I found a lot of problems of bijections like $[0,1) \rightarrow[0, \infty)$. But I don't find ant of these:
$$\text{Is it possible to make a bijection } [0,1] \rightarrow[0, \infty)? \text{ If yes, find at least one bijection.} $$
Can anyone help?
My work: I found one function: $f(x)=\frac{1}{\left( x-\frac{1}{2} \right) ^2}-4$. What about $f(x)=\frac{1}{\left( x-1 \right) ^2}-1$?
Let $$f(x):=\begin{cases} x&\text{if $x\in[0,1)\setminus\{1/n: n\in\mathbb{N}^+\}$}\\ \frac{1}{n+1}&\text{if $x=\frac{1}{n}$ for some $n\in\mathbb{N}^+$}\\ \end{cases}$$ The function $f$ is bijection from $[0,1]\to [0,1)$ which acts as a left-shift in the set $\{1,1/2,1/3,\dots\}$ and it is the identity elsewhere. Now take your bijection $g$ from $[0,1)$ to $[0,+\infty)$ (for example $g(x)=\frac{x}{1-x}$) and consider the composition $g\circ f: [0,1]\to [0,+\infty)$.
P.S. Note that there is no continuous bijection from $[0,1]$ to $[0, \infty)$ because $[0,1]$ is a compact set and $[0, \infty)$ is not a compact set.