Showing that the points of the unit circle have the same infinity of elements as $\mathbb{R}$ I know that using Bernstein's theorem, I simply need to find two one-to-one functions that map from the unit circle to $\mathbb{R}$ and vice versa. 
I thought something like $f(x, y)=\arctan\big(\frac{x}{y}\big)$ could work as a mapping from the unit circle to $\mathbb{R}$, but it seems like isn't one-to-one since and there seems to be problems when $y=0$. I also really can't think of a one-to-one function that maps from the reals to the unit circle.
Is there a better way to do this problem?
 A: Here’s one that I ought to have known when I first started teaching Calculus, but didn’t till several years later:
$$
t\mapsto\left(\frac{2t}{1+t^2}, \frac{1-t^2}{1+t^2}\right)\,.
$$
For $t\in\Bbb R$, it hits all points on the unit circle save $(0,-1)$. I’m sure you know that it’s not possible to make a bicontinuous map between the real line and the circle. But you can take the displayed formula above and hoke it up with the right discontinuous function to make it a fully one-to-one correspondence.
A: Here are two geometric ideas to use. First, cut the circle and lay it onto the strip $[0,2\pi)$ for the injection from the circle to the reals. To go from the reals to the circle, draw a line through the point $(0,1)$ and any point on the x-axis. This will intersect the circle at a unique point so there is a one-to-one mapping from the reals to the circle without the point at $(0,1)$ and you have both injections.
A: Here's a direct construction of a bijection $\mathbb R \to S^1$, which avoids the Schröder-Bernstein theorem.
To start with, the function $f(x) = (\cos x, \sin x)$ is a bijection from $[0,2\pi)$ to $S^1$
So now all you need is another bijection $g : \mathbb R \to [0,2\pi)$, using which we obtain the desired bijection $f^{-1} \circ g : \mathbb R \to S^1$.
To construct $g$, start by subdividing $[0,2\pi)$ into a sequence of half-open subinterals:
$$
[0,2\pi) = \underbrace{[0, \frac{1}{2} \cdot 2\pi)}_{I_1} \cup \underbrace{[\frac{1}{2} \cdot 2\pi, \frac{3}{4} \cdot 2\pi)}_{I_2} \cup \underbrace{[\frac{3}{4} \cdot 2\pi,\frac{7}{8} \cdot 2\pi)}_{I_3} \cup \cdots
$$
Also, subdivide $\mathbb R$ into a bi-infinite sequence of half-open intervals:
$$\mathbb R = \cdots \cup \underbrace{[-2,-1)}_{J_{-2}} \cup \underbrace{[-1,0)}_{J_{-1}} \cup \underbrace{[0,1)}_{J_{0}} \cup \underbrace{[1,2)}_{J_1} \cup \underbrace{[2,3)}_{J_2}\cdots
$$
Now choose your favorite bijection 
$$\sigma : \{1,2,3,\ldots\} \to \{\ldots,-2,-1,0,1,2,\ldots\}
$$
and write it as a sequence, for example
$$(\sigma_1,\sigma_2,\sigma_3,\sigma_4,\sigma_5,\ldots)  =  (0,1,-1,2,-2,\ldots)
$$
For each $i=1,2,3,\ldots$, choose a bijection
$$g_i : J_{\sigma_i} \to I_i
$$
which is possible for any two half-open intervals.
Finally, define
$$g(x) = g_i(x) \quad\text{if $x \in J_{\sigma_i}$}
$$
