10

What is the minimum number of axioms needed for ZFC Set Theory?

I have found that Suppes' Axiomatic Set Theory lists 7 axioms, but I am not sure if this can be reduced or not. Any references on this subject would be much appreciated.

Suppes' list:

  1. Axiom of Extensionality
  2. Sum Axiom
  3. Power Set Axiom
  4. Axiom of Regularity
  5. Axiom of Infinity
  6. Axiom Schema of Replacement
  7. Axiom of Choice
Patrick Clot
  • 465
  • 12
    Note that 6 is an axiom schema, so is really a countably infinite set of axioms. All the others can be combined into one just by putting and between them and making one axiom. – Ross Millikan Jul 01 '17 at 19:47
  • @RossMillikan : I largely disagree with your comment. The axioms are expressed in a language that is amenable to algorithmic proof-checking, and WITHIN that language, item 6 cannot be expressed by only finitely many formulas, but only by infinitely many. It is only in that sense that item 6 is "really a countably infinite set of axioms". And I expect that you know that very well, but I wonder why seemingly so many people, including you, neither see anything wrong with making the assertions you make without the explanation I just gave (especially when the poster seems unaware of that), nor... – Michael Hardy Jul 01 '17 at 19:59
  • ... see anything else wrong with that point of view. As far as making such statements without that explanation goes, Herbert Enderton's book on set theory (at least the edition of which I have a copy) never once in the whole book gives that explanation. – Michael Hardy Jul 01 '17 at 19:59
  • 1
    @MichaelHardy: I said item 6 was a countably infinite set. That was exactly my point. Any finite set of axioms can be combined into one, but an infinite set cannot. I know, but have not pursued the proof, that ZFC is not finitely axiomatizable – Ross Millikan Jul 01 '17 at 20:04
  • 5
    To expand Ross Millikan's comment: Not only is the standard list of axioms of ZFC infinite, but no finite list is possible. This is a consequence of the reflection theorem. – Andrés E. Caicedo Jul 01 '17 at 20:04
  • @RossMillikan : In one very obvious sense, that particular infinite list of axioms is combined into one, on any page of a book in which an account of that "schema" is given. It is only within a certain context --- expressing the axioms in a language amenable to algorithmic proof-checking --- that it cannot be expressed by only finitely many statements. Firstly, that should be made explicit whenever the content of the schema is explained, and you didn't do that. Secondly, it seems insane to try to make that kind of language that is amenable to...... – Michael Hardy Jul 01 '17 at 20:19
  • .....algorithmic proof-checking into the way in which humans think about the subject. – Michael Hardy Jul 01 '17 at 20:21
  • 4
    @Michael: I feel like that's sort of like saying you only need finitely many decimal digits to write $1/3$, because you can write it as $0.\bar{3}$. –  Jul 01 '17 at 20:26
  • @Hurkyl : No, it's "sort of like saying" you can express $1/3$ with only finitely many symbols, by writing $1/3.$ Nobody is denying that within a language amenable to algorithmic proof checking, the axiom schema of replacement requires infinitely many formulas. What I saying is (1) that that should be explained in clear language (Enderton's book, as I said, never mentioned it); and (2) one should not lose sight of the fact that in actual human thought, the "schema" can be expressed as one statement. – Michael Hardy Jul 01 '17 at 20:36
  • I am not familiar with Suppes' book or terminology. Presumably "Sum" means "Union". What happened to the Pairing Axiom ? ... $\forall x,y \exists {x,y}.$ – DanielWainfleet Jul 01 '17 at 22:04
  • @Daniel It is provable from replacement. – Andrés E. Caicedo Jul 01 '17 at 22:17
  • @AndrésE.Caicedo. I dk how that can be done. It seems to me that we have to prove $\forall x,y \exists z; (x\in z\land y\in z)$ to use Replacement to obtain ${x,y}.$ – DanielWainfleet Jul 01 '17 at 23:15
  • @Daniel No, that's not needed. You should probably post a new question asking for details. – Andrés E. Caicedo Jul 01 '17 at 23:38
  • 6
    @MichaelHardy I don't think algorithmic proof-checking is the relevant issue here. One could surely design proof-checkers for a logic that allows schematic predicate symbols and has a rule of inference that lets you substitute arbitrary formulas for atomic formulas built using such a symbol. (Morse's book A Theory of Sets develops such a logic, though in the context of MK rather than ZF.) I think the infinitude of axioms involved in replacement is simply because one wants the underlying logic to be standard first-order logic (and not such a schematic extension). – Andreas Blass Jul 02 '17 at 00:19
  • @AndreasBlass : I would think the reason for wanting it to be standard first-order logic is that only in standard first-order logic can you simultaneously have a system of proof that is simultaneously complete (in the sense that something that is true in every structure in which the axioms are true can be proved from the axioms), sound (in the sense that everything that can be proved from the axioms is true in every structure in which the axioms are true), and effective (i.e. a proof-checking algorithm exists). But even if that's not the reason, is it not the case that it is$,\ldots\qquad$ – Michael Hardy Jul 02 '17 at 02:38
  • $\ldots,$the restriction to first-order logic that makes it impossible to express "replacement" with only one axiom? And Enderton's book doesn't say that, leaving the reader mystified unless the reader has some awareness of the nature of first-order logic. And Enderton isn't the only one; I somehow have acquired an impression that that is commonplace. It doesn't make sense to present an axiom "schema" in that way without mentioning something about that. $\qquad$ – Michael Hardy Jul 02 '17 at 02:42
  • @AndreasBlass : In 2009 a user called Gregbard edited Wikipedia's article titled "theorem" so that it begins with this: "A theorem is an idea, concept or abstraction token instances of which are formed using a string of symbols according to both the syntactic rules of a language (also called its grammar) and the transformation rules of a formal system. Specifically, a theorem is always the last formula of a derivation in some formal system each formula of which is a logical consequence of the formulas which came before it in the derivation." This seemed like an extreme case of$,\ldots\qquad$ – Michael Hardy Jul 02 '17 at 16:35
  • $\ldots,$thinking that first-order logic is everything, rather than that first-order logic is something. $\ldots\qquad$ – Michael Hardy Jul 02 '17 at 16:37
  • 1
    $\uparrow$ It seems to me that a number of errors result from thinking that currently conventional useful ways of thinking about things have been discovered to be the truth. Instead of saying Axiom 6 "is really a countably infinite set of axioms", one could say Axiom 6 is expressible in first-order logic only with infinitely many statements, and that there are reasons for using first-order logic in some contexts. @AndreasBlass – Michael Hardy Jul 02 '17 at 16:39

2 Answers2

12

The reflection principle is a theorem schema in ZFC, meaning that for each formula $\phi(\vec x)$ we can prove in ZFC a version of the principle for $\phi$. In particular, it gives us that if $\phi$ holds (in the universe of sets) then there is some ordinal $\alpha$ such that $V_\alpha\models \phi$.

It follows from this that (assuming its consistency) $\mathsf{ZFC}$ is not finitely axiomatizable. Otherwise, $\mathsf{ZFC}$ would prove its own consistency, violating the second incompleteness theorem. The (standard) list of axioms you presented is actually an infinite list, with replacement being in fact an axiom schema (one axiom for each formula).

It is perhaps worth mentioning that no appeal to the incompleteness theorem is needed: If $\mathsf{ZFC}$ is consistent, and finitely axiomatizable, then it would prove (because of reflection) that there are $\alpha$ such that $V_\alpha\models\mathsf{ZFC}$. It would then follow that there is a least such $\alpha$. But inside $V_\alpha$ there must be some $\beta$ such that $$V_\alpha\models\mbox{``}V_\beta\models\mathsf{ZFC}\mbox{''}$$ (because $V_\alpha$ is a model of set theory, so it satisfies reflection), and easy absoluteess arguments give us that then $\beta<\alpha$ is indeed an ordinal, and $V_\beta$ is really a model of $\mathsf{ZFC}$, contradicting the minimality of $\alpha$.

Asaf Karagila
  • 393,674
Andrés E. Caicedo
  • 79,201
  • 1
    (Of course, if $\mathsf{ZFC}$ is inconsistent, then it is finitely axiomatizable, and one axiom suffices.) – Andrés E. Caicedo Jul 01 '17 at 20:15
  • 1
    A similar argument applies to $\mathsf{PA}$: The (first-order) induction axiom is actually an axiom schema, and $\mathsf{PA}$ is not finitely axiomatizable, because it can prove the consistency of any of its finite fragments. – Andrés E. Caicedo Jul 01 '17 at 20:17
5

It seems to me the OP is asking whether any of the 7 assertions (specifically 6 axioms and a schema) listed are unnecessary, i.e. whether we can remove any one of them and still derive all of ZFC (similar to how Suppes removed pairing and comprehension because they can be proved from the other axioms). And the answer to that is no, none of these 7 assertions can be proven from the other 6. E.g., ZFC-Infinity+ $\neg$Infinity holds in the model $HF,$ and is thus relatively consistent to ZFC.

Elliot Glazer
  • 1,395