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Let $(A_i \mid i \in I)$ be a family of non-empty set. Then $A_i \neq \emptyset$ for all $i \in I$, thus $\exists a_i \in A_i$ for all $i \in I$. We define $f:I \to \bigcup A_i$ by $f(i)=a_i$.

From reading other posts in MSE, I'm sure that my argument is wrong without Axiom of Choice, but I can not understand why it's wrong. The usual answer is that you can not appeal to Existential Instantiation infinite many times. However, I'm unable to understand this explanation under set theory. Maybe the problem is related to underlying logic, with which I'm not familiar.

Please explain me why you can not appeal to Existential Instantiation infinite many times?

Akira
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    If each $A_i$ has two elements, which of them is $a_i$? – Asaf Karagila Sep 08 '18 at 09:13
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    Perhaps familiarize yourself with the underlying logic? At least if all signs point to that being the root cause of your argument failing. It's like asking "Why can't I prove that $\sqrt2$ is rational? It seems to have to do with mathematics, and I'm not very familiar with that..." – Asaf Karagila Sep 08 '18 at 09:18
  • @AsafKaragila Your second comment really touches my heart :) – Akira Sep 08 '18 at 09:19
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    @LeAnhDung Yes, familiarize yourself with how formal proofs in first order logic work (with natural deduction, ideally). The reason is simple but it might not do much good if you don't have a frame of reference. Each application of existential instantiation is one line of a proof. If you did it infinitely many times, your proof would never end. – spaceisdarkgreen Sep 08 '18 at 09:37
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    @AsafKaragila I think a better analogy would be "Why do I need induction in this proof of $\forall \Bbb N\colon \phi(n)$ (for some nice $\phi$)? After all $\phi(0)$ is obvious and I can see that I can always show $\phi(n+1)$ from $\phi(n)$. Joining all those proofs together, I have a proof for all $n$, even without the principle of induction." -- Here, the problem is that the "concatenated proof" would have forbidden infinitely length, just as in the case of AC, the (even uncountably) infinitely many invocation of $\exists$-elimiation break the solid concept of proof – Hagen von Eitzen Sep 08 '18 at 10:02

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