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If I have sets $A_0$, $A_1$, $A_2$, $A_3$,... How can I prove the existence of the set $\lbrace A_0, A_1, A_2, A_3,...\rbrace$?

To be more precise, if there exists set $A_n$ for each $n\in\mathbb{N}$, does the set $\lbrace A_n:n\in\mathbb{N}\rbrace$ exists?

What I got so far is that I have to use the Axiom of Infinity somehow, otherwise we could prove that $\mathbb{N}$ exists without using the Axiom of Infinity. (we can define 0,1, 2 ,3,... without using the Axiom of Infinity, but to get the set $\lbrace 0,1,2,3,...\rbrace$ requires the Axiom of Infinity).

What I guess is that I also have to use the Axiom Schema of Replacement, but the book I use (Jech's Introduction to Set Theory) use the notation $\bigcup_{n=0}^\infty A_n$ even before introducing the axiom. (To have something like $\bigcup_{n=0}^\infty A_n$ you have to guarantee that the set $\lbrace A_n:n\in\mathbb{N}\rbrace$ exists in the first place so that you can apply the Axiom of Union).

Thank you in advance.

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  • $\begingroup$ You're right that Replacement in addition to Infinity is required. When you say that you "have" sets $A_0, A_1, ...$ what actually does that mean? Probably it means that for some formula $\varphi(n, x)$, $A_n = \{x \mid \varphi(n, x)\}$, and you can prove that $\forall n\,(n \in \mathbb{N} \to \exists y\, y = A_n)$. Yes? Then the Replacement schema applied to $\mathbb{N}$ alias $\omega$ lets you conclude that the collection you want exists (is a set). $\endgroup$ – BrianO Oct 19 '15 at 19:37
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Well, what does it mean that you have the sets $A_1,\ldots$? It means that there is a function $A$ whose domain is $\Bbb N$ and $A(n)$ is the set you denote by $A_n$.

So the set $\{A_n\mid n\in\Bbb N\}$ is just the range of the function $A$. So you just need axioms which guarantee that the range of a function exists. The exact specifics depends on how you define a function in the first place.

One major caveat: It might be that there is no actual function, just a "definable sequence". In this case the axiom of replacement assures that there is in fact a function. In weaker set theories (e.g. Zermelo's set theory) this indeed might end up differently. Namely, there will be no function object in the universe, and there will be no such set $\{A_n\mid n\in\Bbb N\}$.

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