Exercise 1.4

A different metric on $$\mathbb{R}^2$$ is given by $$d((x_1,y_1),(x_2,y_2)) = \max\{|x_1-x_2|,|y_1-y_2|\}$$. Check that $$d$$ is a metric. This is sometimes called the square metric.

Olomana
We verify property 1 by inspection and property 2 by symmetry.

For property 3, we look at the projection of our 3 points on the X-axis or Y-axis and use 1.3 to bootstrap our argument.

Suppose $$|x_1 - x_3| \ge |y_1 - y_3|$$. By 1.3, on the X-axis:

$$|x_1 - x_2| + |x_2 - x_3| \ge |x_1 - x_3|$$

We can modify the right-hand side without making it any larger:

$$|x_1 - x_2| + |x_2 - x_3| \ge \max\{|x_1 - x_3|, |y_1 - y_3|\}$$

and modify the left-hand side without making it any smaller:

$$\max\{|x_1 - x_2|, |y_1 - y_2|\} + \max\{|x_2 - x_3|, |y_2 - y_3| \ge \max\{|x_1 - x_3|, |y_1 - y_3|\}$$

The inequality still holds, giving us property 3. If $$|x_1 - x_3| < |y_1 - y_3|$$, we use 1.3 and start with the Y-coordinates.

'''I like it. --Steven.clontz 07:28, March 2, 2011 (UTC)'''