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As I understand, the axiom of infinity in ZFC theory gives us an infinite set from which the natural numbers can be extracted. (See "Axiom of infinity".)

Might the following proposed axiom be a weaker version (using everyday functional notation)?

$\exists X,S,x_0( S:X\to X\land \forall a,b\in X(S(a)=S(b)\to a=b) \land x_0\in X \land \forall a\in X(S(a)\ne x_0 ))$

In words, there exists $X,S$ and $x_0$ such that $S$ is a an injective function on $X$, and $x_0\in X$, and $x_0$ has no pre-image in $X$ under $S$.

Note: It can be shown that $X$ is Dedekind-infinite, and that it is possible to extract the natural numbers (as defined by Peano's Axioms) from $X$ where $S$ is the usual successor function and $x_0$ is the $0$ (or $1$).

Peter Mortensen
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Dan Christensen
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    Dan, as indicated in the answer, $\mathsf{ZF}$ without the axiom of infinity suffices to deduce the usual axiom of infinity from this version, so they are indeed equivalent. The key idea in that answer is the use of the recursion theorem. First, you prove that you can argue by induction on ordinals. Second, you verify (using induction) that for every natural number $n$ there is a unique function $f_n$ with domain ${m\mid m<n}$ and such that $f_n(0)=x_0$ and $f_n(t+1)=S(f_n(t))$ for all $t$ such that $t+1$ is in the domain of $f_n$. – Andrés E. Caicedo Aug 28 '13 at 20:50
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    Third, you check that if $m\ne n$, then $f_m$ and $f_n$ agree on their common domain. This allows you to fourth, extract a "function" (perhaps a proper class) by "pasting together" all the $f_n$. The range of this function is in bijection with its domain, $\omega$, since $S$ is injective. The range is a definable subclass of $X$, so it is a set. By replacement, the domain of this function is a set as well, so $\omega$ exists. The answer indicates that there are even ways to avoid the use of replacement. – Andrés E. Caicedo Aug 28 '13 at 20:54
  • @AndresCaicedo It is interesting to know that my "axiom" is indeed equivalent to AOI in ZF. It really just postulates the existence of what might be called a partial Peano system (one without induction). I have shown that, from this stripped down system, a full Peano system (with 2nd order induction) can be extracted. In a way, it justifies 2nd order induction -- but only if you assume the existence of this other quite similar system postulated by my "axiom." – Dan Christensen Aug 29 '13 at 03:01
  • Yes, I think it is interesting that this apparently very different approach ends up being equivalent. It also ties up nicely with your other question. – Andrés E. Caicedo Aug 29 '13 at 03:15

1 Answers1

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Your axiom is actually equivalent to the Axiom of Infinity modulo ZFC-{Infinity}. This is very easy to see via the replacement axiom, but it follows even just using Extensionality, Pairs, Unions, Specification and Powers: you can prove a version of the recursion theorem, from which you can extract the usual set $\omega$ of von Neumann numerals.

walcher
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  • What does "modulo ZFC" mean? Are you saying the proposed axiom is not weaker? – Dan Christensen Aug 20 '13 at 14:26
  • "Modulo ZFC" means that the axioms of ZFC (without Infinity) prove their equivalence. And yes, I am saying the proposed axiom is not weaker. – walcher Aug 20 '13 at 14:30
  • Perhaps I don't have a proper understanding of the notion relative weakness, but the proposed axiom allows for $S(a)=a$ for some $a\in X$. Isn't this is not the case in ZFC? If so, doesn't this make the proposed axiom weaker in some sense? – Dan Christensen Aug 20 '13 at 14:38
  • @DanChristensen: Suppose we use the Axiom of Replacement to isolate the subset of $X$ for which $S(a) \neq a$. The injectivity and lack of "predecessor" for $x_0$ would be conserved, hence the Axiom of Infinity would be provable. – hardmath Aug 20 '13 at 14:51
  • No, this is possible in ZFC: take $X=\mathscr{P}(\omega)$, then $\omega \subset \mathscr{P}(\omega)$, so we can define $S:X\to X$ by $S(n)=n+1$ if $n\in \omega$ and $S(x)=x$ if $x\notin \omega$, then for example $S(\omega )=\omega $. The point is that the $X$ in 'weak infinity' is not the infinite set, but only some infinite set, which contains the inductive set $I={x_0,S(x_0),...}$ for which it is true that $S(a)\neq a$ for all $a\in I$. – walcher Aug 20 '13 at 14:59
  • In fact, you don't even need to go all the way to the powerset; $\omega+1$ works just fine, with $S(n)=n+1$ and $S(\omega)=\omega$. – Steven Stadnicki Aug 20 '13 at 15:12
  • The successor function in the ZFC Axiom of Infinity is often given as $S(x)=x\cup {x}$. I understand $x\cup{ x }\ne x$. So, you could not have $S(x)=x$. – Dan Christensen Aug 20 '13 at 15:13
  • @Dan Christensen I see what you mean. The problem with your $S$ is that it is not explicit about its behaviour on $X$, it is just defined to be some extension of the successor function from the base set $I={x_0,S(x_0),...}$ to the larger set $X$. But in a similar way we could extend the successor function on $\omega$ to a larger set $A$ by taking any injective map $f:A\setminus \omega \to A\setminus \omega$. – walcher Aug 20 '13 at 15:31
  • @walcher My $S$ is just an arbitrary injective function on the set $X$, at least one element $x_0$ of which has no pre-image under $S$. That seems less restrictive than the successor function defined in ZFC. It's not very easy to compare them since the $S$ in ZFC is not defined as a function in the usual sense with a domain and codomain. – Dan Christensen Aug 20 '13 at 16:16
  • Actually your $S$ is in a sense more restrictive, because the successor "function" defined in ZFC is prima facie just an operation forming a new set out of an old one, whereas your $S$ is actually a set of ordered pairs (this is how we define functions in set theory). This means that the existence of $S$ is dependent upon the existence of an infinite set $X$ (its domain), whereas the operation $x\mapsto x\cup {x}$ exists even if there are no infinite sets. The Axiom of Infinity says that there is a set consisting of $\emptyset$ and every finite iteration of the successor operation. – walcher Aug 20 '13 at 17:34
  • That the successor operation is also a function (in the sense of "set of ordered pairs") is a consequence of the non-trivial recursion theorem. – walcher Aug 20 '13 at 17:35
  • @walcher If we are to determine the relative weakness of the two alternatives, I think we have to look at the structure of the set that is postulated by each. Both must be infinite, but my guess is that the structure postulated in ZFC is somehow embedded in (a proper subset of) that postulated in the proposed axiom, thus making the proposed axiom weaker or less restrictive. – Dan Christensen Aug 20 '13 at 17:55
  • Relative weakness is determined by logical consequences. Each axiom taken together with the axioms of Extensionality, Pairs, Unions, Powers and Specification implies the other (and if we add Replacement as well it becomes plain obvious). Therefore neither axiom is 'weaker' or 'less restrictive'. – walcher Aug 20 '13 at 18:02
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    Shouldn't you say "modulo $\mathsf{ZFC}-\text{Infinity}$"? Any tautology is equivalent to the Axiom of Infinity modulo $\mathsf{ZFC}$. – Trevor Wilson Aug 20 '13 at 18:23
  • @walcher can you answer this related question? https://math.stackexchange.com/questions/4538740/equivalence-of-two-formulations-of-the-axiom-of-infinity – Emanuele Paolini Sep 25 '22 at 14:01
  • I don't think the latter half of your answer is right - I believe that you do need some amount of replacement to prove even the weak version of the recursion theorem you have in mind here. (See here.) – Noah Schweber Sep 26 '22 at 02:20