# 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.

• Yes, this is correct. Good job! – Stefan Mesken Nov 23 '16 at 8:33
• IIRC some presentations of ZFC have (1), but most don't. I don't see how it is avoidable since the Axiom of Infinity refers to the empty set. See en.wikipedia.org/wiki/Axiom_of_infinity – Dan Christensen Nov 23 '16 at 17:02

By the empty set axiom (or infinity), $\emptyset$ exists. Now letting $x=y= \emptyset$, the pairing axiom implies the existence of the set $z = \{x,y\} = \{\emptyset, \emptyset \} = \{ \emptyset \}$. Q.E.D.
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.
• Even that may depend on the exact formulation of INF. For example, if inductive might be defined as $(\exists z\,z\in A\land \forall t\,t\notin z)\land\ldots$ or as $(\forall z\,(\forall t\,t\notin z)\to z\in A)\land\ldots$, i.e., "There is a set in $A$ that is empty and is closed under successor" or "$A$ contains 'all' empty sets and is closed under successor" – Hagen von Eitzen Nov 23 '16 at 9:44
• You're right, of course. But I've never seen Infinity formulated as "Every empty set is an element of $A$", but rather "The empty set is an element of $A$", which in turn is translated to "There exists an elemnt of $A$ which is the empty set". – Asaf Karagila Nov 23 '16 at 11:34