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}<br /> |x_2| + |y_2| + |x_1 - y_1| &amp; \text{if } x_1 \neq y_1 \\<br /> |x_2 - y_2| &amp; \text{if } x_1 = y_1<br /> \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}<br /> |x_2| + |y_2| + |x_1 - y_1| &amp; \text{if } x_1 \neq y_1 \\<br /> |x_2 - y_2| &amp; \text{if } x_1 = y_1<br /> \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!