I need to prove that the set $\{A^n : n \in \mathbb{N} \}$ using the ZFC axioms (without Replacement).

My (rough) plan would be to construct some set containing "more" sets than necessary, then use the comprehension axiom to remove the undesirable sets. Something like: for some $Y$, $\{ X \in Y : \exists n \in \mathbb{N}, X = A^n\}$.

I know how to prove that any one of these sets (e.g., $A^2$), exists, but I have no idea how to prove that the set containing all of these powers exists.

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    $\begingroup$ @tomasz: When one declare "no replacement", then your argument doesn't work anymore. To prove it's a duck, you must provide genetic analysis. You can't replace the definition of a duck with that one. That is the whole point of replacement! $\endgroup$ – Asaf Karagila Mar 6 '19 at 20:21
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    $\begingroup$ Yes, that would resolve the problem. That is what I meant by being dependent on the definition. Because iterated Cartesian product, with the standard definition of ordered pairs, won't do. $\endgroup$ – Asaf Karagila Mar 6 '19 at 20:24
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    $\begingroup$ @tomasz: People really don't get how much these things break down without replacement. Replacement is literally the reason we are allowed to say "we have this property, and we don't care about the implementation of it". I smell a blog post brewing once I get home. $\endgroup$ – Asaf Karagila Mar 6 '19 at 20:26
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    $\begingroup$ @tomasz: karagila.org/2019/in-praise-of-replacement if I may toot my own horn... :P $\endgroup$ – Asaf Karagila Mar 6 '19 at 22:24
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    $\begingroup$ @AsafKaragila That escalated quickly... ;-) $\endgroup$ – Stefan Mesken Mar 6 '19 at 23:35

If you define $A^n$ as the set of functions from the ordinal $n$ into $A$, then the answer is positive.

Note that a function $n\to A$ is a subset of $n\times A$, and hence also a subset of $\omega\times A$. Thus, every function $n\to A$ is an element of $\mathcal P(\omega\times A)$, so any family of such functions (in particular, every family of all such functions for a fixed $n$) is an element of $\mathcal P(\mathcal P(\omega\times A))$.


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