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In a quite informal way the axiom schema of replacement says that

Let $X$ be a set.

Let $\phi(x,y)$ a formula s.t. for each $x$ in $X$ there exists an unique set $y$ satisying $\phi$.

Then there exists the family $Y$ composed by all such $y$.

Therefore even if I know that any single set $y$ exists, without this axiom I cannot conclude that the family $Y$ exists.

In the finite case this should not be true, because from single sets $a$, $b$ and $c$ I can build $\{a,b,c\}$ by means the axiom of pair and the union axiom.

Therefore I suppose that we need this axiom schema just for the infinite case.

Am I wrog?

Thanks in advance.

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kavsnzre
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1 Answers1

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That depends very much on your axiomatization of $\sf ZF$.

Specifically, the axiom of pairing is a consequence of replacement, power set, and infinity (one could use "empty set" instead of infinity, but that would be redundant, since infinity implies the empty set exists directly).

If minimality is what your heart desires, then pairing is a theorem, not an axiom, and then I don't see why things work out in the finite case either.

But let's just put this aside, and let pairing be part of our system. Indeed, then in that case you need the axiom of replacement for infinite collections.

The classic example is $V_{\omega+\omega}$. We start with $V_0=\varnothing$, and for $n$, $V_{n+1}=\mathcal P(V_n)$. When we reach $\omega$, $V_\omega=\bigcup\{V_n\mid n<\omega\}$, and so we continue again with power sets and unions.

It is not hard to check that $V_{\omega+\omega}$ actually satisfies all the axioms of $\sf ZF$ except Replacement. Including pairing, just to be clear. But now consider the function given by $f(0)=V_\omega$ and $f(n+1)=\mathcal P(f(n))=V_{\omega+n+1}$. The range of $f$ is exactly $\{V_{\omega+n}\mid n<\omega\}$, and it is easy to see that this is not an element of $V_{\omega+\omega}$.

So indeed Replacement fails, and since pairing and union hold, it fails for infinite sets.

Asaf Karagila
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  • Firstly, thank you Asaf!

    I try to recap: (1) yes, we need the replacement for putting togheter an infinite number of sets; (2) since replacement implies pairing, in a minimal setting we need it even for putting togheter a finite number of sets.

     – kavsnzre Dec 05 '17 at 11:19
  • Yeah. While I can't quite sign off on that (2) point, I don't see any naive way to prove pairing without a whole bunch of axioms. So let's say that it seems like Replacement might even be needed for finite sets in minimal settings. – Asaf Karagila Dec 05 '17 at 11:27
  • Ok, my mistake was that I haven't in mind the actual set of axioms that I used in the reasoning. We could formulate the point (2) in this way: (2') the possibility of build a set of two elements can be implied by different other axioms of my theory other than replacement (e.g. infinite+power set); but replacement might be needed in some cases, whenever other axioms are not assumed – kavsnzre Dec 05 '17 at 11:35
  • In order to show that the schema of replacement is necessary, you need to show that ${ V_{\omega + n} | n < \omega }$ is not necessarily a set. However, how can I conclude this from it is not an element of $V_{\omega + \omega}$. – XT Chen Jun 11 '19 at 08:10
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    @YukinoshitaisMyWaifu: Being a set literally means being an element of $V_{|omega+\omega}$ (in this context). – Asaf Karagila Jun 11 '19 at 08:11
  • Why? And I just find that ${ V_{\omega + n } | n < \omega } \subseteq \mathcal P(V_{\omega + \omega})$. – XT Chen Jun 11 '19 at 08:16
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    @YukinoshitaisMyWaifu: When you talk about a model of set theory, $M$, then sets are the elements of $M$. Here the model is $V_{\omega+\omega}$. The fact that that itself is a set in a large universe is irrelevant for this proof. – Asaf Karagila Jun 11 '19 at 08:17
  • Got your points. Thanks. – XT Chen Jun 11 '19 at 08:22