How to sort out different cases in a proof related to a metric space

Question

Problem:

Let $M = (X, d)$ be a metric space.

Show that $e(x, y) = min(1, d(x, y))$ is a metric.

To prove the triangle inequality for metric $e$, three cases are considered.

Let $x, y, z \in X$.

(A) $d(x, y) \le 1$ and $d(y, z) \le 1$

(B) $d(x, y) > 1$

(C) $d(y, z) > 1$

Why the above three cases are exhaustive?

Is there a systematic method to find the different cases?

Answer 1

This is specific to this exercise: the case (B) $\ d(x,y) >1$ [or (C) $\ d(y, z) >1$] means $e(x, y) =1$ [or $e(y, z) =1$], and if neither holds, that just means $d(x, y)\le1$ and $d(y, z)\le1$, i.e. case (A), when $e(x, y) =d(x, y)$ and $e(y, z) =d(y, z) $.
So these 3 scenarios are indeed exhaustive.

Nevertheless, to provide full details, one also has to consider subcases $d(x, z) \le1$ and $d(x, z) >1 $.

Answer 2

You have a metric space $X$ with a metric $d$, and you are given three points $x,y,z$. You are told to consider $d(x,y)$ and $d(y,z)$.

Your statement simply asserts that there are only certain possibilities. There's no extra trickery here. You either have that

• Both $d(x, y) \le 1$ and $d(y, z) \le 1$,
• Only one of them is less than (or equal to) $1$,
• Neither of them are less than (or equal to) $1$.