Prove that $(0,1) =c\mathbb R$

Question

How do I prove this knowing that $f(x) = \tan(x\pi/2)$ is a bijection between $(0,1)$ and $(0, \infty)$? We also have a bijection between $(-1,1)$ and $(0,1)$.

Answer 1

Use what you have to define the bijection you need. You have a bijection: $$ f\colon (0,1) \stackrel{\thicksim}\to (0,+\infty) $$ Now you can define a bijection $g\colon (-1,1)\stackrel{\thicksim}\to \Bbb R$ by: $$ g(x) = \begin{cases} f(x)&\text{if $x>0$,} \\ 0&\text{if $x=0$,} \\ -f(-x)&\text{if $x<0$.} \\ \end{cases} $$ Finally, compose this with a bijection $h \colon (0,1)\stackrel{\thicksim}\to (-1,1)$ (for example, $x\mapsto 2x-1$) to get a bijection $$ g\circ h\colon (0,1)\stackrel{\thicksim}\to \Bbb R. $$