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The Wikipedia article on ZFC insists that the empty set exists since it suffices for any set to exist, since the Axiom of Specification for which we always specify "false" will construct the empty set.

I agree that in the end everything works out because the Axiom of Infinity guarantees the existence of $\mathbb{N}$, but apparently this is not needed.

Directly paraphrasing the article, ZFC is "formalized," the "domain of discourse" must be nonempty, and therefore $\exists x\,\ x = x$. I don't completely understand this. Why is the assertion $\exists x$ even true (reflexivity I can accept)?

VF1
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    I think you have misinterpreted what $\exists x\ x=x$ means. It means "there exists a set $x$ such that $x=x$". It is not two separate statements: "$\exists x$" on its own is meaningless. – Clive Newstead Jan 28 '14 at 01:58
  • @CliveNewstead I don't see why "there exists a set $x$" is meaningless, but, even supposing it is, I still don't see why I can accept $\exists x,\ x=x$. – VF1 Jan 28 '14 at 02:00
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    The (informal) translation into English isn't meaningless, but "$\exists x$" is formally meaningless. In any case, $\forall x\ x=x$ is an axiom of first-order logic, so if a set exists at all then it must satisfy $x=x$. Since the empty set axiom asserts that a set exists, it thus follows that $\exists x\ x=x$ is true. – Clive Newstead Jan 28 '14 at 02:02
  • You might want to try and check out these probable duplicates: http://math.stackexchange.com/questions/278863/the-existence-of-the-empty-set-is-an-axiom-of-zfc-or-not and http://math.stackexchange.com/questions/457251/axiom-of-empty-set and probably there are one or two more hiding out there. – Asaf Karagila Jan 28 '14 at 02:04
  • @AsafKaragila thanks for the links. I'd argue that this isn't a duplicate however. I see how the existence of the empty set follows from the existence of a set, but my question isn't about the former. – VF1 Jan 28 '14 at 02:07
  • @AsafKaragila I have found the answer to my specific question directly in the first link, thanks. – VF1 Jan 28 '14 at 02:16
  • Yes, I was just going to suggest that first link. – Asaf Karagila Jan 28 '14 at 02:17
  • Note that $\exists x ,,, x=x$ is not a well-formed formula. But $\exists x(x=x)$ is. – André Nicolas Jan 28 '14 at 03:44

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The formula $x=x$ (where $x$ is a variable) is an axiom of first-order logic, in which set theory is formulated. By generalisation it follows that $\forall x\ x=x$ is true... so if a set exists at all then it must satisfy $x=x$. Since the axiom of infinity asserts that a set exists, it must follow that $\exists x\ x=x$ is a true statement.

As for the empty set, its existence follows from the axioms of infinity and separation. The axiom of infinity gives you that a set exists... let's call it $\omega$. The axiom of separation tells you that the following is true $$\exists x[y \in x\ \leftrightarrow\ y \in \omega\ \wedge\ y \ne y]$$ but any witness $x$ to this sentence must be the empty set.

Clive Newstead
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  • If I recall correctly, the quantifiers are traditionally omitted from the axioms of first-order logic. – Asaf Karagila Jan 28 '14 at 02:05
  • @AsafKaragila: Fixed. – Clive Newstead Jan 28 '14 at 02:05
  • @CliveNewstead I would agree that this would be true if I accept the empty set axiom, but I thought the point was that we didn't need one. Without it, $\exists x,\ x = x$ wouldn't follow from the axiom $\forall x,\ x = x$. – VF1 Jan 28 '14 at 02:06
  • @CliveNewstead I found what I was looking for in Andres Caicedo's answer here - all one has to do is negate the existence theorem and prove by contradiction! – VF1 Jan 28 '14 at 02:10
  • @VF1: I've updated my answer again. – Clive Newstead Jan 28 '14 at 02:10
  • @CliveNewstead I apologize that my question may have not been clear - the empty set's existence can be proven from the axiom of specification and infinity, it's just that Wikipedia asserted that $\omega$ was not needed, and I was asking why. – VF1 Jan 28 '14 at 02:12
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It is a convention that domains of discourse should always nonempty (or syntactically $\exists x: \top$ should be a tautology).

If you reject this convention, then ZFC with the axioms of infinity and empty set omitted does indeed have an empty model.

It is by no means necessary to adopt this convention; in fact, there are quite good reasons to reject it. In my experience, texts either adopt or reject this convention without comment, so it is easy enough to go for many years without even being aware that there are split opinions on this. To wit, my participation in the talk page for the very article you link was the first time I had ever heard of it.

It is wikipedia's convention, as far as I can tell, to reserve the phrase "first-order logic" to refer to first-order logic with the adoption this convention, and use the phrase "free logic" refer to first-order logic without this convention. (that talk page was also the first time I had ever heard the phrase "free logic")

  • Thank you for the thorough answer, it does make sense. However, as I learned here, it seems that no convention is necessary. $\exists x,\ x=x$ must be true, since its negation implies $\forall x,\ x\neq x$, which is the opposite of an axiom of first-order logic. – VF1 Jan 28 '14 at 03:16
  • @VF1: It is not an axiom of first-order logic! It is an axiom of first-order logic with the convention I mentioned. $\neg \forall x: x \neq x$ is *not* a tautology if you develop first-order logic without adopting this convention: in fact, the empty set models $\forall x: x \neq x$, demonstrating that its negation cannot a tautology if you develop first-order logic so as to allow empty domains of discourse. –  Jan 28 '14 at 04:40
  • OK - that answers everything, thanks! – VF1 Jan 28 '14 at 05:58
  • It is pretty standard to include the axiom of universal instantiation (i.e. $\forall x\phi \to \phi[x/y]$) in first-order logic and to call systems which weaken this axiom free logics. I don't think it's fair to say that this is merely a convention on Wikipedia. –  Jan 28 '14 at 08:12
  • @GME: Isn't the problem is $\phi(y) \vdash \exists x \phi(x)$, rather than $\forall x \phi(x) \vdash \phi(y)$? At the very least, with category semantics, this is clear for interpretations into the empty set, since $\phi(y)$ is satisfied by any interpretation of $y$ as a generalized element of $\varnothing$. –  Jan 28 '14 at 15:40
  • @GME: I mention Wikipedia specifically, since that's the topic of discussion, and that's the extent of my knowledge on the topic. When I learned of it, all of the references I had used the phrase "first-order logic" for the weakened version rather than "free logic", and I had never felt inclined to do a survey . –  Jan 28 '14 at 15:42
  • @Hurkyl Right, but that rule follows from the contrapositive of the axiom of universal instantiation. The term 'free logic' goes back to the 60's (in a paper by Karl Lambert), and it's pretty widely used in philosophical logic. It surprises me that people were using "first-order logic" for free logics (or for some free logic in particular). If it's not too much trouble, I'd genuinely be interested to get references for this. Obviously, no worries if not! –  Jan 28 '14 at 16:41
  • @GME: I bet there is a simple explanation: e.g. that people get the idea that omitting the empty domain is merely meant as a simplifying assumption and think nothing of leaving the assumption out; that emphasizing the distinction between adopting this convention or not is needlessly nitpicky; or maybe even that category-theoretic (or other) concerns suggest that free logic is the primary object of study of first-order logic, and it is the convention of rejecting the empty domain that requires extra qualification, rather than the other way around. –  Jan 28 '14 at 17:58
  • @GME: I have neither handy, but one of the topics in Sheaves in Geometry and Logic (MacLane, Moerdijk) is about first-order intuitionistic logic and geometric logic (primary aimed at studying interpretations in toposes), and does not assume non-empty domains; and IIRC, the various logics that come up in Categories, Allegories (Freyd, Scedrov) also do not omit the empty domain. I do not recall what else I had handy to look up at the time. –  Jan 28 '14 at 18:02
  • @Hurkyl Thanks for the references - I'm not familiar with the category theoretic literature. There are interesting connections between the simplest free logic and classical logic for languages without function or constant symbols - in particular, $\phi$ is provable in CL iff $Ex_0,...,Ex_n \wedge \exists x(x=x) \to \phi$ is provable in FL (where $\phi$'s free variables are $x_0,...,x_n$). But working in FL directly is more fiddly in my experience. In any case, this is all terminological - it's just my impression that it's classical, not free logic, which is usually meant by first-order logic. –  Jan 28 '14 at 19:38
  • @Hurkyl And is what is presented in most standard logic text books. Anyway, thanks for the discussion! –  Jan 28 '14 at 19:39
  • @Hurkyl Actually, it seems that this convention may be necessary to the avoid the possibility of an empty model. According to this formulation of the Axiom of Infinity, we must presuppose the existence of $\emptyset$ in order for it to be an element of $\mathbb{N}$. – VF1 Jan 29 '14 at 01:48
  • @VF1 : $;;;$ While that may be true, the existence of an empty set follows from Specification plus any of the three formalizations of the Axiom of Infinity that I can think of (and uniqueness of the empty set follows from Extensionality). $:$ Let "isempty(x)" represent "(forall y)(not (y in x)). $:$ The "three formalizations ... I can think of" are obtained by replacing "EmptySet in $\mathbf{I}$" with one of the following: "(exists x)(isempty(x) and x in I)" or "(forall x)(isempty(x) implies x in I)" or "((exists x)isempty(x) implies (exists x)(isempty(x) and x in I))". $;;;;;;$ –  May 10 '14 at 19:57
  • @RickyDemer So, basically, the difference between your formalization and the one in Wikipedia is that there's an additional existential qualifier $\exists\emptyset$ in the Axiom of Infinity? – VF1 May 10 '14 at 22:10
  • @VF1 : $;;;$ One of my formalizations does not have that qualifier, and one of them has that qualifier on the left side of its implication. $:$ (Also I don't know if you realized this point: "the one in Wikipedia" is not quite formalized, because $\emptyset$ is not in the language of ZF.) $;;;;;;$ –  May 11 '14 at 00:20