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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.

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    $\begingroup$ Yes, this is correct. Good job! $\endgroup$ – Stefan Mesken Nov 23 '16 at 8:33
  • $\begingroup$ 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 $\endgroup$ – Dan Christensen Nov 23 '16 at 17:02
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Here is an alternative proof, demonstrating that the power set axiom is not needed (this is sometimes useful).

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.

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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.

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    $\begingroup$ 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" $\endgroup$ – Hagen von Eitzen Nov 23 '16 at 9:44
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    $\begingroup$ 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". $\endgroup$ – Asaf Karagila Nov 23 '16 at 11:34

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