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This question already has an answer here:

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...

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marked as duplicate by Asaf Karagila, Yiorgos S. Smyrlis, anomaly, Mark Fantini, apnorton Sep 17 '14 at 21:35

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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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.

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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.)}$

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  • $\begingroup$ A minor point, it may be worth mentioning that you are excluding cases where the $ x_i $ are not eventually all equal to 9. $\endgroup$ – Epsilon Sep 17 '14 at 0:45
  • $\begingroup$ @NickR This is a good point -- I was going to add that to my answer when I got back to my computer. $\endgroup$ – user84413 Sep 17 '14 at 20:13
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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, $\mathbb{R} \times \mathbb{R} \cong 2^{\mathbb{N}} \times 2^{\mathbb{N}} \cong 2^{\mathbb{N} \sqcup \mathbb{N}} \cong 2^\mathbb{N} \cong \mathbb{R}$.

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