3
$\begingroup$

For a set $X$, let $\mathbb{R}^{X}$ be the set of all maps from $X$ to $\mathbb{R}$.

For $f,g\in\mathbb{\mathbb{R}}^{X}$, define $d(f,g) = \sup_{x\in X}\frac{|f(x)-g(x)|}{1+|f(x)-g(x)|}.$

I am trying to show that $(\mathbb{R}^{X},d)$ is a metric space but I can't get the bounds in the right way.

  • 2
    @Yury These functions may not be bounded, so $\rho$ is not really a metric.2012-11-22

1 Answers 1

5

Edit: this answer was ridiculously complicated. Here is a shorter argument.

Consider the following function $f:[0,+\infty)\longrightarrow [0,+\infty)$ $ \phi(t):=\frac{t}{1+t}. $

Since it is bounded by $1$, $d(f,g)$ is a well-defined nonnegative real number for every $f,g\in\mathbb{R}^X$. It is straightforward to see that $d$ satisfies separation and symmetry.

For the triangular inequality, observe that $\phi$ is increasing on $[0,+\infty)$ by checking its derivative is positive. Since $|t-s|\leq |t-u|+|u-s|$, it follows that $ \frac{|t-s|}{1+|t-s|}\leq \frac{|t-u|+|u-s|}{1+|t-u|+|u-s|}=\frac{|t-u|}{1+|t-u|+|u-s|}+\frac{|u-s|}{1+|t-u|+|u-s|} $ $ \leq \frac{|t-u|}{1+|t-u|}+\frac{|u-s|}{1+|u-s|}\quad\forall s,t,u\in\mathbb{R}. $ Applying this $t=f(x)$, $s=h(x)$ and $u=g(x)$ yields $ \frac{|f(x)-h(x)|}{1+|f(x)-h(x)|}\leq \frac{|f(x)-g(x)|}{1+|f(x)-g(x)|} + \frac{|g(x)-h(x)|}{1+|g(x)-h(x)|}\qquad\forall x\in X. $ So $ \frac{|f(x)-h(x)|}{1+|f(x)-h(x)|} \leq d(f,g)+d(g,h) \qquad\forall x\in X. $ And finally $ d(f,h)\leq d(f,g)+d(g,h). $

  • 0
    You can ask on meta.math.SE2013-04-19