# Injective, Surjective, and Cardinality

Let $f:(0,\infty)\to (0,1)$ given by $f(x) = \frac{x}{x+1}$. Decide whether $f$ is injective and whether it is surjective? What does this say about the cardinality of $R$ and $(0,1)$?

I am not how to set up the functions considering that the function is between two coordinates it seems?

• What do you mean the function is between "two coordinates"? $(0,1)$ is notation for the open interval $0<x<1$. Also, $(0, \infty) = \mathbb{R_+}$.This may be the source of your confusion. – MathematicsStudent1122 May 3 '16 at 21:20

As $f$ is continuous, $\lim_{x\rightarrow 0}f(x)=0$ and $\lim_{x\rightarrow\infty}f(x)=1$, then $f$ is surjective. Now, as
$f'(x)=\dfrac{1}{(x+1)^2}$
is positive for all $x\geq 0$, $f$ is strictly increasing and so injective. Therefore, as $f$ is a bijection of $(0,\infty)$ onto $(0,1)$, the two intervals have the same amount of elements, which implies the same cardinality. The same relation is valid for $\mathbb{R}$ and $(0,1)$, but with this function I don't see how to prove it.