Is this set bounded with metric d? Let $d:\mathbb{R}\times \mathbb{R}\to \mathbb{R}$ with $d(x,y)=|e^{x}-e^{y}|$ metric function on $\mathbb{R}$. How can we check if A is bounded on $(\mathbb{R},d)$ if :

*

*$A=(-\infty,0)$

*$A=(-1,1)$

*$A=(0,+\infty)$ ? 
From the mean value theorem for example we have in 1. that $|e^{x}-e^{y}|\leq |x-y|$, but can we use it to prove the problem ?

 A: Hint: $(-\infty,0)$ is contained in the open ball around $-1$ with radius $r$ if $r >1+\frac  1 e$.  So the first set is bounded.
A slight modification of the ball shows that the second set is also bounded.
To prove that the third one is not bounded assume that $(0,\infty)$  is contained in the open ball with center $x$ and radius $r$. Show that $n \in (0,\infty)$  but $n$ is not in this ball if the integer $n$ is large enough.
A: *

*You have

$$d(x, - 1) = \vert e^x - e^{-1} \vert \le e^0 - \lim\limits_{x \to - \infty}e^x = 1$$ as $e^x$ is continuous increasing. Proving that $d$ is bounded on $A=(-\infty , 0)$.


*By Mean Value Theorem you have
$$d(x,0)= \vert e^x - e^0 \vert = e^c \vert x \vert \le e \vert x \vert\le e$$ where $c$ in $(-1,1)$. Therefore $d$ is again bounded on that interval.


*Finally $$\lim\limits_{x \to \infty} d(x,1) = \lim\limits_{x \to \infty} \vert e^x - e^1 \vert = \infty.$$ hence, $d$ is not bounded on $(0, \infty)$.
A: Hint: it might be helpful to consider the set $B=\{e^x : x \in A\}$. If $B$ is bounded under the usual metric $d'(u,v) = |u-v|$, then $A$ is bounded under your metric.
