove that
$$d(a,c)\leq d(a,b)+d(b,c)$$
for all points $a,b,c$. For any given points $a,b,c$, just work through the possibilities:


*

*$a=b$ and $b=c$

*$a=b$ and $b\neq c$

*$a\neq b$ and $b=c$

*$a\neq b$ and $b\neq c$


What happens in each case?
A: I trust you've covered the first two properties of the metric, which follow immediately from the definition of the trivial metric space:


*

*$d(a, b) \geq 0$ for all $a, b$ 

*$d(a, b) = 0 \iff a = b$ (by definition)


Now, for proving triangle inequality holds:
What are the possible cases for $d(a, b), d(b, c), d(a, c)$?
Either:


*

*$a = b, \; b = c,\;\implies a = c$; 

*$a = b,\; b\neq c,\; \implies\; a \neq c$;

*$a \neq b,\;b = c,\; \implies a \neq c$;

*$a \neq b,\;b \neq c,\; a = c;\;\;$ or

*$a \neq b,\; b\neq c,\;a \neq c$.


Show that whatever the pairwise relationship between $a, b, c,\;$ it follows that:
$$\large d(a, c) \leq d(a, b) + d(b,c)\tag{triangle inequality}$$
I.e., To prove the triangle inequality holds, prove that it holds for in each of the above cases.
