# Show that ($\mathbb{R}$, $d$) is a metric space

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?

• $e^x = e^y \Rightarrow x = y$ as the exponential function is one-to-one. What isn't asked in your question but is interesting to think about is whether this metric space is complete, as it would be with the usual metric. Commented Oct 8, 2015 at 22:15
• $\pm$ is redundant Commented Oct 8, 2015 at 22:19

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!
Non-degeneracy: $e^x$ is one to one. So $e^x = e^y iff x = y$ so $d(x,y) = 0 iff x=y$.
Triangle inequality:$e^x$ is convex.