# Why isn't the inductive set _the_ set of natural numbers?

ZFC's axiom of infinity states:

$$\exists x (\varnothing \in x \wedge \forall y \in x (y\cup \left \{y \right \} \in x))$$

Isn't this set $x$ really $\mathbb{N}$? It wouldn't be $\mathbb{N}$ if x would contain some set $z$ that is different from $\varnothing$, $S (\varnothing)$, $S (S (\varnothing))$, etc. But we can't really prove that such a set $z$ is an element of $x$, or can we? I also know that this inductive set $x$ is not unique, but that, in the light of my question, doesn't make any sense to me either. I feel like I'm missing a very peculiar technical point, and I don't know what it is. Any help is appreciated.

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@drhab With your $x$, $y:=\mathbb N$ is $\in x$ but $y\cup \{y\}$ is not. You do need a limit ordinal (or at least add something that starts at some $S\notin \mathbb N$) – Hagen von Eitzen Feb 15 '14 at 19:06
@HagenvonEitzen thank you, I will delete my comment. – drhab Feb 15 '14 at 19:18
Another example from NFU (plus Infinity): the universal set $V$ contains $\emptyset$ and forall $x$ it contains $x\cup\{x\}$. But the universe is hardly just the natural numbers. – Malice Vidrine Feb 15 '14 at 19:25
I'm no expert on ZF, but the axiom does not rule out the possibility of other "junk" being in $x$ in additional to elements of $N$. To obtain $N$, you must obtain the intersection of all sets like $x$. – Dan Christensen Mar 19 '14 at 21:32

Write $$\phi(x):\Leftrightarrow (\emptyset \in x\land \forall y\in x\,(y\cup\{y\}\in x)).$$ Then the Axiom of Infinity states $$\tag1 \exists I\,\phi(I).$$ From such a set $I$ you can derive (using the other axioms) the existence of $$\tag2 \omega := \bigcap\{\,x\in\mathcal P(I):\phi(x)\,\}$$ and then show that this $\omega$ is what you want (and does not depend on the choice of $I$). However, trying to pack the minimality construction of $(2)$ into the axiom makes it look much more complicated than $(1)$ - and unnecessarily so, as we can prove just what we need.

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Another set which satisfies the axiom is $N''=\mathbb N \cup N'$ where $N'$ is the set containing $x_0 = \mathbb N$ (a set which is not in $\mathbb N$) and all the sets $x_{n+1} := x_n \cup \{ x_n\}$. If you define $n+1 := n \cup \{n\}$, $0 := \{\}$ and $\omega := \mathbb N$ you have $$N'' = \{0, 1, 2, \dots, \omega, \omega+1, \omega+2,\dots\}.$$

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Axiom INF tells you that some set $x$ exists with property $\phi\left(x\right)$ as defined in the answer of Hagen von Eitzen.

Applying axiom SEP you find that $\omega:=\left\{ y\in x\mid\forall I\left[\phi\left(I\right)\Rightarrow y\in I\right]\right\}$ is a set.

Here $\omega=\left\{ 0,1,2,\dots\right\}$

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