Give a bijection $f: (c,d) \to \Bbb R$ (f no trigonometric) to prove every open interval has the same cardinality of R

Question

I want to prove that every open interval has the same cardinality of R.

I've proved that $|(a,b)|=|(c,d)|$ so I may find a bijection $f: (a,b) \to (c,d)$.

I need a bijection $f: (c,d) \to \Bbb R$ (a bijection defined by an interval to $\Bbb R$) The problem is that the only function like this that I know is tan(x) that is bijective in ($- \pi/2, \pi/2)$ but I am not allowed to use trigonometric functions.

Does anybody know another function like this?

Answer 1

Are you allowed to use fractions? In that case, you could use $$ \frac{1}{c-x} + \frac{1}{d-x} $$ Of course, you would have to do some work to prove that it is monotonous on $(c, d)$, but that shouldn't be too hard (you are allowed to use differentiation, are you not?)

Answer 2

You already know that there is a bijection $(c,d) \to (-1,1)$ and it thus suffices to find a bijection $f \colon (-1,1) \to \mathbb R$. The following will do:

For all $n \in \mathbb N_0$ we let $f \restriction_{[1- 2^{-n},1-2^{-n-1}]}$ be the linear function from $(1- 2^{-n} ; 2^n)$ to $(1-2^{-n-1} ; 2^{n+1})$ and likewise $f \restriction_{[-1+ 2^{-n-1},-1+2^{-n}]}$ is the linear function from $(-1+ 2^{-n-1} ; -2^{n+1})$ to $(-1+ 2^{-n} ; -2^{n})$.

The picture below demonstrates what's going on: