Counterexample to a Set that cannot be Written as a Cross Product Let $f:\mathbb{R} \to \mathbb{R}$ be a function, and let $S = \{(x,f(x)):x \in \mathbb{R}\}$. Then for any $A \subseteq \mathbb{R}$ and $B \subseteq \mathbb{R}$, $S\neq A \times B$. 
My first thought was that this statement was false and I was searching for a counterexample. However the more I think about it I believe it is true because $A \times B$ would cause each $x \in A $ to be paired with multiple $f(x)\in B$. Thus $f$ would not be a function.
Any thoughts regarding this question would be appreciated.

$S$ is of the form $A \times B$ iff $f$ is a constant: suppose $B$ has only one point $b_0$. Then, for any $x$, $(x,f(x)) \in A \times B$ so $f(x)=b_0$ and $f$ is a constant. If $B$ has two distinct points $b_1$ and $b_2$ pick any $a \in A$. Then $(a,b_1)\in A\times B=S$ so $b_1=f(a)$. Similarly, $b_2=f(a)$ so we get a contradiction to the fact that $b_1 \neq b_2$. Thus you can take any non-constant $f$ to get a counterexample. 
