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If I need the existence of an infinite set for a proof, for instance the smallest set $x$ such that $(\emptyset\in x)\land(\forall y,z\in x)(y\cup\{z\}\in x)$, how do I know such set exists in a model of ZF? It is not a subset of $\omega$ so I can't just use the infinity axiom with separation.

I think I can define intersection from separation so it would probably enough to show that some set satisfying above formula exists, then using intersection of all such sets to get the smallest one. But how do I know some set exists?

  1. How does it work for the set mentioned above?
  2. Is there a general (algorithmic?) way of checking this?
akkarin
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  • why do you need such a set ?, from where does this problem come from ? – Renart Feb 23 '16 at 19:48
  • Hint: for (1): use the axiom of replacement (it might be overkill for your $x$, but replacement will definitely do the job). For (2), I am not quite sure what you are looking for: $\exists x\phi(x)$ isn't decidable in ZF for arbitrary formulas $\phi$, so you can't check automatically whether some formula $\phi(x)$ defines a set that can be proved to exist in ZF. Maybe that wasn't what you meant in (2). – Rob Arthan Feb 23 '16 at 20:17

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Given that $\omega$ exists, and given that you've used replacement etc. to justify definitions by recursion, you can define the following function $f$ on $\omega$: $$\begin{align} f(0) & = X_0 = \{\emptyset\}, \\ f(n+1) &= X_{n+1} = X_n\cup \{ y\cup\{z\}\mid y,z\in X_n \}. \\ \end{align}$$ Now, by Replacement and Union, $X = \bigcup range(f) = \bigcup_{n<\omega} X_n$ exists, and is the set you want:

Clearly $\emptyset\in X$ and $X$ is closed under $(y,z)\mapsto y\cup\{z\}$; and clearly any other set with those properties contains each $X_n$, by induction on $\omega$, and so contains $X$.

BrianO
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  • How exactly is $f$ defined? Does it map $n\mapsto X_n$? Is "justify definitions by recursion" the recursion theorem? – akkarin Feb 25 '16 at 18:30
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    Yes that's the function I meant. (oops) I just edited to make it clear what the heck $f$ is. No it is not "the recursion theorem" if by that you mean a result from recursion theory (What do you mean? Kleene's 1st recursion theorem? his 2nd recursion theorem? In any case, neither of those.) In ZF you can prove a theorem schema that given a class Function $G\colon \omega \times V\to V$, and any set $a_0$, there is a unique function $f\colon\omega\to V$ such that $f(0) = a_0$ and for $n\in \omega$, $f(n+1) = G(n, f(n))$. There are more general forms, but this covers the $f$ in the answer. – BrianO Feb 25 '16 at 19:42
  • I remember the statement that recursive definitions are allowed as recursion theorem. Apparently some people call it this way but it doesn't seem to be standard. What would $G$ look like in your example above? Would it be $(n,X)\mapsto X\cup{y\cup{z}|y,z\in X}$ for $n>1$ and $(0,X)\mapsto\emptyset$? It seems a bit weird that $G(n,X)$ doesn't depend on $n$ given $n>0$. – akkarin Feb 25 '16 at 20:30
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    Yes, exactly (the first part), $G\colon (n,X)\mapsto X\cup {y\cup{z}\mid y,z\in X}$, for every $n$ — this the value of $f(n+1)$. – BrianO Feb 25 '16 at 21:30
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    It isn't so weird that G doesn't use $n$. Another statement (Kunen's) of The Recursion Theorem is: if $G\colon V\to V$ is a class Function, then there is a unique $F\colon \mathsf{Ord}\to V$ such that $F(\alpha) = G(F!\restriction ! \alpha)$ for all $\alpha\in\mathsf{Ord}$. Note that $G$ can 'recover' $\alpha$: $\alpha = domain(F!\restriction! \alpha)$. If $\alpha = n+1 < \omega$, then $G$ can access $n = \bigcup \alpha$, so it can access $F(n)$ too. ¶ In my answer, I'm using a simpler form than I stated in my first comment: all we need is $f(n+1) = G(f(n))$. – BrianO Feb 25 '16 at 21:40