$\mathbb{R}^2$ and the axiom of choice Is choice required to guarantee that $\mathbb R^2 := \mathbb R \times \mathbb R$ – or $\mathbb R^n := \displaystyle\prod_{k=1}^n \mathbb R$ in general – isn't the empty set $\varnothing$? If not, what's an example of a Cartesian product of two non-empty sets not guaranteed to be non-empty (without choice)?
 A: If $X$ is nonempty and $\kappa$ is an arbitrary cardinal, then you do not need $AC$ to prove that $X^{\kappa}$ (the cartesian power of $X$) is nonempty: since $X$ is nonempty, there exists $x_0\in X$. Then define $f\colon\kappa\to X$ by $f(i) = x_0$ for all $i\in\kappa$. This is an element of $X^{\kappa}$, so the latter is not empty. No choice needed. 
(If $\kappa=0$, you get the empty function, which is the unique element of $X^{\varnothing}$)
A: The ability to make (and remember) finitely many choices during a proof is built into the rules of first-order logic. You don't need a specific set-theoretic axiom to do that.
By induction, you can then prove in general that choice functions for a family of sets you know has finite cardinality must also exist.

Independently of this, $A^B$ is always nonempty when $A$ is. This also does not require any axiom of choice -- you can just choose one element $a\in A$ and then consider the function that maps every element of $B$ to this $a$.
A: Choice is not required for finite products of sets.
In the particular case of $\mathbb{R}^2$ it's even easier: $(0,0)\in\mathbb{R}^2$.
