Equivalency of real numbers and points in the plane? I understand that the set of real numbers is equivalent to the set of real numbers in the interval $(-1,1)$ by simply using $arctan$ function. However, I do not know how to find a one-to-one mapping from real numbers (or any equivalent set to real numbers) to the set of all points in a plane in $\mathbb{R}^2$. Can anyone help? 
 A: The interval $(-1,1)$ and plane $\mathbb{R}^2$ are in bijective correspondence with the interval $A=(0,1)$ and open square $A\times A =(0,1) \times (0,1)$, respectively. These can be generated in a variety of ways, e.g.
$$(-1,1)\to (0,1); x \mapsto (x+1)/2$$
and by using $\arctan$ followed by a linear map for $\mathbb{R} \to (0,1)$ on each each factor of $\mathbb{R}^2 = \mathbb{R} \times \mathbb{R}$. So we're done if we can find a bijection $A \times A \to A$. To that end, view every element of $A$ via a decimal expansion $0.x_1x_2x_3\ldots=x$. The map $A \times A \to A$ given by
$$(0.x_1x_2x_3\ldots,0.y_1y_2y_3\ldots)\mapsto 0.x_1 y_1 x_2 y_2 x_3 y_3\ldots$$
provides the desired bijection. (Note that $0.x_1 x_2 x_3 \ldots$ is the image of $(0.x_1 x_3 x_5\ldots,0.x_2 x_4 x_6,\ldots)$.)
A: Map the two points of the plane $$\begin{align} 0.a_1a_2a_3a_4\ldots \\
0.b_1b_2b_3b_4\ldots \end{align}$$ to $$0.a_1b_1a_2b_2a_3b_3\ldots$$
Okay, it is only a bijection from $[0,1]^2\to[0,1]$ and you have to find solutions for thinks like $0.29999...=0.3$. But you should get an idea how the bijection might work.
