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I'm trying to show that the metric $(\mathbb{R}^2,d_2)$ is indeed a metric. Here $$d_2((x_1,y_1),(x_2,y_2))=\max\{\lvert x_1-x_2\rvert, \lvert y_1-y_2\rvert\}.$$

I've got everything up to the triangle inequality, and I've got $$\max\{\lvert x_1-x_2\rvert , \lvert y_1-y_2\rvert\}\leq \max\{\lvert x_1-x_3\rvert , \lvert y_1-y_3\rvert\}+\max\{\lvert x_2-x_3\rvert , \lvert y_2-y_3\rvert\}$$ written down, but I'm not sure where to go from here.

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  • $\begingroup$ Show that $\lvert x_1-x_2\rvert \leqslant d_2((x_1,y_1),(x_3,y_3)) + d_2((x_3,y_3),(x_2,y_2))$ and the same for the $y$ coordinates. $\endgroup$ – Daniel Fischer Sep 3 '13 at 21:04
  • $\begingroup$ It's not immediately obvious to me... $\endgroup$ – Trancot Sep 3 '13 at 21:09
  • $\begingroup$ Maybe this becomes easier by generalizing. Assume you have some set $X$ with a metric $d$ and some set $Y$ with a metric $d'$. Now build a metric $d_2$ on $X \times Y$ by $d_2((x_1,y_1),(x_2,y_2)) = \max(d(x_1,x_2),d'(y_1,y_2))$. $\endgroup$ – Magdiragdag Sep 3 '13 at 21:13
  • $\begingroup$ @Trancot, $|x_1-x_2| \le |x_1-x_3| + |x_2-x_3| \le \dots$ $\endgroup$ – njguliyev Sep 3 '13 at 21:13
  • $\begingroup$ You know the triangle equality $|a+b|\le |a| + |b|$ and you also know that $a\le \max(a,b)$ and that $a\le c \land b\le c\implies max(a,b)\le c$ $\endgroup$ – xavierm02 Sep 3 '13 at 21:13
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HINT: You know the triangle inequality for the usual metric on $\Bbb R$:

$$|x_1-x_2|\le|x_1-x_3|+|x_3-x_2|\;.$$

Apply that separately to the $x$-coordinates and the $y$-coordinates of $\langle x_1,y_1\rangle,\langle x_2,y_2\rangle$, and $\langle x_3,y_3\rangle$. Then use the fact that if $a\le b$ and $c\le d$, then $\max\{a,c\}\le\max\{b,d\}$. (You should of course justify that fact.)

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