Prove cardinality of R2 without creating a map to R1

Question

I think we can prove $|\mathbb{R}^2|=\aleph_1$, by creating a bijection between $\mathbb{R}$ and $\mathbb{R}^2$. But this map is difficult to construct. Is there any easier way to show $|\mathbb{R}^2|=\aleph_1$?

Answer 1

Let $2$ denote the set $\{0,1\}$, and $Y^X$ denote the set of maps $X\to Y$.

$$\begin{align}|\Bbb R^2|&=|(2^{\Bbb N})^2|\\&=|2^{\Bbb N\times2}|\\&=|2^{\Bbb N}|\\&=|\Bbb R|.\end{align}$$

The central argument, $$|(A^B)^C|=|A^{B\times C}|,$$ is proved for instance here.

Answer 2

Maybe we can apply Schröder-Bernstein Theorem to make it "slightly" easier:

Theorem (Schröder-Bernstein): If $|A|\leq|B|$ and $|B|\leq|A|$, then $|A|=|B|$.

Note that $|A|\leq |B|$ $\Leftrightarrow$ $\exists$ injection $A\rightarrow B$ $\Leftrightarrow$ $\exists$ surjection $B\rightarrow A$. So in order to show $|A|=|B|$ we only need, instead of a bijection, one of the followings:

• An injection $A\rightarrow B$ and a surjection $A\rightarrow B$.
• An injection $B\rightarrow A$ and a surjection $B\rightarrow A$.
• Two injections $A\rightarrow B$ and $B\rightarrow A$.
• Two surjections $A\rightarrow B$ and $B\rightarrow A$.

The projection $(x,y)\mapsto x$ is clearly a surjection $\mathbb{R}^2\rightarrow\mathbb{R}$. An injection $\mathbb{R}^2\rightarrow\mathbb{R}$, sadly however, still requires some technique you may have encountered:

For $(x,y)\in\mathbb{R}^2$, write the decimals $$x=\pm\overline{x_nx_{n-1}\cdots x_1x_0.x_{-1}x_{-2}\cdots}$$ $$y=\pm\overline{y_ny_{n-1}\cdots y_1y_0.y_{-1}y_{-2}\cdots}$$ Here we add a least number of $0$'s before the decimal of $x$ or $y$ so that they have the same number of digits before the decimal dot. We also require that decimal representations cannot end in a series of $9$'s. Now we define $f(x,y)\in\mathbb{R}$ to be $$\overline{ax_ny_nx_{n-1}y_{n-1}\cdots x_1y_1x_0y_0.x_{-1}y_{-1}x_{-2}y_{-2}\cdots}$$ where $$a=\begin{cases}1,\quad x\geq 0, y\geq 0,\\ 2,\quad x\geq 0,y<0,\\ 3,\quad x<0,y\geq 0,\\ 4,\quad x<0,y<0.\end{cases}$$

It is easy to show $f:\mathbb{R}^2\rightarrow\mathbb{R}$ is an injection since you can reconstruct $x$ and $y$ from the decimal representation of $f(x,y)$. Just read every digit.