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I've been thinking about set theory recently, and one of the things I noticed was that if we restrict to finite sets, then the power set can be constructed through repeated application of the axioms of separation, pairing, and union. And that made me wonder how ZFC looks when we replace the Axiom of Infinity with its negation.

So I have a few different questions. For this question, "finite set theory" refers to the Axioms of Extensionality, Empty Set, Pairing, Union, Separation, and the negation of Infinity.

  • Can finite set theory to prove that no infinite sets exist? The negation of infinity cuts off the cumulative hierarchy at $V_\omega$, which might be enough given Foundation, but it's not entirely clear to me that the nonexistence of an inductive set implies the nonexistence of any infinite set.

  • Of course, the cumulative hierarchy requires the power set operation, so this raises my initial question. Is the Axiom of Power Set a theorem of finite set theory?

  • Are the Axioms of Replacement and Choice theorems of finite set theory? Since they both involve functions, a related question is whether functions can be constructed in finite set theory without the power set operation.

  • Given responses below, it seems like the Axiom of Foundation may be more important than I thought. How does it affect the answers to these if added to "finite set theory"?

eyeballfrog
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  • Related: http://math.stackexchange.com/questions/107639/what-are-the-consequences-if-axiom-of-infinity-is-negated – templatetypedef Jan 03 '17 at 00:04
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    In ZF-Inf the axiom of Foundation is equivalent to V=$\cup_{x\in On}V_x. $ In ZF-Inf+$\neg$ Inf we have $\omega$ =On and V=V${\omega}.$ And Choice is satisfied by (V${\omega},\epsilon$) as we can define, in (V$_{\omega},\epsilon$), a canonical well-order of V..... I dk whether you can prove Power (P) in ZF-Inf-P+$\neg$ Inf....... Note the example in Eric Wofsey's answer does not satisfy every instance of Replacement. – DanielWainfleet Jan 07 '17 at 06:04

2 Answers2

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Here is a model of your "finite set theory" (including Foundation) in which there is an infinite set and Power Set and Replacement fail. Let $A=\{\emptyset,\{\emptyset\},\{\{\emptyset\}\},\{\{\{\emptyset\}\}\},\dots\}$ and let $M$ be the closure of $V_\omega\cup\{A\}$ under Pairing, Union, and taking subsets (so if $X\in M$ and $Y\subseteq X$ then $Y\in M$). It is clear that $M$ satisfies all of your axioms except possibly the negation of Infinity. To prove the negation of Infinity, note that if $X\in M$, then the transitive closure of $X$ contains only finitely many elements of cardinality $>1$ (since this is true of $A$ and for every element of $V_\omega$, and is preserved by taking pairs, unions, and subsets). So $M$ cannot contain any inductive set.

However, $M$ does contain an infinite set, namely $A$. It is also clear that $M$ fails to satisfy Power Set, since $M$ contains every subset of $A$ but $\mathcal{P}(A)\not\in M$ (either by the criterion mentioned above, or by noting that every element of $M$ is countable). Replacement also fails, since the usual recursive definition of the obvious bijection $A\to\omega$ can be implemented in $M$, so Replacement would imply $\omega$ is a set.

This model does satisfy Choice in the form "if $X$ is a set of disjoint nonempty sets then there is a set that contains one element from each of them" (since $M$ contains all subsets of $\bigcup X$). It does not satisfy Choice in the form "if $X$ is a set of nonempty sets then there exists a choice function $X\to\bigcup X$", basically because it is very hard to construct functions as sets in $M$ (for instance, if $X=A\setminus\{\emptyset\}$, the unique choice function for $X$ would have infinitely many 2-element sets in its transitive closure). Probably it is possible to build a model where any reasonable form of Choice fails, but I don't know how exactly to do that at the moment.

Eric Wofsey
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  • When you say it satisfies choice, what exactly do you mean? AC has infinitely many equivalents, so which one do you think of when you say that choice holds? – Asaf Karagila Jan 02 '17 at 21:51
  • Right, that's a good point (I've clarified the version I have in mind). Probably this model doesn't satisfy any version of Choice that involves talking about functions, since it is not closed under forming cartesian products. – Eric Wofsey Jan 02 '17 at 21:57
  • Very nice - +1! – Noah Schweber Jan 02 '17 at 22:04
  • Well that settles that. Thanks! It does raise bunch of other questions that I'll have to think about (but that's math for you, I guess). – eyeballfrog Jan 02 '17 at 22:17
  • Here's a a paper about that: https://doi.org/10.1002/malq.19930390138 – quiliup Aug 15 '21 at 18:41
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Since you don't include Foundation, we can get a very strange model indeed. Namely, consider an $\omega$-model $M$ of ZF-Foundation+"there is an infinite set $A$ of Quine atoms" (a Quine atom is a set satisfying $x=\{x\}$, and the existence of many Quine atoms is consistent with ZF-Foundation). Now look at the structure $W$ which is the closure of $$V_\omega^M\cup A\cup\{[A]^n: n\in\mathbb{N}\}$$ under Pairing, Union, and Separation. Here "$[X]^k$" denotes the set of $k$-element subsets of $X$.

$W$ satisfies finite set theory, but not Powerset ($\mathcal{P}(A)$ doesn't exist) or Choice (there is no choice function for the set of two-element subsets of $A$).

I am not sure whether $W$ satisfies Replacement, but I suspect it does, and I don't immediately see how to add a failure of Replacement. Meanwhile, incorporating Foundation seems difficult.

If we define $W$ as the Pairing/Union/Separation-closure of $$V_\omega^M\cup A\cup\mathcal{P}(A)\cup\{[\mathcal{P}(A)]^n: n\in\mathbb{N}\}$$ instead, where $M$ satisfies "$A$ is an amorphous set of Quine atoms", then I believe we get a model of finite set theory together with the negation of Replacement and Powerset; powerset fails since we don't have $\mathcal{P}^2(A)$, and Replacement fails since the image of the map sending each subset of $A$ to its cardinality if it is finite, or the cardinality of its complement otherwise, does not exist.

Noah Schweber
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  • The imagine of an amorphous set is amorphous. So I suspect Replacement holds. But if you look at the power set of the amorphous set, this is probably not the case, since you could define the map sending a finite subset of the amorphous set to its size. – Asaf Karagila Jan 02 '17 at 20:59
  • @AsafKaragila Good point, I've edited. – Noah Schweber Jan 02 '17 at 21:02
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    Interesting. I have difficulty getting a handle on the implications of Foundation, as it's not as "obvious" to me what it's doing for set theory. Given your answer, I think I should have included Foundation in the original question. Should I make an edit about that? – eyeballfrog Jan 02 '17 at 21:04
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    @eyeballfrog I think since Foundation adds a layer of difficulty, I'm happier with Foundation not included, but it's up to you. – Noah Schweber Jan 02 '17 at 21:06
  • @Noah Schweber I've made the edit, as I think Foundation gets it closer to what I was envisioning when I asked the question. I'd like this answer to stay up, though, as it's really interesting. – eyeballfrog Jan 02 '17 at 21:11
  • Your $W$ does not satisfy Union, since the only infinite sets it contains are subsets of $A$. So for instance, $A\cup{\emptyset}$ is not in $W$, even though ${A,\emptyset}$ is. Also, as far as I can tell your $W$ does satisfy Choice, since you only need Choice to hold for sets, and the collection of 2-element subsets of $A$ is not a set in $W$. – Eric Wofsey Jan 02 '17 at 21:12
  • @EricWofsey Dangit, of course you're right. I think I've patched it; thoughts? – Noah Schweber Jan 02 '17 at 21:16
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    I think that should work, though "closure under Separation" is a bit slippery (in what structure are you going to evaluate the formulas you use to define subsets?). But it should work if you transfinitely iterate adjoining all definable subsets. You should then be able to prove that the Choice function in question does not exist because any element of your model must have a definition that only involves finitely many elements of $A$, and so is invariant under the automorphism given by any permutation of $A$ that fixes those elements. – Eric Wofsey Jan 02 '17 at 21:31