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.

  • $\begingroup$ 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

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 $\{A_n\mid n\in\Bbb N\}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.