Using ZFC axioms, prove that the set $\{ \emptyset \}$ exists. As the title describes, I want to prove that the set $\{ \emptyset \}$ exists, using ZFC axioms. I have an answer that I wish to check if I understood ZFC correctly.
Is it that simple as:
1) The empty set axiom - There is a set having no elements.  we get $\{ \}$.
2) The Power Set axiom - For every set $A$, there is a set $B$ whose elements are the subsets of $A$. We get $\{ \emptyset \}$.
Thanks.

This really depends on what axioms you have at your disposal.
Your suggestion works, if you have the empty set axiom and power set. You can also use the empty set and pairing. You can also use just the axiom of infinity (well, an extensionality, you can't do anything without extensionality!):
Let $A$ be an inductive set whose existence is guaranteed by the axiom; then by the axiom $\varnothing\in A$, and therefore $\varnothing\cup\{\varnothing\}\in A$. In particular, $\{\varnothing\}$ exists.
