Is it a metric?

Q11. Basic Topology (Rudin)

For $x \in R^1$ and $y \in R^1$ define

(a) $d_1 (x, y) = (x - y)^2$

Determine whether it is a metric or not.

Solution: $d_1 (x, y) = (x - y)^2$ satisfies the condition that distance function is positive for all x, y and 0 when x = y since square is always non negative.

Also $d_1 (x, y) = d_1 (y, x)$ as $(x - y)^2 = (y-x)^2$

Finally we check the triangular inequality.  We have to show that $(x-y)^2 + (y-z)^2 \ge (x-z)^2$ . If this is true then we will have $2y^2 - 2xy - 2yz  + 2zx \ge  0 \implies (y-x)(y-z) \ge 0$ . But this is not true for all real numbers. For example if we have x < y < z then (y - x)(y - z) becomes negative. Thus x = 2, y = 4 and z = 7 violates the triangular inequality as $(2-4)^2 + (4-7)^2 \le (2-7)^2$.

Hence $d_1$ is not a metric.

(b) $d_2 (x, y) = \sqrt {|x -y| }$

Determine whether it is a metric or not.

Solution:  $d_2 (x, y) = \sqrt {|x -y| }$ satisfies the condition that distance function is positive for all x, y and 0 when x = y since square root of |x-y| is always non negative and equals 0 only when x=y.

Also $d_2 (x, y) = d_2 (y, x)$ as $\sqrt {|x -y|} = \sqrt {|y-x|}$

Finally we check the triangular inequality. We have to show that $\sqrt {|x -y|}  + \sqrt {|y-z|} \ge  \sqrt {|x-z|}$