Bijection from $[0,1]$ to $(1, \infty)$ I've come across many different versions of this question on here, but not any that map the $[0,1]$ to $(1, \infty)$. 
I was thinking that it must be piece-wise defined, since the endpoints 0 and 1 will be the trickiest part of defining the bijection... The only method of doing this that I could come up with would be to possibly show a bijection from $[0,1]$ to $(1,2)$, then construct another bijection from $(1,2)$ to $(1, \infty)$, and then the composition will be from $[0,1]$ to $(1, \infty)$, but I haven't been able to come up with any function that can do this... Any help is much appreciated.
 A: Hint: There is a simple (continuous) bijection from $(0,1)\to (1, \infty)$. So if you can find a bijection from $[0,1] \to (0,1)$ then you're done. One way to do this is in two steps: find a bijection $[0,1]\to(0,1]$ and then find another from $(0,1]\to(0,1)$.
To find a bijection from $[0,1]\to(0,1]$, you intuitively want to fix "most of" $[0,1]$, but you need to send $0$ to somewhere that isn't $0$. And wherever you send $0$ can't be sent to itself, so has to be sent somewhere else. And wherever you send that to also has to be sent somewhere else... and so on.
Let's try sending $0$ to $\frac12$. And we can send $\frac12$ to $\frac14$, and $\frac14$ to $\frac 18$, and so on, and fix everything that isn't of the form $\frac1{2^n}$ for an integer $n \ge 2$. Can you show that this gives a bijection?
Can you do something similar to find a bijection from $(0,1]\to(0,1)$?
A: You can consider a piece-wise function defined where 0 maps to some value y1 not equal to 1 (say 2). Then you can manipulate the period and height of tan(x) such that  tan(0) = 1 and tan(1) is infinite.
