Assume that $\forall s (\emptyset \in s)$. Let $f(\hat{s})$ denote $\{x|x$ is not a set$\textup{ is not a set.}\}$. By definition, $f(\hat{s})$ is a set and thus the first axiom holds, so that $\emptyset \in f(\hat{s})$. By definition, the empty set $\emptyset$ is a set and thus $\emptyset \in f(\hat{s})$ is not the set of all non-set items, which is a contradiction.

Therefore, are all mathematical objects sets? Or does the set of all non-set objects simply not exist?