Show that $(0, 1)$ is uncountable if and only if $\mathbb{R}$ is uncountable. 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.  
 A: 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)$?
A: We can take the function $f: (0,1) \to \mathbb{R}$ by $x \mapsto \tan\left(x-\dfrac{1}{2}\right)\pi$. This function is bijective, thus proving $(0,1) \cong \mathbb{R}$
A: Scalings of $\arctan$ tend to be useful here. It also shows any interval of $\mathbb{R}$ is homeomorphic to $\mathbb{R}$.
A: Since $(0,1)$ is infinite, its cardinality is equal to the cardinality of $[0,1)$.
Hint: $\mathbb{R}$ is the countable union of intervals of same cardinality as $[0,1)$: namely, $[n, n+1)$ for $n \in \mathbb{Z}$.
A: Here's a proof that doesn't involve any trigonometry: assume $(0, 1)$ is countable. Then any interval $(i, i + 1)$ is also countable (if $(0, 1) = \{a_1, a_2, \ldots\}$ then $(i, i + 1) = \{i + a_1, i + a_2, \ldots \}$). But then $\Bbb{R}= \Bbb{Z} \cup \bigcup_{i \in \Bbb{Z}}(i, i + 1)$ is a countable union of countable sets and hence is countable. So if $\Bbb{R}$ is uncountable, then so is $(0, 1)$.
A: If I can find an invertable 1-1 function between $(0,1)$ to $\mathbb R$ they have the same cardinality.  So I figure I need to "stretch" $(0,1)$.  The obvious way to stretch is to do $x \rightarrow 1/x$.  That will stretch (and flip) $(0,1)$ to $(1, \infty)$.
THis is almost okay but I really to extend it to the whole reals. I figure I can actuall do this in pieces:  $(0,1/4)\rightarrow (1, \infty)$, $[1/4,1/2]\rightarrow [0,1]$, $[1/2, 3/4]\rightarrow [-1,0]$ and $(1/4,1)\rightarrow (-\infty, -1)$.  I figure each step of those is easy.
To make life easier, I can shift and readjust and do $(-2,2)\rightarrow \mathbb R$.  $(1,2)\rightarrow (1,\infty)$ is just like I did above: $x \rightarrow 1/(x-1)$.   $(-2,-1)\rightarrow(-\infty, -1)$ via $x \rightarrow 1/(x + 1)$ is just the same thing but in the negative direction.  $[-1,1]\rightarrow [-1,1]$ can simply be $x \rightarrow x$ 
So $f(x) = \begin{cases}\frac 1{x+1} & \text{ if } -2<x < -1\\
 x & \text{ if }  -1\le x\le1\\
\frac1{x-1} & \text{ if } 1 \le x\\
 \end{cases}$
is a 1-1 mapping from $(-2,2)\rightarrow \mathbb R$.
To map from $(0,1)\rightarrow \mathbb R$ I simply stretch $(0,1)$ to $(-2,2)$.
So $f(x) = \begin{cases}\frac 1{(4x-2)+1} & \text{ if } 0<x < 1/4\\
 (4x-2) & \text{ if }  1/4\le x\le 3/4\\
\frac1{(4x-2)-1} & \text{ if } 3/4 \le x < 1\\
 \end{cases}$
is an inverable 1-1 function from $(0,1) \rightarrow \mathbb R$.  
It's hacky and cludgy but it works (even if it isn't continuous) and it is one that one can come up with directly by onself without any clever insight or intimate familiarity with the much better trig and/or exponent functions.
(although the trig and/or exponent functions are well-worth becoming intimate with.)
A: The tangent function was used in one posted answer and a piecewise defined function in another.  Here's a simpler one:
$$
f(x) = \frac{x - \frac 1 2}{x(1-x)} \text{ for } 0<x<1.
$$
This takes the interval $(0,1)$ to the interval $(-\infty,\infty)$ and requires no transcendental functions such as trigonometric functions --- you just have to add, subtract, multiply, and divide.
