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This question is my attempt to hone in on the problem of proving the strong form of the Axiom of Regularity $A\neq\emptyset\to\exists x\in A(x\cap A=\emptyset)$ (where $A$ is an arbitrary class) from the weak form $\forall a\!:a\neq\emptyset\to\exists x\in a(x\cap a=\emptyset)$. In ZF with the negation of the Axiom of Infinity (which I abbreviate as $\sf ZF^{\neg\infty}$), all sets have finitely many elements, and this allows us to prove that sets like $x=\{\{\{x\}\}\}$ do not exist, but I don't see how to prove that the infinite family $x_n=\{x_{n+1}\}$ does not exist, since I can't collect all the elements together and use regularity on that, since the resulting class is (provably) not a set.

My ultimate goal is to prove that every set has a transitive closure (that is a set), but these sets will foil my attempt if I can't prove their non-existence. Conversely, supposing I couldn't prove this, this would mean that $\infty$ is required for the proof, so could I turn it around and prove, given that every set has a transitive closure, that $\omega\in V$?

Mario Carneiro
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    What is $ZF\lnot\infty$? – Asaf Karagila Jan 20 '13 at 17:45
  • @AsafKaragila $ZF\neg\infty$ is ZF with the Axiom of Infinity replaced with its negation "$\omega\notin V$". – Mario Carneiro Jan 20 '13 at 17:48
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    You might want to write that explicitly, and use a reader friendly $\sf Inf$ for denoting the axiom of infinity rather than the $\infty$ symbol. I recommend writing something like "Let $ZF^\times$ denote the axioms of $ZF$ when we replace the axiom of infinity by its negation", and just work with $ZF^\times$ rather than the clumsy $ZF\lnot\infty$. – Asaf Karagila Jan 20 '13 at 17:50
  • You cannot hope to prove that the existence of transitive closures implies Inf, since $V_\omega$ models ZFC-Inf and has transitive closures of all its elements. – Miha Habič Jan 20 '13 at 17:55
  • @AsafKaragila I like to work with models using the negations of infinity, choice, or regularity, so I use $ZF\neg\infty$, $ZF\neg C$ and $ZF\neg Reg$ as unambiguous shorthand. I've added clarification, though. – Mario Carneiro Jan 20 '13 at 17:57
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    Mario, note that the axiom of choice is not an axiom of ZF, therefore $ZF\lnot C$ is completely reasonable. For ZF without regularity and ZF without infinity there are relatively common and accepted notation (of the form $\sf ZF_0, ZF^\times$ and so on). It's usually a good practice to stick with pre-existing notation. – Asaf Karagila Jan 20 '13 at 17:59
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    @MihaHabič You are correct about that. I am hoping to clarify the difference between ZF without infinity, ZF with transitive closures, and ZF with infinity, and whether these are indeed each a proper strengthening of the previous system. – Mario Carneiro Jan 20 '13 at 18:01
  • @AsafKaragila I'll admit I've never seen those notations. Does $\sf ZF^\times$ mean ZF without infinity or ZF with the negation of infinity? I am trying to say the negation of infinity, not its absence (and I think $\sf ZF\neg\infty$ does that job). – Mario Carneiro Jan 20 '13 at 18:04
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    @MihaHabič Actually, my hope is to prove that $\sf ZF\neg\infty$ implies that $V=V_\omega$. Obviously we can prove that $V_\omega\subseteq V$, but to prove equality do we need transitive closures as an axiom? – Mario Carneiro Jan 20 '13 at 18:09
  • I'm not seeing the difference between the two forms you've presented for the Axiom of Regularity, aside from capitalization, so I presume that is important somehow. What is the difference between objects $A$ and objects $a$? – Cameron Buie Jan 20 '13 at 18:13
  • Well, I have to admit not to find any of the papers I've read in which those theories were referenced to. I do recommend adopting a sub-/superscript notation when omitting axioms from ZF, for example $\sf ZF^{\lnot\infty}$ or $\sf ZF^{\lnot Reg}$ or something similar. It's easier on the eyes I think. :-) – Asaf Karagila Jan 20 '13 at 18:15
  • @CameronBuie $A$ is a class, $a$ is a set. The first one is also not quantified, since you can't quantify over classes; it's essentially a theorem schema for every possible class. – Mario Carneiro Jan 20 '13 at 18:19
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    You should be able to use a compactness argument to show that after appending to the language of set theory the constant symbols $c_0 , c_1 , \ldots$ and adding as new axioms the statements "$c_n = { c_{n+1} }$" the resulting theory is consistent (if the original one was). – user642796 Jan 20 '13 at 18:40
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    @ArthurFischer Yes, but this is irrelevant, because the sequence you build this way cannot be seen from within the model. – Andrés E. Caicedo Jan 20 '13 at 19:07
  • @MarioCarneiro Your sequence $(x_n)_{n<\omega}$ is given also as a class, right? Anyway, you can still prove that every set has a rank, so what is the problem? – Andrés E. Caicedo Jan 20 '13 at 19:09
  • @AndresCaicedo The sequence itself (if I gather all elements together into a class) is a proper class, but the elements themselves (i.e. $x_n$ for some specific $n$) are all sets, and if they exist, then they contradict the statement that every set has a transitive closure, because $x_0$ exists, but $\operatorname{TrCl}(x_0)={x_n|n\in\omega}$ does not. – Mario Carneiro Jan 20 '13 at 19:31
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    Mario, you can still prove that the set theoretic rank of each set exists (and is a natural number). This solves the problem, as there can be no sequence as you fear, or its elements would have no rank. – Andrés E. Caicedo Jan 20 '13 at 19:37
  • @AndresCaicedo But that's just the problem. The proof that the sets $V_{\alpha+1}={\cal P}(V_\alpha)$ cover $V$ requires the strong form axiom of regularity, applied to $V-\bigcup_{\alpha\in{\sf On}}V_\alpha$, in order to show that sets of the type I've described don't exist. Without it, since they occupy the space outside the rank hierarchy, they have no rank, but this is no contradiction, unless you prove that all sets have a rank without using strong regularity. (By the way, I never said this, but you can prove strong regularity using transitive closures and weak regularity.) – Mario Carneiro Jan 20 '13 at 20:07
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    Oh, I see, you are right! You are just assuming weak regularity. The standard formulation of $\mathsf{ZF}^{\lnot\infty}$ replaces it with the schema of $\epsilon$-induction, from which of course it all follows easily. – Andrés E. Caicedo Jan 20 '13 at 20:10
  • Can someone make sense of the paper http://projecteuclid.org/euclid.ndjfl/1054837937 ? It looks to be an answer to my question, but I don't understand the proofs (it doesn't help that they are using weird notation). Specifically, Lemma 2.3 and 2.4 seem to be proving that if my $x_n$ exist, then there is an infinite ordinal, which proves my problem by contraposition (since $\omega\notin V$ is an axiom). – Mario Carneiro Jan 20 '13 at 20:56

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Mario,

I finally found a reference that addresses the issue you have identified.

("I found", he says. I asked at the Set Theory community on Google+, and was provided with a perfect reference.)

The theory you call $\mathsf{ZF}^{\lnot\infty}$ is not strong enough to prove the existence of transitive closures. This is the reason why the theory is usually formulated with foundation (regularity) replaced by its strengthening of $\in$-induction, namely the scheme that for every formula $\phi(x,\vec y)$ adds the axiom $$ \forall \vec y\,(\forall x\,(\forall z\in x\,\phi(z,\vec y)\to\phi(x,\vec y)\to\forall x\,\phi(x,\vec y)). $$

Over the base theory $\mathsf{BT}$ consisting of (the empty set exists), extensionality, pairing, union, comprehension, and replacement, one can prove that for any $x$, there is a transitive set containing $x$ iff there is a smallest such set, that is, there is a transitive set containing $x$ iff its transitive closure exists.

Over $\mathsf{BT}$ plus foundation, the scheme of $\in$-induction is equivalent to the statement $\mathsf{TC}$ that every set is contained in a transitive set.

Unfortunately, as I claimed above, $\mathsf{ZF}^{\lnot\infty}$ cannot prove the existence of transitive closures. To see this, start with $(V_\omega,\in)$, the standard model of $\mathsf{ZF}^{\lnot\infty}$. Let $$ \omega^\star=\{\{n+1\}\in V_\omega\mid n\in\omega\}. $$ Define $F:V_\omega\to V_\omega$ by $$ F(n)=\{n+1\},F(\{n+1\})=n $$ for $n\in\omega$, and $F(x)=x$ for $x\notin\omega\cup\omega^\star$. Now define a new "membership" relation by $$ x\in_F y\Longleftrightarrow x\in F(y) $$ for $x,y\in V_\omega$. It turns out that $$ (V_\omega,\in_F)\models\mathsf{ZF}^{\lnot\infty}+\lnot\mathsf{TC}. $$ In fact, $\emptyset$ is not contained in any transitive set in this structure. (Note that $\emptyset$ is not the empty set of this model.)

All this and much more is discussed in an excellent little paper

Richard Kaye and Tin Lok Wong. On interpretations of arithmetic and set theory, Notre Dame Journal of Formal Logic, 48 (4), (2007), 497–510. MR2357524 (2008i:03075).

(The pdf linked to above has some additional information than the version published at the Journal.) The result that $\mathsf{ZF}^{\lnot\infty}$ is consistent with $\lnot\mathsf{TC}$, and the proof just sketched, are due to Mancini. For additional details, Kaye-Wong refers to

Antonella Mancini and Domenico Zambella. A note on recursive models of set theories, Notre Dame Journal of Formal Logic, 42 (2), (2001), 109-115. MR1993394 (2005g:03061).

Andrés E. Caicedo
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    Talk about incentives to join Google+... :-) – Asaf Karagila Jan 31 '13 at 08:33
  • Showing that your model satisfies ${\sf ZF}^{\neg\infty}$ looks to be difficult, but to show that it satisfies $\neg{\sf TC}$, note that $F(n)={n+1}$ implies that $n+1\in_F n$, so we get $0\ni_F 1\ni_F 2\ni_F\dots$, and so the transitive closure is $\omega\notin V_\omega$. Nice references! – Mario Carneiro Jan 31 '13 at 09:14
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    @MarioCarneiro It is actually not difficult: If $F:V_\omega\to V_\omega$ is any bijection, then defining $\in_F$ as above gives a model of $\mathsf{ZF}^{\lnot\infty}$, except for foundation. Care is needed only to ensure that foundation holds (as in the example given above). Many other examples, and related topics, are discussed in "$\omega$-models of finite set theory", by Enayat, Schmerl, and Visser. – Andrés E. Caicedo Feb 27 '13 at 16:45