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Consider the set of real numbers $\mathbb{R^2}$, and define the function $d: \mathbb{R^2} \times \mathbb{R^2} \to [0,\infty)$ as $d((u_1,u_2),(v_1,v_2):=max\{|u_1-v_1|,|u_2-v_2|\}$.

M1) $\forall (x,y) \in \mathbb{R^2}, d((x,y),(x,y))=max\{|x-x|,|y-y|\}=max\{|0|,|0|\}=0$.

M2) Let $(a_1,a_2),(b_1,b_2) \in \mathbb{R^2} \ni d((a_1,a_2),(b_1,b_2))=0$. Then $max\{|a_1-b_1|,|a_2-b_2|\}=0$.

Because $\forall z \in \mathbb{R^2}, |z| \ge 0$, then $max\{|a_1-b_1|,|a_2-b_2|\}=0 \iff |a_1-b_1|=|a_2-b_2|=0$.

This means that $a_1-b_1=0$ and $a_2-b_2=0$. Thus $a_1=b_1 \wedge a_2=b_2$, and hence $(a_1,a_2)=(b_1,b_2)$.

M3) Consider $d((a_1,a_2),(b_1,b_2))=max\{|a_1-b_1|,|a_2-b_2|\}$.

Note that $|a_1-b_1|=|-(a_1-b_1)|$ and $|a_2-b_2|=|-(a_2-b_2)|$. This means that $max\{|a_1-b_1|,|a_2-b_2|\}=max\{|b_1-a_1|,|b_2-a_2|\}$.

Thus $d((a_1,a_2),(b_1,b_2))=d((b_1,b_2),(a_1,a_2))$

I'm stuck on M4 (i.e. the triangle inequality). Right now my approach is to somehow use the following two lemmas:

  1. If $b \in \mathbb{R}$ s.t. $b \ge 0$, then $|a| \le b \iff -b \le a \le b$

  2. Let $d: X \times X \to [0,\infty)$ be a function which satisfies M1 and M3. Then for all $x,y,z \in X$ M4 holds $\iff \forall$ distinct points $x,y,z \in X$, M4 holds.

Any insight would be appreciated.

Karam
  • 741

1 Answers1

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If $(x_1,x_2),(y_1,y_2),(z_1,z_2)\in\Bbb R^2$, then$$|x_1-z_1|\leqslant|x_1-y_1|+|y_1-z_1|\quad\text{and}\quad|x_2-z_2|\leqslant|x_2-y_2|+|y_2-z_2|.$$Now, $\max\{|x_1-z_1|,|x_2-z_2|\}$ is one of the numbers $|x_1-z_1|$ and $|x_2-z_2|$. If it is $|x_1-z_1|$, then\begin{align}\max\{|x_1-z_1|,|x_2-z_2|\}&=|x_1-z_1|\\&\leqslant|x_1-y_1|+|y_1-z_1|\\&\leqslant\max\{|x_1-y_1|,|x_2-y_2|\}+\max\{|y_1-z_1|,|y_2-z_2|\},\end{align}and the other case is similar.

José Carlos Santos
  • 427,504