- #1

tylerc1991

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## Homework Statement

Prove the triangle inequality for the following metric [itex]d[/itex]

[itex] d\big((x_1, x_2), (y_1, y_2)\big) = \begin{cases}

|x_2| + |y_2| + |x_1 - y_1| & \text{if } x_1 \neq y_1 \\

|x_2 - y_2| & \text{if } x_1 = y_1

\end{cases}, [/itex]

where [itex]x_1, x_2, y_1, y_2 \in \mathbb{R}.[/itex]

## Homework Equations

We may assume the triangle inequality for real numbers. That is, [itex]|x + y| \leq |x| + |y|[/itex] for [itex]x, y \in \mathbb{R}[/itex]

[itex]d[/itex]

## The Attempt at a Solution

We wish to show that

[itex]d\big( (x_1, x_2), (z_1, z_2) \big) \leq d\big( (x_1, x_2), (y_1, y_2) \big) + d\big( (y_1, y_2), (z_1, z_2) \big)[/itex]

We may write

[itex]d\big( (x_1, x_2), (z_1, z_2) \big) \leq |x_2| + |z_2| + |x_1 - z_1|[/itex]

[itex]d\big( (x_1, x_2), (y_1, y_2) \big) \leq |x_2| + |y_2| + |x_1 - y_1|[/itex]

[itex]d\big( (y_1, y_2), (z_1, z_2) \big) \leq |y_2| + |z_2| + |y_1 - z_1|[/itex],

although I am not sure if this is helpful.

There are essentially two cases to consider: (i) [itex]x_1 = y_1[/itex] and [itex]y_1 = z_1[/itex], and (ii) [itex]x_1 \neq y_1[/itex] or [itex]y_1 \neq z_1[/itex]

I can do the first case, since [itex]|x_2 - z_2| = |x_2 - y_2 + y_2 - z_2| \leq |x_2 - y_2| + |y_2 - z_2|[/itex], which completes this case.

I have problems with the second case though. I feel like there is some trick that I am missing. Or maybe I shouldn't be handling cases at all? Any help would be appreciated!