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I have to prove that $d(x,y)=\left|\frac{x}{1+\sqrt{1+x^2}}-\frac{y}{1+\sqrt{1+y^2}}\right|$ is a metric on $\mathbb{R}$. I managed to prove the non-negativity, symmetry and triangle inequality, but I am stuck on proving $d(x,y)=0\Leftrightarrow x=y$. In my textbook I have the following indication: "prove that $f(x)=\frac{x}{1+\sqrt{1+x^2}}$ is a strictly increasing function". How this indication helps me, why do I have to prove that the given function is strictly increasing?

Melissa
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    When you have that a function is strictly increasing it's injective, hence from $$f(x) = f(y)$$ it follows that $$x = y$$. Additionally you have $$d(x,y) = 0 \iff f(x) = f(y)$$ so alltogether you get $$d(x,y) = 0 \iff x=y$$ – Gono Feb 23 '20 at 13:30

2 Answers2

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Since $f(x) = \frac{x}{1+\sqrt{1+x^2}}$ is strictly increasing, suppose without loss of generality that $x<y$.

$\implies f(x) < f(y) \implies d(x,y)=|f(x)-f(y)|>0 \qquad (1)$

The proof where $f(x)=f(y) \implies d(x,y) = |f(x)-f(y)|=|f(x)-f(x)|=0$ is trivial. In the other direction, given that $|f(x)-f(y)|=0$, if either $x<y$ or $y<x$ but $d(x,y)= 0$ then this presents a contradiction to (1).

Adam
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Consider $f(x)=\dfrac{x}{1+\sqrt{1+x^2}}$. Its derivative is $f'(x)=\dfrac{1}{\sqrt{1+x^2}(1+\sqrt{1+x^2})}$ is clearly positive then $f$ is increasing and consequently injective. So $f(x)=f(y)\Rightarrow x=y$ which proves the not obvious side of the separation axiom $d(x,y)=0\iff x=y$. The symetry $d(x,y)=d(y,x)$ is clear. It remains the triangular inequality.

We have $$\left|\dfrac{x}{1+\sqrt{1+x^2}}-\dfrac{y}{1+\sqrt{1+y^2}}\right|+\left|\dfrac{y}{1+\sqrt{1+y^2}}-\dfrac{z}{1+\sqrt{1+z^2}}\right|\hspace{10mm}(*)$$ By the triangular inequality in $(\mathbb R,|\space|)$ one has that $(*)$ is greater or equal than $$\left|\dfrac{x}{1+\sqrt{1+x^2}}-\dfrac{z}{1+\sqrt{1+z^2}}\right|$$ so we have prove that $$d(x,y)+d(y,z)\ge d(x,z)$$

Piquito
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