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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.

Julien
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elich
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  • What part of showing that $(\mathbb{R}^x , d)$ that is a metric space are you stuck on? What have you tried? – Elchanan Solomon Nov 22 '12 at 00:14
  • What is the connection between $\mathbb R^x$ and $X$? Or do you mean $\mathbb R^X$? – joriki Nov 22 '12 at 00:17
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    Yo can first prove that $\rho(f,g) = \sup_{x\in X} |f(x) - g(x)|$ is a metric. Then prove that if a function $\rho(x,y)$ is a metric then $d(x,y) = \rho(x,y)/ (1 + \rho(x,y))$ is a metric. – Yury Nov 22 '12 at 00:20
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    @Yury These functions may not be bounded, so $\rho$ is not really a metric. – Michael Greinecker Nov 22 '12 at 01:01

1 Answers1

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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 $$ d(f,h)\leq d(f,g)+d(g,h). $$

Julien
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