Union on the empty set and the set containing the empty set I'm trying to get a clearer sense of some of the consequences the axiom of unions has on the empty set. I understand that $\emptyset = \{\} \not= \{\emptyset\}$. 
But assuming the following identities are correct, I don't understand why $\bigcup\emptyset = \bigcup \{\} = \bigcup \{\emptyset\}$.
It's likely that I'm floundering on some minutiae of set theory, but it's making me uncomfortable, and I'd like to know what I'm missing. 

$z \in \bigcup A$ iff there exists $y \in A$ for which $z \in y$. No such $z$ exists for $A = \emptyset$ or $A = \{ \emptyset \}$.
Indeed, for the former, we have no $y$; for the latter, there is a $y$, but it's empty, so there's no $z$.
