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The axiom of infinity is usually presented as follows: $$\exists x((\exists y(\forall z\,\neg(z\in y))\land y\in x)\land\forall s(s\in x\implies\bigcup\{s,\{s\}\}\in x))$$ Now, this does in fact state the existence of the empty set. However, I have often seen people say you can prove the existence of the empty set using infinity and separation: by applying existential instantiation to the axiom of infinity, you have an infinite set $a$, so that you can define

$$\varnothing=\{x\in a:\neg(x=x)\}$$

But why would you need this, when the existence of such set has already been stated in the axiom of infinity?

Anne Bauval
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Elvis
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    They may be working from this. – J.G. Dec 21 '24 at 17:48
  • @J.G. why wouldn't people wanna use this simpler version, if it's equivalent? – Elvis Dec 21 '24 at 19:01
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    Why? I can only guess at individuals' psychology. Some people like to state what looks like a weaker axiom, then show it's equivalent to the stronger-looking version. Such people may prefer this because, with some slightly different supporting axioms, the weaker-looking version could be genuinely weaker, in which case it's worth highlighting which version we're committing to. – J.G. Dec 21 '24 at 19:48
  • I don't understand why this question has been marked as a duplicate of that question. As far as I can tell, the OP is asking: why do people derive that the empty set exists using one approach, when there is another approach that seems easier? (Please correct me if I am wrong.) It's not about the axiom of the empty set being redundant in ZFC (indeed, the axiom of the empty set is not even mentioned in the question). – Joe Dec 22 '24 at 20:47
  • @Joe yes, that's exactly whay I was asking. Thank you. – Elvis Dec 23 '24 at 02:12
  • I've updated my answer, since I've thought of a better reason why one might prefer the approach using separation. – Joe Dec 23 '24 at 17:39
  • @Elvis: Is there anything I can do to improve my answer? – Joe Dec 29 '24 at 13:50
  • @Joe no, sorry, I forgot to mark it as accepted. – Elvis Dec 29 '24 at 14:32

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As you say, the existence of the empty set seems to be contained in the statement of the axiom of infinity. However, to formally prove the existence of the empty set in this manner, you still need to take a few steps. First, you would need to use the inference rule that allows you to conclude $\exists x(P(x))$ from $\exists x(P(x)\land Q(x))$. In this case, you would deduce that $\exists x\exists y(\forall z(z\not\in y)\land y\in x)$. Then, you would have to do further work to deduce the statement $\exists y(\forall z(z\notin y))$. None of this is particularly difficult, of course, but neither is the approach using separation.

One reason to prefer the argument using separation is that it is more easily generalisable. Suppose, for instance, that you are interested in working in a version of set theory without the axiom of infinity. Perhaps the axioms you have chosen can prove $\exists x(x=x)$, but they can't prove the existence of an inductive set. Then, you can prove the existence of the empty set as follows: let $x$ be such that $x=x$. Assuming that separation is one of your axioms, we can conclude that there is a set $y$ such that for all $z$, $z\in y$ if and only if $z\in x$ and $z\neq z$. Then, we can insantiate $y$, and prove that $y$ satisfies the definition of the empty set.

Ultimately, though, it doesn't matter what path you take to proving the empty set exists; indeed, some authors adopt it as a (redundant) axiom when they are introducing ZFC. There are many, many variations on how the axioms of ZFC are presented, and all we care about in the end is that your set of axioms generates the same theory as my set of axioms.

Joe
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  • If there is something wrong with my answer, then I'd like to know. – Joe Dec 21 '24 at 19:36
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    That logical inference rule is already assumed in the background machinery of FOL. – Alex Dec 21 '24 at 22:58
  • @Alex: Yes, agreed – my point is that is that you still have to use an inference rule like that to conclude that the empty set exists, much like the “standard argument” also uses a few steps. The OP seemed to be suggesting in their question that the existence of the empty set is immediate from the axiom of infinity using their argument, and that’s not the case. Therefore, I don’t see a strong reason to prefer one approach or the other. – Joe Dec 22 '24 at 00:19
  • @Alex: Assuming that you downvoted this post, do you still judge it to be incorrect/misleading? I tried to clarify what I meant with an edit. – Joe Dec 22 '24 at 20:57
  • No, it is correct now. I removed the downvote and upvoted. – Alex Dec 23 '24 at 01:46