# Proving $\max (d_1, d_2)$ is a metric

Let $d_1 \&\ d_2$ be 2 metrics on a set M. So I am trying to prove that the triangle inequality holds. I did it differently than the proof in my notes, and want to know what is wrong with it.

My Proof: Let $d_3$ be another metric in M. So we want to show $\max\ (d_1, d_2) \le \max (d_1, d_3) + \max(d_3, d_2)$ Without loss of generality, we can assume $d_1$ is the max between $d_1 \&\ d_2$. If $\max(d_1, d_3) = d_3$ then we are done. However, $\max(d_1, d_3) = d_1$ still satisfies, since $\max(d_3, d_2) \ge 0$. $\ \therefore\ , d_1 \le d_1 + \max(d_3, d_2)$ .

Proof provided by professor: $x,y,z \in M$. Without loss of generality, $d(x,y) = d_1(x,y)$ I am unsure of why this okay to say, are we not just defining the new metric to be equal $d_1$, which is a specific case? Then we have $d(x,y) = d_1 (x,y) \le d_1(x,z) +d_1(z,y) \le d(x,z) + d(z,y)$. Not sure why it is less than, I thought $d = d_1$.

It is incredibly likely that I copied down the proof wrong since I was behind in notetaking, so if someone could intimate me as to how the proof above should continue (I wish to solve it myself) and whether my attempt was in vain or not, would be much appreciated.

## 1 Answer

You have not been asked to prove that $\max:\{\text{metrics on }X\}\times\{\text{metrics on }X\}\to\Bbb R^X$ is a metric. In point of fact, such a statement doesn't even make sense. You've been asked to prove that, for all $d_1$ and $d_2$ metrics on $X$, the map $X\times X\to\Bbb R$, $(x,y)\mapsto\max\{d_1(x,y),d_2(x,y)\}$ is a metric on $X$.

• Oh... okay. I see. That makes a lot more sense. Thank you. – zodross Aug 29 '18 at 23:07