Increasing function induces a metric in $\mathbb{R}$

Question

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!

Use that the absolute value satisfies the triangle inequality. — Martin R, Sep 03 '20 at 17:15

score 0 · Answer 1 · answered Sep 03 '20 at 17:15

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answered Sep 03 '20 at 17:15

OnoL

But how can one prove this assertion? Why the fact that f is increasing makes that statement valid? – Ralph Sep 03 '20 at 17:25
@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. – TheSilverDoe Sep 03 '20 at 17:49

score 0 · Answer 2 · answered Sep 03 '20 at 17:16

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

answered Sep 03 '20 at 17:16

TheSilverDoe

score 0 · Answer 3 · answered Sep 03 '20 at 17:34

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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 $x=y \implies d(x,y)=0$)

answered Sep 03 '20 at 17:34

M-S

3 Answers3