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Basically the problem is that given an inductive set $X$ we can define the successor function on $X$ such that $S:X\longrightarrow X$ and for all $x\in X$, $S(x)=x\cup \{x\}$. So, one of Peano axioms says that such a function is injective on $\omega$. My attempt in trying to prove that $S$ is injective uses the axiom of regularity:

Let's suppose that for some $x$ and $y$ in $X$ we have $S(x)=S(y)$, meaning $x\cup \{x\}=y\cup \{y\}$. As $x\in x\cup \{x\}$ then $x\in y$ or else $x=y$. If $x=y$ then $S$ is injective. If $x\in y$ but $x\neq y$ then since $y\in y\cup \{y\} $ and $y\cup \{y\}=x\cup \{x\}$, so $y\in x$. Then $x\in y$ and $y\in x$, which is a contradiction. Therefore $S$ is injective.

Another attempt that I'm trying is by means of induction since $X$ is inductive but till now the only thing that comes to my mind is using the same argument.

How can I prove it without using the axiom of regularity? Is it possible?

Asaf Karagila
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Daniela Diaz
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    On ordinals, the successor function is injective without any appeal to regularity. – Andrés E. Caicedo Dec 02 '13 at 22:02
  • @AndresCaicedo I prefer to say that on ordinals regularity is a theorem, not an axiom. – Hagen von Eitzen Dec 02 '13 at 22:04
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    @Andres: But $X$ is just an inductive set, not $\omega$. – Asaf Karagila Dec 02 '13 at 22:05
  • @HagenvonEitzen Go ahead. It is kind of silly to call it a theorem, though: $\in$ is well-founded since it is well-ordered. It holds by definition. – Andrés E. Caicedo Dec 02 '13 at 22:06
  • @AndresCaicedo Well, as you need to say "since", it is a (very shallow) theorem. – Hagen von Eitzen Dec 02 '13 at 22:07
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    @AsafKaragila The question seems to be to show that $S$ is injective on $\omega$. – Andrés E. Caicedo Dec 02 '13 at 22:07
  • @HagenvonEitzen Yes. $(p\land q)\to p$. – Andrés E. Caicedo Dec 02 '13 at 22:08
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    @Andres: It's unclear. I read it as asking if the successor function is injective on every inductive set $X$, whereas the fact it is inductive on $\omega$ is one of the Peano axioms (which is incorrect, Daniela, the Peano axioms are in fact theorems when working in $\sf ZF$). – Asaf Karagila Dec 02 '13 at 22:09
  • My reading is the same as @Asaf’s. – Brian M. Scott Dec 02 '13 at 22:14
  • @AsafKaragila The argument I give is for any inductive set $X$, in particular it works for $\omega$ definided as the smallest inductive set. What I'm trying is to deduce from this definition of $\omega=\mathbb{N}$ the Peano axioms. – Daniela Diaz Dec 02 '13 at 22:18
  • Daniela, I don't understand your last comment. – Asaf Karagila Dec 02 '13 at 22:19
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    @AsafKaragila The idea is that having defined $\mathbb{N}$ as the smallest inductive set, then the successor function on it is injective, which is one of the Peano axioms (I'm trying to prove, just like you said, that Peanos axioms are theorems that can be deduced from the axioms of set theory). I took any inductive set $X$ to prove that because in particular $\mathbb{N}$ is inductive. – Daniela Diaz Dec 02 '13 at 22:25
  • I see. Let me write an answer addressing this (and you might want to edit this information into your question as well). – Asaf Karagila Dec 02 '13 at 22:43

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If we abandon regularity, we may indeed face some weird sets $a\ne b$ such that $a=\{b\}$ and $b=\{a\}$. Then if $a,b\in X$ (e.g. $X=\omega\cup\{a,b\}$), injectivity fails indeed. But if you want to work only with the case $X=\omega$ (the smallest infinite ordinal), all is fine: Ordinals are regular automatically.

Hagen von Eitzen
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In the comments Daniela mentions that the goal is to show that indeed the successor function is injective on $\omega$. As Hagen shows, if we take an arbitrary inductive set then the axiom of regularity is needed.

But if we reduce to linearly ordered inductive sets, i.e. $X$ is inductive and for every $x,y\in X$ either $x=y$ or $x\in y$ or $y\in x$, then we can prove that $S$ is injective without an appeal to the axiom of regularity. (Note that we are talking about a sharp linear order, i.e. irreflexive, transitive and total. These properties imply that the order is strongly antisymmetric: $\forall x\forall y(\lnot(x<y\land y<x))$. However the following proof can be easily modified to accommodate the case where we take reflexive, antisymmetric and transitive.)

Suppose that $X$ is linearly ordered and inductive, and $x,y\in X$ such that $x\cup\{x\}=y\cup\{y\}$, if $x=y$ we are done, otherwise $x\in y$ and $y\in x$, however since $(X,\in)$ is [strongly] antisymmetric this cannot happen so it is a contradiction.

Asaf Karagila
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  • Well, here I have this question: how can you prove without using the axiom of regularity that the antisymmetryc property follows in the order of $\mathbb{N}$. – Daniela Diaz Dec 02 '13 at 23:08
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    You just have to show it is linearly ordered by $\in$. It is irreflexive and transitive. Either you do that by taking the intersection of all inductive sets and showing that it has to be a linearly ordered set; or that you use the construction of $\omega$ as the closure of ${\varnothing}$ to the successor function. – Asaf Karagila Dec 02 '13 at 23:12
  • @DanielaDiaz See p.43-44 of Hrbacek and Jech's Introduction to Set Theory for a proof that $(\Bbb{N},<)$ is a linearly ordered set without assuming axiom of regularity. – 19021605 Sep 17 '25 at 03:32