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In many mathematical proofs dealing with mathematical objects, we begin by saying something along the lines of: Let $X$ be a such and such mathematical object. Let $y \in X$.

And then we proceed to exploit the properties of $X$ to prove what we want using $y$. This is so commonplace, and what I would like to know is:

How is this different from assuming the axiom of choice? We are inherently assuming that we can choose an element from some sort of set based object. Often, this element $y$ is taken to be arbitrary, but this is often said near the end of the proof. Is it important that we are doing this with one "set based object", instead of a set of "set based objects"?

Thank you

zyx123
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    Can you cite an example? This sounds like a method of proving something for all elements of $X$, which isn't actually a choice at all. – aschepler Jun 11 '22 at 13:04
  • Yes, like I said it is arbitrary, but it's as if we are kind of pretending that we have chosen something. Any undergrad group theory/ rings /module theorem would have many examples. – zyx123 Jun 11 '22 at 13:10
  • Also, in the axiom of choice, the choice that we make is also arbitrary. – zyx123 Jun 11 '22 at 13:13
  • "Arbitrary" doesn't have a precise meaning, though. Are you familiar with predicate logic written with $\forall$ and $\exists$? – aschepler Jun 11 '22 at 13:15
  • In the axiom of choice, the statement is that there exists a choice function. So that means that there exists an element from each set for the choice function. – zyx123 Jun 11 '22 at 13:24
  • Neither "Let $y\in X$" nor the axiom of choice is about what "we can choose" or about us in any way. The axiom of choice asserts the existence of sets that satisfy certain conditions. It is not needed when we don't want to form a set but merely talk about an element $y$ of $X$. – Andreas Blass Jun 11 '22 at 13:25
  • The condition they satisfy is that they are a collection of nonempty sets. What else do they satisfy? – zyx123 Jun 11 '22 at 13:27

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