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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)?

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    $\begingroup$ 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. $\endgroup$ Jan 28, 2014 at 1:58
  • $\begingroup$ @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 $\exists x\,\ x=x$. $\endgroup$
    – VF1
    Jan 28, 2014 at 2:00
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    $\begingroup$ 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. $\endgroup$ Jan 28, 2014 at 2:02
  • $\begingroup$ You might want to try and check out these probable duplicates: math.stackexchange.com/questions/278863/… and math.stackexchange.com/questions/457251/axiom-of-empty-set and probably there are one or two more hiding out there. $\endgroup$
    – Asaf Karagila
    Jan 28, 2014 at 2:04
  • $\begingroup$ @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. $\endgroup$
    – VF1
    Jan 28, 2014 at 2:07

2 Answers 2

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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")

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  • $\begingroup$ 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. $\endgroup$
    – VF1
    Jan 28, 2014 at 3:16
  • $\begingroup$ @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. $\endgroup$
    – user14972
    Jan 28, 2014 at 4:40
  • $\begingroup$ OK - that answers everything, thanks! $\endgroup$
    – VF1
    Jan 28, 2014 at 5:58
  • $\begingroup$ 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. $\endgroup$
    – user104955
    Jan 28, 2014 at 8:12
  • $\begingroup$ @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 $\varnothing$. $\endgroup$
    – user14972
    Jan 28, 2014 at 15:40
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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.

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  • $\begingroup$ If I recall correctly, the quantifiers are traditionally omitted from the axioms of first-order logic. $\endgroup$
    – Asaf Karagila
    Jan 28, 2014 at 2:05
  • $\begingroup$ @AsafKaragila: Fixed. $\endgroup$ Jan 28, 2014 at 2:05
  • $\begingroup$ @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 $\forall x\,\ x = x$. $\endgroup$
    – VF1
    Jan 28, 2014 at 2:06
  • $\begingroup$ @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! $\endgroup$
    – VF1
    Jan 28, 2014 at 2:10
  • $\begingroup$ @VF1: I've updated my answer again. $\endgroup$ Jan 28, 2014 at 2:10

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