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Im trying to check if $d_1(x,y)=2|x-y|$ and $d_2(x,y)=|x-y|^2$ are metric spaces.

Im just not sure how to proceed with checking the triangle inequality property $d(x,y)\le d(x,z)+d(z,y)$. Is what I did below sufficient?

$d(x,y) = 2|x-y| =2|x-z+z-y| \le 2|x-z|+2|z-y| = d(x,z)+d(z,y)\\ \text{and} \\ d(x,y)=|x-y|^2=|x-z+z-y|^2\le |x-z|^2+|z-y|^2=d(x,z)+d(z,y)?$

Also is $d(x,y)=|\arctan(x)-\arctan(y)|$a metric space? Can I write

$d(x,y)=|\arctan(x)-\arctan(y)| = |\arctan(x)-\arctan(z)+\arctan(z)-\arctan(y)| \le |\arctan(x)-\arctan(z)|+|\arctan(z)-\arctan(y)|$

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    @Steven: Substituting appropriate values can show you that something is *not* a metric. It will not show you that something *is* a metric, though it may give useful information towards a proof.2012-05-18

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Clearly the function $d(x,y)=|\arctan(x)-\arctan(y)|$ is symmetric and non-negative. The arctangent function is a bijection from $\Bbb R$ to $\left(-\frac{\pi}2,\frac{\pi}2\right)$, so in particular it’s injective, and $d(x,y)=0$ iff $x=y$. Thus, it only remains to check the triangle inequality, i.e., to check that

$|\arctan(x)-\arctan(y)|\le|\arctan(x)-\arctan(z)|+|\arctan(z)-\arctan(y)|$ for all $x,y,z\in\Bbb R$. But this is just the triangle inequality for the absolute value, so it’s true.

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    @Steven: I don’t know of a nice general test. With something like $d_2$ in the original problem, I’d notice that $|x-y|^2$ gets big very fast compared with $|x-y|$ and that would make me suspicious. Then I’d probably look either at -x<0 and see right away that (2x)^2=4x^2>2x^2, or at 0 and see right away that x^2+(y-x)^2=y^2+2x^2-2xy=y^2+2x(x-y). I’d look at these because they’re substantially simpler than the general case, since one of the points is $0$, but still general enough that I might find my counterexample.2012-05-19
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To check d(x,y)=|arctan(x)-arctan(y)| you can proceed as follows: symmetry follows since arctan is an injective function. To show triangle inequality note that:

$d(x,z) = |arctan(x) - arctan(z)| = |(arctan(x)-arctan(y)) + (arctan(y) - arctan(z))|$.

Apply now triangle inequailty for real numbers:

$d(x,z) \leq |arctan(x)-arctan(y)| + |arctan(y) - arctan(z)| = d(x,y) + d(y,z)$ as desired.

For fun: draw a picture and interpret this metric geometrically.

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    Thanks, I drew a picture of the $\arctan$ and it does help.2012-05-17