I have shown the forward direction by assuming $(0,1)$ is countable and constructed a set containing all real numbers between $(0,1)$ and tried to list them in $x_n$ numbers were each $x_n$ = 0.$a_{n1}$$a_{n2}$$a_{n3}$$\dots$ and so on and showed that there exist a number b between (0,1) that is not in the function thus creating a contradiction (short explanation of where I am).
now how do I prove the other way? (the set (0,1) is uncountable if R is uncountable) I thought of stating that since R is uncountable then (0,1) is too since (o,1) $\subseteq$ of R. but that is a false statement since Q $\subseteq$ R and Q is countable.
Best Answer
Hint: Note that the restriction of $\tan$ to its principal domain and its inverse $\arctan$ provide a bijection between $(-\pi/2,\pi/2)$ and $\mathbb{R}$. Can you find a bijection $(-\pi/2,\pi/2) \cong (0,1)$?