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In notes I have come across, it is stated that for a Hausdorff space induced by countably many semi-norms $p_n$ that,

$$d(x,y) = \sum_{n=1}^{\infty}2^{-n}\frac{p_n(x-y)}{1+p_n(x-y)}$$ is a metric.

But I can't see why this metric satisfies the triangle inequality. As I believe it satisfies the triangle inequality iff $p_n(x-y)(1+p_n(x-z)p_n(z-y)) \leq p_n(x-z) + p_n(z-y)$

If you shed some light on this, it would be much appreciated!

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up vote 1 down vote accepted

For triangle inequality, use non-decreasingness of $x\mapsto \frac x{1+x}$, and the triangle inequality for $p_n$.

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Sorry but the fact that the function $u:x\mapsto x/(1+x)$ is nondecreasing is irrelevant. Instead, one should check that $u(x+y)\leqslant u(x)+u(y)$ for every nonnegative $x$ and $y$. – Did Apr 14 '13 at 15:54
$\frac{p_n(x-y)}{1+p_n(x-y)} \leq \frac{p_n(x-z)+p_n(z-y)}{1+p_n(x-z)+p_n(y-z)} \leq \frac{p_n(x-z)}{1+p_n(y-z)} + \frac{p_n(z-y)}{1+p_n(y-z)}$ – user58514 Apr 14 '13 at 16:00
With your observation, and the fact that the $p_n$ are positive gives the result? I'm still a little puzzled as to why the other inequality in my question holds though. As it seems to be a re-arrangement of this inequality. And doesn't hold if $2p_n(x-y) = p_n(x-z) = p_n(y-z) = 2$ – user58514 Apr 14 '13 at 16:05

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