How to Prove $\mathbb R\times \mathbb R \sim \mathbb R$? How to prove $\mathbb R\times \mathbb R \sim \mathbb R$?
I know you have to split the problem up into two claims, for each direction to prove that it is a bijection, but i don't know how to go much further than that...
 A: By using the continuous and increasing function $f(x)=\frac{1}{2}+\frac{1}{\pi}\arctan(x)$ we can see that $\mathbb{R}\sim(0,1)$. Hence we just need to show that $(0,1)^2\sim (0,1)$. Obviously, there is an injective map from $(0,1)$ to $(0,1)^2$, given by $g(x)=(x,1/2)$. Hence we just need to show that there is an injective map from $(0,1)^2$ to $(0,1)$, then apply the Cantor-Bernstein theorem. Consider that any real number $z$ in $(0,1)$ can be represented by the sequence $\{a_n\}_{n\in\mathbb{N}^*}$ in $\{0,1\}^{\omega}$ such that
$$ z = \sum_{n\geq 1}\frac{a_n}{2^n}$$
and $\{a_n\}_{n\in\mathbb{N}^*}$ is not eventually equal to one. The map $z\to\{a_n\}_{n\in\mathbb{N}^*}$ is injective.  
Given $(z,w)\in(0,1)^2$, map $z\to\{a_n\}_{n\in\mathbb{N}^*}$ and $w\to\{b_n\}_{n\in\mathbb{N}^*}$, then "zip" the binary representations of $z$ and $w$ by taking:
$$u = \sum_{n\geq 1}\frac{c_n}{2^n},\quad c_{2m}=a_m,\quad c_{2m+1}=b_m.$$
The map $(z,w)\to u$ is injective. Done.
A: One way to do this is to use the Cantor-Schroder-Bernstein theorem.
Since $\mathbb{R}\sim(0,1)$, it is enough to show that $(0,1)\times(0,1)\sim(0,1)$.
1) Define an injection $f:(0,1)\rightarrow(0,1)\times(0,1)$ by $f(x)=(x,x)$.
2) Define an injection $g:(0,1)\times(0,1)\rightarrow(0,1)$ by
$g(x,y)=z$ where $x=.x_1x_2x_3x_4\cdots$, $y=.y_1y_2y_3y_4\cdots$, and $z=.x_1y_1x_2y_2x_3y_3x_4y_4\cdots$.
$\text{(To ensure that $g$ is well-defined, we do not allow decimal expansions that end in an infinite string of 9's.)}$
A: We have $\mathbb{R} \cong 2^{\mathbb{N}}$. In fact, $\mathbb{R}$ embeds into $P(\mathbb{Q})$ via $r \mapsto \mathbb{Q}_{<r}$, and conversely $2^{\mathbb{N}}$ embeds into $\mathbb{R}$ via $(a_n \in \{0,1\}) \mapsto \sum_{n=1}^{\infty} \frac{a_n}{2^n}$. Now we use Cantor-Schröder-Bernstein to conclude $\mathbb{R} \cong 2^{\mathbb{N}}$.
Hence, $\mathbb{R} \times \mathbb{R} \cong 2^{\mathbb{N}} \times 2^{\mathbb{N}} \cong 2^{\mathbb{N} \sqcup \mathbb{N}} \cong 2^\mathbb{N} \cong \mathbb{R}$.
