Originally Posted by

**Tikoloshe** Sure would. Now there is a small point that I think shenanigans87 initially tried to make. That is the definition of a function includes its domain. In some contexts of discrete mathematics there is a distinction between the domain and first set of your cartesian product.

That is, let $\displaystyle f:A\to B$ be a function from $\displaystyle A$ to $\displaystyle B$. Then $\displaystyle f\subset A\times B$ such that $\displaystyle \forall a\in\mathrm{Dom}(f)$ there is a unique $\displaystyle b\in B$ such that $\displaystyle (a,b)\in f$.

My point is the (possible) distinction between $\displaystyle A$ and $\displaystyle \mathrm{Dom}(f)$. Depending on what definition you choose to find useful, you might demand that the two sets are always equal for a function. In this case, $\displaystyle f=\{(0,3),(2,2)\}$ is **not a function on** $\displaystyle \mathbb{R}$, but **is a function on the set **$\displaystyle \{0,2\}$.