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If $X=\mathbb{R}^2$. For $x=(x_1,x_2)$, $y=(y_1,y_2)$ define

$$d_{1/2}(x,y)=\left(|x_1-y_1|^{1/2}+|x_2-y_2|^{1/2}\right)^2\;.$$

Prove or disprove $(X,d_{1/2})$ is a metric space.

Attempt at a solution: After multiplying it out, it seems to boil down to $2|x_1-y_1|^{1/2}|x_2-y_2|^{1/2}$ satisfying the triangle inequality. However, I can't seems to prove or disprove this one way or another.

Intution, however, is leading me towards the fact that it is in fact not a metric space.

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  • $\begingroup$ It might be easy to prove one way or the other if the distance were defined in a readable way. $\endgroup$ – André Nicolas Sep 12 '12 at 4:17
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Hint:

try $x = (1,0), y = (0,1)$.

Another way of gaining an intuition for this is thinking about the unit ball defined by this "norm", i.e. the set of points whose distance to the origin is $1$. For various $p$, if the norm is $( \sum_{i=1}^2 |x_i - y_i|^p )^{1/p}$, the balls look like this (from Wikipedia):enter image description here

What do all the balls for legitimate norms have in common, and what does this have to do with the triangle inequality?

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Try computing the triangle inequality on each pair from $(0,0), (0,d), (d,d)$.

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