Question
Let $\mathbb{R}$ be the set of real numbers and define $d$ : $\mathbb{R}$ $\times$ $\mathbb{R}$ $\rightarrow$ $\mathbb{R}$ by $d(x, y) = \mid e^{x} – e^{y} \mid $.
i) Show that ($\mathbb{R}$, $d$) is a metric space
ii) What are the properties of the exponential function that allows one to deduce
that d is a metric? Formulate a generalization of the metric d based on this
observation.
Attempted solution
In order to show that this is a metric space I must show that $d$ satisfies the definition of a metric (symmetry, triangle inequality, non-degeneracy). For the symmetry part we have that:
$d(x, y) = \mid e^{x} – e^{y} \mid = \mid e^{y} – e^{x} \mid = d(y, x)$
For the triangle inequality, we have for some number $z$ $\in$ $\mathbb{R}$ that:
$\mid e^{x} – e^{y} \mid \leq \mid e^{x} – e^{z} \mid + \mid e^{z} – e^{y} \mid$ $\Rightarrow$ $d(x, y) \leq d(x, z) + d(z, y)$
For non-degeneracy we must show that $d(x, y) = 0$ iff $x=y$. Showing this:
$\mid e^{x} – e^{y} \mid = 0$ $\Rightarrow$ $\pm(e^{x} – e^{y}) = 0 $ $\Rightarrow$ $x = \ln (e^{y}) = y$
However, I have a feeling (based on the second part of the question) that I haven't quite shown that ($\mathbb{R}$, $d$) is metric. Could anyone explain to me what I've done wrong? What are the properties of the exponential function that are key to deducing that $d$ is a metric?
Thanks in advance.
Best Answer
Hint: it's all in the last step. What's an equivalent condition to $|e^x-e^y|=0$? What is special about the exponential function that allows you to conclude from this that $x=y$? Try to answer without mentioning logarithms!