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.

  • 1
    $\begingroup$ $e^x = e^y \Rightarrow x = y$ 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. $\endgroup$
    – Simon S
    Commented Oct 8, 2015 at 22:15
  • $\begingroup$ $\pm$ is redundant $\endgroup$ Commented Oct 8, 2015 at 22:19

2 Answers 2


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!

  • 3
    $\begingroup$ That the exponential function is injective? $\endgroup$
    – Joey
    Commented Oct 8, 2015 at 22:20
  • 2
    $\begingroup$ @Joey We have a winner. :) $\endgroup$ Commented Oct 8, 2015 at 22:22

Symmetry: subtraction under absolute value is always symmetric so there's nothing about exponentiation needed.

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.


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