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Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a strictly increasing function. Show that $d(x,y) = |f(x) - f(y)|$ is a metric in $\mathbb{R}$.

The first two properties (non-negativity and symmetry) are straightforward to prove, but the triangular inequality is a lot harder for me!

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    $\begingroup$ Use that the absolute value satisfies the triangle inequality. $\endgroup$ – Martin R Sep 3 '20 at 17:15
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You can simply use the fact that $D(x,y)=|x-y|$ for $x,y\in\Re$ is a metric on reals. So $$|f(x)-f(y)|\le |f(x)-f(z)|+|f(z)-f(y)|$$ which is what you wanted.

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  • $\begingroup$ But how can one prove this assertion? Why the fact that f is increasing makes that statement valid? $\endgroup$ – Ralph Sep 3 '20 at 17:25
  • $\begingroup$ @Ralph The fact that $f$ is increasing does not matter for the triangular inequality. It does matter for the last point of the definition of a distance. $\endgroup$ – TheSilverDoe Sep 3 '20 at 17:49
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$$d(x,z) = |f(x)-f(z)| \leq |f(x)-f(y)|+|f(y)-f(z)| = d(x,y)+d(y,z)$$

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  • $d(x,z) = |f(x)-f(z)| \leq |f(x)-f(y)|+|f(y)-f(z)| = d(x,y)+d(y,z)$

  • $d(x,y)=0 \iff |f(x)-f(y)|=0 \iff f(x)=f(y) $, or $f$ is strictly increasing so it's injective, i.e: $x=y$ ( for the other implication $x=y \implies d(x,y)=0$)

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