This is problem 5 from section 1.4. of the book Real Analysis and Probability by R.M. Dudley:
Let $X$ be a nonempty set of cardinality less than $c$ ( $c$ is defined as the cardinality of $2^\mathbb{N}$), and let $Y$ have cardinality $c$. Show that $X\times Y$ has cardinality $c$. Hint: Reduce to the case where $X$ has cardinality $c$.
I proved the reduced case by showing that $2^\mathbb{N}\times 2^\mathbb{N}$ has the same cardinality as $2^\mathbb{N}$, by considering the sequence $(z_n)$ formed by alternating the terms of two sequences $(x_n)$ and $(y_n)$. But I can't figure out how to reduce the original problem to this case.
Any ideas?
Best Answer
As Alphie says himself in the comments, it’s simply a case of Schroeder-Bernstein (picking any $x_0 \in X$):
$$\mathfrak{c} = |Y| = |\{x_0\} \times Y| \le |X \times Y| = |X| \times |Y| \le \mathfrak{c} \times \mathfrak{c} = \mathfrak{c}$$
(That last step is really the “crucial” fact here: we have an AC-free proof of that equality while without choice we would not know $\kappa\times \kappa = \kappa$ for an arbitrary cardinal number, but this proof does not use AC).