Metric on space of functions 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.
 A: 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
$$
d(f,h)\leq d(f,g)+d(g,h).
$$
