I'm just begining to build the systems of numbers based on the axioms of set theory ($\mathsf{ZF}$). Accordingly the axiom of infinity is no more than assuming the existence of $\mathbb{N}$ (of course the axiom is formulated in terms of the existence of an inductive set). Now the original statement in terms of logic is

$\exists I(\emptyset \in I\wedge(\forall x(x\in I \longrightarrow x^{+}\in I)))$, understanding $x^{+}:=x\cup\{x\}$.

This is the way I learnt it first. After some days I have learnt it I just worried about understanding the concept. But now that I try to use it again I just came up (I don't know why, maybe it's because of the similarity to the notation of the other axioms in my attempt to remember the original notation) with this apparently equivalent statements:

$1) \exists I\forall x(x\in I\longrightarrow (x=\emptyset \vee x^{+}\in I))$

$2) \exists I\forall x(x\in I\longrightarrow (\emptyset \in I\wedge x^{+}\in I))$

If I put the original statement in the equivalent form

$\exists I\forall x(\emptyset \in I\wedge(x\in I \longrightarrow x^{+}\in I))$

Then to prove the equivalence I probably should deal with the main part that is bound with the quantifiers. So I want to ask how to prove that my statements are equivalent or not, logically speaking.

Note: Intuitively both statements are false because I could form I=$\emptyset$ and this is not and inductive set.


You are correct in your note. Both the formulations are susceptible to fail because taking $I=\varnothing$ satisfies the inner formula vacuously.

The first formulation is incorrect as well because $\{\varnothing\}$ also satisfies it.

The second formulation is correct, but note that $\varnothing\in I$ can be moved out of the implication and we have the same formulation as the last form of the axiom of infinity.

  • $\begingroup$ I see. So, just to confirm, basically you say that I necessarily need to use the other axioms to prove the equivalence? I think this is the point of my confusion. $\endgroup$ – Daniela Diaz Jul 13 '13 at 10:12
  • $\begingroup$ You already use a lot of the other axioms when you write $x^+$ and $\varnothing$. $\endgroup$ – Asaf Karagila Jul 13 '13 at 10:21
  • $\begingroup$ Does it make sense if I consider the statements just as pure logical symbols, like for example $\in$ is just some binary relation, $\emptyset$ as a constant and also $^{+}$ as a unary relation, all without meaning attached, and without considering the other axioms? $\endgroup$ – Daniela Diaz Jul 13 '13 at 10:31
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    $\begingroup$ You can't completely detach the meaning from this. Replace this by $\prec$ for example, then $y\prec\{x\}$ if and only if $y=x$. But why would something like that exist? Suppose we have a universe with only two objects $x,y$ and $x\prec y$, and define $\varnothing=x, x^+=y^+=y$. Then we have that $y=\{\varnothing\}$ and therefore the first formulation holds. Even worse, if you instead take $y^+=x,x^+=y$ and $y\prec y,x\prec y$ then $y$ satisfies all the formulations given above. $\endgroup$ – Asaf Karagila Jul 13 '13 at 10:45
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    $\begingroup$ You can't prove some syntactical equivalence here because we rely, a lot, on the fact that $\in$ is extensional, and the existence of unions, power sets, and replacement. $\endgroup$ – Asaf Karagila Jul 13 '13 at 10:46

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