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Which of the following define a metric?

a. $d((x, y), (x’, y’)) = \min\{|x – x’|, |y – y’|\}$ on $\mathbb{R}^2$.
b. $d((x, y), (x’, y’)) = |x| + |y| + |x’| + |y’|$ on $\mathbb{R}^2$..
c. $D((x, y), (x’, y’)) = d(x, x’) + d(y, y’)$ on $X \times X$, where $(X, d)$ is a metric space.

I have found that (a) is not true, but not sure about the others.

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Usually the easiest thing to check when considering possible metrics is that $d(x, x) = 0$. The next easiest is that $d(x, y) = d(y, x)$ and maybe non-negativity. It is often trickier to determine whether $d(x, y) = 0 \implies x=y$ and the triangle inequality are true. What have you determined about the proposed metric in b? – Michael Albanese Jan 9 '13 at 7:13
Try $(x,y)=(1,1),(x',y')=(1,1)$ for b). Try proving c) from the axioms. – copper.hat Jan 9 '13 at 7:13
-1: As pointed out in the answer below the only thing you have to do is to go through a list of axioms and try to verify them. Surely a little bit of thinking results in more than "I have found that (a) is not true, but not sure about the others." Also see: – Martin Jan 9 '13 at 7:18

To show that a distance formula is a metric, you need to show that

1) $d(x, y) \geq 0$ (non-negativity)
2) $d(x, y) = 0 \iff x=y$
3) $d(x, y) = d(y, x)$ (symmetry)
4) $d(x, z) \leq d(x, y) + d(y, z)$ (triangle inequality)

Properties 1, 2, 3 are often easily established or checked.

Conversely, to show that a distance formula is not a metric, you need to show that one of these properties does not hold. Likewise, it is often property 4 that tricky to violate.

On to your question:

a) This is not a metric, which you stated. Which property does it violate?

b) This is not a metric. Which property does it violate?

c) This is known as the taxicab metric, or Manhattan metric, when $X =\mathbb{R}$. Prove that it is a metric.

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Also known as the Manhattan length. – copper.hat Jan 9 '13 at 7:20
@copper.hat: I face the taxicab metric while I teach Geometry. Why is it really called like that? Taxi?? Thanks. – Babak S. Jan 9 '13 at 7:45
@BabakSorouh: – Martin Jan 9 '13 at 7:55
@Babak Because that is how taxis in Manhattan calculate distance. The streets of Manhattan are laid out in a (almost) perfect grid, and so they will say "I'm X blocks away", where X is the sum of the horizontal and vertical distance. It doesn't really matter what path they take to get there, the distance travelled is (almost) the same. – Calvin Lin Jan 9 '13 at 7:55
@Martin: Thanks. Thanks Calvin. Nice illustration. – Babak S. Jan 9 '13 at 7:57

For (b) try seeing what happens when $(x,y)=(x',y')$.

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