Cartesian product I have a question. How do I prove that the cardinality of the Cartesian  product of $\mathbb{R}$ with $\mathbb{R}$ (reals) equals to the cardinality of the continuum?
 A: Since $|\mathbb{R}| = 2^{\aleph_0}$, you easily have:
$$|\mathbb{R} \times \mathbb{R}| = |\mathbb{R}|\cdot|\mathbb{R}|=\left(2^{\aleph_0}\right)^2 = 2^{2 \cdot \aleph_0} = 2^{\aleph_0}$$
In fact, for every infinite cardinal $\kappa$, you have $\kappa\cdot\kappa=\kappa$ and for infinite cardinals $\kappa$ and $\lambda$ more generally that $\kappa \cdot \lambda = \mbox{max}\left\{ \kappa , \lambda\right\}$.

Without cardinal arithmetic, you can show $|\mathbb{R} \times \mathbb{R}| = |\mathbb{R}|$ by giving injective functions:


*

*$f :\mathbb{R} \to \mathbb{R}^2$ showing that $|\mathbb{R} \times \mathbb{R}| \ge |\mathbb{R}|$

*$f :\mathbb{R}^2 \to \mathbb{R}$ showing that $|\mathbb{R} \times \mathbb{R}| \le |\mathbb{R}|$
The first one is trivial. I can add a sketch for the second injection if you want.

Since $\mathbb{R}$ can be bijectively mapped to $(0,1)$, you can bijectively map $\mathbb{R}^2$ to $(0,1) \times (0,1)$, an "open square". Now pick an element $(x,y) \in (0,1) \times (0,1)$ and write down the unique decimal expansions (excluding infinite repeating 9's) of $x$ and $y$.
Construct a new number $z \in (0,1)$ by picking the decimals of $x$ and $y$ in an alternating way, then different pairs $(x,y)$ are mapped to different numbers $z$.
(Reference: The Real Numbers: An Introduction to Set Theory and Analysis
by John Stillwell)
