Cardinality of Cartesian Product of Uncountable Set with Countable Set Is it true that if $I$ is an infinite set, then $I\times \mathbb{N}$ has the same cardinality as $I$? I believe it, but I have minimal background in set theory. My guess is that we can construct an injection from $I\times \mathbb{N}$ to $I,$ but I don't see an obvious way to do so.
 A: As mentioned in the comments, the equality of cardinals $\alpha \cdot \beta = \max\{\alpha,\beta\}$ holds when at least one of $\alpha$ and $\beta$ is an infinite cardinal. In particular, if $I$ is infinite with cardinality $\alpha$, then $|I\times\Bbb N| = \alpha\cdot\aleph_0 = \max\{\alpha,\aleph_0\} = \alpha$.
The proofs I've seen of $\alpha \cdot \beta = \max\{\alpha,\beta\}$ all go through the simpler-looking special case $\alpha \cdot \alpha = \alpha$ (again for $\alpha$ infinite). Unfortunately, I don't know a proof of this special case that avoids well-ordering and ordinals, although it's still simpler to prove since one can use transfinite induction on $\alpha$. Fortunately, the derivation of the full case from this special case is easy: if $\alpha > \beta > 1$ then $\alpha \le \alpha + \beta \le \alpha + \alpha = \alpha\cdot2 \le \alpha\cdot\beta \le \alpha\cdot\alpha = \alpha$ (where all inequalities follow easily from definitions); hence we must have $\alpha\cdot\beta=\alpha$.
