Cartesian power of an infinite set

Question

Let $X$ be an infinite set, i.e., there is an injection $\mathbb N \hookrightarrow X$. There is an obvious injection $X\hookrightarrow X\times X$ (just take $a\in X$ and send any $x \in X$ to $(x,a)$). Is it possible to find (explicitly) an injection $X \times X \hookrightarrow X$?

Don't know if this is helpful: $f: (m,n) \mapsto 2^m(2n+1)-1$ is a bijection from $\mathbb N \times \mathbb N$ to $\mathbb N$.

Answer 1

It's not possible: indeed, it is the case that if $|S^2| = |S|$ for all infinite $S$, then the axiom of choice holds.

(If there were an injection $X \times X \to X$ for every $X$, then by the Cantor-Schröder-Bernstein theorem, $|X| = |X \times X|$ for all $X$.)