The empty set is a function. Let A $\neq$ $\emptyset$. We know that the following is true:
0) $f$: $\emptyset$ $\rightarrow$ A is a function & $g$: A $\rightarrow$ $\emptyset$ is not a function.
We also know the following are true:
1) $g$ $\subseteq$ A $\times$ $\emptyset$ = $\emptyset$. So $g$ = $\emptyset$. Hence: A $\times$ $\emptyset$ = $g$.
2) $f$ $\subseteq$ $\emptyset$ $\times$ A = $\emptyset$. So $f$ = $\emptyset$. Hence: $\emptyset$ $\times$ A = $f$.
Some obvious consequences of this data: $f$ is a function, $g$ is not a function, $f$=$\emptyset$ & $f$=$g$ (by extensionality). This gives us:
3) $\emptyset$ is a function & $\emptyset$ is not a function (since $f$ and $g$ are the same object).
This is a contradiction.
Question: It's clear that $f$ vacuously satisfies the conditions required for a set to be a function and $g$ fails to satisfy those conditions. But since $f$ and $g$ just are the same set (by 1) & 2) they are the empty set), I'm having difficulty articulating why this doesn't lead to a contradiction. Where have I gone wrong?