If $S$ is a non-finite set, meaning that $|S| \notin \mathbb{N}$, then is it always true that there is a bijection between $S \times S$ and $S$, where $\times$ is the Cartesian product of two sets?
It is true for $|S| = |\mathbb{N}|$ and $|S| = |\mathbb{R}|$, the first one can be proven by the "criss-crossing" bijection between $\mathbb{N}$ and $\mathbb{Q}$.
The second one follows from the bijection between $[0,1[$ and $[0,1[ \times [0,1[$:
If $a \in [0,1[$, then $a = 0.a_1 a_2 a_3 a_4 a_5 \dots$ is its decimal expansion with $a_1,a_2,a_3,a_4,a_5,\ldots \in \{0,\dots,9\}$.
With this, we can uniqely build a value in $[0,1[ \times [0,1[$ by considering only its even and only its odd decimal places: $(0.a_1 a_3 a_5 \dots,0.a_2 a_4 a_6 \dots) \in [0,1[ \times [0,1[$, and this process is entirely reversible, therefore
$$|\mathbb{R}| \stackrel{(*)}{=} |[0,1[| = |[0,1[ \times [0,1[| \stackrel{(**)}{=} |\mathbb{R} \times \mathbb{R}|$$
$(*)$ $x \leftrightarrow \tan(2\pi x)$ is a bijection.
$(**)$ Placing the $[0,1[ \times [0,1[$ blocks in a spiral starting from $(0,0)$ towards any direciton gives the bijection.
But is this also true if let's say $|S| = \mathcal{P}(\mathbb{R})$ or $|S| = \mathcal{P}(\mathcal{P}(\mathbb{R})), \dots$ (where $\mathcal{P}(A)$ denotes the powerset of $A$), or if $S$ is any other infinite set?
