Proving triangle inequality for: $(X,d)$ is a metric space. Let $d′(x,y) = \min \{1, d(x,y)\}$.Prove that $(X,d′)$ is also a metric space.

Question

$(X,d)$ is a metric space. Let $d′(x,y) = \min \{1, d(x,y)\}$.Prove that $(X,d′)$ is also a metric space.
Could somebody please explain how to prove the triangle inequality here? Showing symmetry and non-negativity is straightforward, but I don't understand how to prove the triangle inequality. I don't understand my tutor's notes…

Answer 1

It is easy to do this case-wise:

Start with $d'(x,z) + d'(z,y) = min(1,d(x,z))+min(1,d(z,y))$

1. Now suppose both $d(x,z)<1$ and $d(z,y)<1$ so we take $d'(x,z) + d'(z,y) = d(x,z) + d(z,y) \geq d(x,y) \geq min(1,d(x,y))$

2. If $d(x,z)<1$ and $d(z,y)>1$, then $d'(x,z) + d'(z,y) = d(x,z) + 1 \geq 1 \geq min(1,d(x,y))$

3. $d(x,z)>1$ and $d(z,y)>1$, then $d'(x,z) + d'(z,y) = 1 + 1 \geq 1 \geq min(1,d(x,y))$