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The axiom in infinity in ZF states the following:

$$\exists x ((\emptyset \in x )\land \forall y (y \in x \implies y\cup \{y\} \in x) $$

My notes then defines

$$\mathbb{N}:= \bigcap\{z \in \mathcal{P}(x): \emptyset \in z \land \forall y(y \in z \implies y \cup \{y\} \in z)\}$$

But this seems to depend on the set $x$, so is this actually well-defined? I.e. if I choose another set $x'$ satisfying the same property and I define $\mathbb{N}$ w.r.t. $x'$, then will I even get the same set?

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No, this does not depend on the set $x$.

If $x$ and $x'$ both have this property, then $x\cap x'$ also has this property. It follows that $\Bbb N$ is defined by intersecting over a subset of both $x$ and $x'$, which means that it does not depend on the choice of $x$.

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  • $\begingroup$ How could I have missed this! Thanks! $\endgroup$ – user745578 Mar 30 at 19:57
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    $\begingroup$ Because you haven't answered this question several times in the past... teaching set theory is helpful to notice obvious things. :-) $\endgroup$ – Asaf Karagila Mar 30 at 19:59
  • $\begingroup$ I'm sorry. I searched on the site but couldn't find anything. $\endgroup$ – user745578 Mar 30 at 20:00
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Yes you will, because the first property ensures that $x$ (or any set satisfying the property) contains at least all finite ordinals (it contains $\emptyset$, but then it contains $\{\emptyset\}=1$, $\{\{\emptyset\},\emptyset\}=2$, etc.).

So whatever $x$ you start from it contains $\mathbb{N}$.

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    $\begingroup$ This looks circular. $\endgroup$ – user745578 Mar 30 at 19:57
  • $\begingroup$ I intended it more as an explanation than a logical demonstration. $\endgroup$ – Captain Lama Mar 30 at 19:59
  • $\begingroup$ @user745578 It is not necessarily. If you define "finite" ordinal as an ordinal, all of whose elements are successors or zero, then this shows by induction that every finite ordinal is in every inductive set. $\endgroup$ – Jonathan Mar 30 at 20:36

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