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I am trying to understand the Axiom of Choice. However, I found a proof of the Axiom. Let $A_i, i\in I$ be a set of mutually disjoint nonempty sets. We will construct a set that contains exactly one element in common with each of the sets as follows. Consider arbitrary $i\in I$ and fix it. Since the set $A_i$ is nonempty, $\exists x$ such that $x\in A_i$. Now we use the existential instantiation and write $x(i)$ for a new symbol (just a symbol) such that $x(i)\in A_i$. Consider one-element sets $\{x(i)\}$ and let $A=\bigcup_{i\in I}\{x(i)\}$. Then $A$ is a set and $A$ contains exactly one element in common with each $A_j$.

Where am I wrong?

Please help me.

Ivan Feshchenko
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    You are assuming AC in your proof. You know how to make any one choice of $x(i)$. It's easy to iterate this to get any finite number of choices $x(1), \ldots, x(n)$. But how do you do this to make infinitely many choices? Unless you have a consistent way to make the choices, you would need to write down each of your $x(i)$ individually, which would be an infinite thing to try and do. AC (and its variants countable choice, dependent choice, etc) give you the power to make all of these choices at once. – Chris Grossack Mar 28 '21 at 21:31
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    +1 no idea why this was downvoted, it's an excellent question, OP is trying to understand why the axiom of choice isn't just obviously true, something that every mathematician believed to be the case implicitly for a very long time. – hunter Mar 28 '21 at 21:32
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    I'm not sure the $i$-symbol can be the same $i$ as the $i$-object or. More generally, the $x(i)$-symbol does not have the property of referring to an object that is an element of $A_i$. It is if it is inside a FOL predicate where $x(i)\in A_i$ is specified. –  Mar 28 '21 at 21:35
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    This is a good question, but one which has been asked many times on this site - see e.g. here. (If there's something which the answers to that question don't clear up, add that to the question and it should be reopened.) – Noah Schweber Mar 28 '21 at 21:36
  • Thanks for the dupe find @NoahSchweber – Caleb Stanford Mar 28 '21 at 21:39
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    As I tried to explain in the comment to your MO question. You can write anything that makes sense. You can write $x(i)$ to denote an element of $A_i$. But you cannot conclude that there exists a single function $x\colon I\to\bigcup A_i$ which satisfies $x(i)\in A_i$. Think of it as context in programming. The $x(i)$ has a context inside "pick an element of $A_i$", once you move to a different $i$, or even "come back to the same $i$ again", there's no guarantee that the value is the same. And as I said, you can write it like that but it will be confusing to the reader. And also to you. – Asaf Karagila Mar 28 '21 at 21:42

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