# 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.

• The union of the empty set with any set whatsoever is the other set. That makes sense, right? You have a bucket full of stuff, and you add to it the contents of an empty bucket. The result is the same bucket full of stuff you started with. – user4894 Oct 2 '17 at 22:29

$z \in \bigcup A$ 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 $z$.

• Thanks for a very instructive response that helped me get an even better sense of the axiom of union. – innumeratus Oct 2 '17 at 22:42

We can view $\bigcup$ as a (class) function $$\bigcup \colon \mathrm{Set} \to \mathrm{Set}$$
and viewed as such, your question is in instance witnessing that this function is not injective. There are many such examples, e.g. $$\bigcup \{ \{a\}, \{a,b\} \} = \bigcup \{ \{a,b\} \} = \{a, b \}$$ which may, at first, not be very intuitive but readily follows from the definition. With a little practice, this will become second nature in no time.