Triangle Inequality for a Metric

[tylerc1991]

Homework Statement

Prove the triangle inequality for the following metric $d$

$d\big((x_1, x_2), (y_1, y_2)\big) = \begin{cases} |x_2| + |y_2| + |x_1 - y_1| & \text{if } x_1 \neq y_1 \\ |x_2 - y_2| & \text{if } x_1 = y_1 \end{cases},$

where $x_1, x_2, y_1, y_2 \in \mathbb{R}.$

Homework Equations

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

The Attempt at a Solution

We wish to show that

$d\big( (x_1, x_2), (z_1, z_2) \big) \leq d\big( (x_1, x_2), (y_1, y_2) \big) + d\big( (y_1, y_2), (z_1, z_2) \big)$

We may write
$d\big( (x_1, x_2), (z_1, z_2) \big) \leq |x_2| + |z_2| + |x_1 - z_1|$
$d\big( (x_1, x_2), (y_1, y_2) \big) \leq |x_2| + |y_2| + |x_1 - y_1|$
$d\big( (y_1, y_2), (z_1, z_2) \big) \leq |y_2| + |z_2| + |y_1 - z_1|$,
although I am not sure if this is helpful.

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

I can do the first case, since $|x_2 - z_2| = |x_2 - y_2 + y_2 - z_2| \leq |x_2 - y_2| + |y_2 - z_2|$, 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!

 

[sunjin09]

Homework Statement

Prove the triangle inequality for the following metric $d$

$d\big((x_1, x_2), (y_1, y_2)\big) = \begin{cases} |x_2| + |y_2| + |x_1 - y_1| & \text{if } x_1 \neq y_1 \\ |x_2 - y_2| & \text{if } x_1 = y_1 \end{cases},$

where $x_1, x_2, y_1, y_2 \in \mathbb{R}.$

Homework Equations

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

The Attempt at a Solution

We wish to show that

$d\big( (x_1, x_2), (z_1, z_2) \big) \leq d\big( (x_1, x_2), (y_1, y_2) \big) + d\big( (y_1, y_2), (z_1, z_2) \big)$

We may write
$d\big( (x_1, x_2), (z_1, z_2) \big) \leq |x_2| + |z_2| + |x_1 - z_1|$
$d\big( (x_1, x_2), (y_1, y_2) \big) \leq |x_2| + |y_2| + |x_1 - y_1|$
$d\big( (y_1, y_2), (z_1, z_2) \big) \leq |y_2| + |z_2| + |y_1 - z_1|$,
although I am not sure if this is helpful.

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

I can do the first case, since $|x_2 - z_2| = |x_2 - y_2 + y_2 - z_2| \leq |x_2 - y_2| + |y_2 - z_2|$, 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!

For your case (ii) there are 3 cases
1. x1≠y1≠z1, where
d1+d2=|x1-y1|+|y1-z1|+|x2|+|y2|+|y2|+|z2|
≥|x1-y1+y1-z1|+|x2|+0+0+|z2|
=d
2. x1=y1≠z1, where
d1+d2=|x2-y2|+|y1-z1|+|y2|+|z2|
=|x2-y2|+|y2|+|x1-z1|+|z2|
≥|x2-y2+y2|+|x1-z1|+|z2|
=|x2|+|x1-z1|+|z2|
=d
3. x1≠y1=z1, same as 2

 

[tylerc1991]

For your case (ii) there are 3 cases
1. x1≠y1≠z1, where
d1+d2=|x1-y1|+|y1-z1|+|x2|+|y2|+|y2|+|z2|
≥|x1-y1+y1-z1|+|x2|+0+0+|z2|
=d
2. x1=y1≠z1, where
d1+d2=|x2-y2|+|y1-z1|+|y2|+|z2|
=|x2-y2|+|y2|+|x1-z1|+|z2|
≥|x2-y2+y2|+|x1-z1|+|z2|
=|x2|+|x1-z1|+|z2|
=d
3. x1≠y1=z1, same as 2

That makes much more sense. Thank you so much!