Is there a bijection between $[0, \infty)$ and $(0,1)$ I tried $\tan(x)$, and $\log(x)$, but seems it does not work, so I wonder is there a bijection or not?
 A: Consider that 
$$
[0,\infty) = \{0\} \cup (0,1) \cup \{1\} \cup (1,2) \cup \{2\} \cup \cdots
$$
and
$$
(0,1) = (0,1/2) \cup \{1/2\} \cup (1/2,3/4) \cup \{3/4\} \cup (3/4, 7/8) \cup \{7/8\} \cup \cdots
$$
So we can make a bijection between then so that $0$ maps to $1/2$, $1$ maps to $3/4$, $2$ maps to $7/8$, etc., and the interval $(0,1)$ maps to the interval $(0,1/2)$ and the interval $(1,2)$ maps to the interval $(1/2,3/4)$, etc.
You cannot find a bijection with a single continuous function, because of topological reasons which are more advanced than the construction just described. In particular, $[0,\infty) \setminus \{0\}$ is connected but $(0,1)$ minus any point is disconnected, and the image of a connected set under a continuous map is connected. 
A: $$
f(x) = \begin{cases} \dfrac x {x+1} & \text{if } x\notin\{0,1,2,3,\ldots\}, \\[8pt]
\dfrac{x+1}{x+2} & \text{if } x\in\{0,1,2,3,\ldots\}.
\end{cases}
$$
$f : [0,\infty) \to (0,1)$ is one-to-one and onto.
A: Yes. You can find an injection from $(0,1)$ to $[0,\infty)$ by using $f(x) = 1/x$. You can find an injection from $[0,\infty)$ to $(0,1)$ by $\frac{1}{10} \arctan(x)+\frac{1}{10}$. 
Since you have injections from each set to the other, there exists a bijection between them (via cantor-schroder-bernstein). 
