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The Cartesian product of any nonempty collection of nonempty sets is nonempty. In other words, if $I$ is any nonempty (indexing) set and $A_i$ is a nonempty set for all $i\in I$, then there exists a choice function from $I$ to $\cup_{i\in I}A_i$.

This is a text from Dummit and Foote. Can someone please explain what these sentences mean? I find the text very confusing. Is this a theorem or axiom?

Further, later in the text there is a theorem that says

Assuming the usual axioms of set theory, the following are equivalent:

(1) Zorn's Lemma (2) the Axiom of Choice (3) the Well Ordering Principle

What are axioms of set theory and how are they related?

jk001
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    The fact that it is called the Axiom of Choice should be a clear indication that it is very likely to be an axiom and not a theorem. – Arturo Magidin Aug 31 '21 at 18:05
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    Thank you for your explanation. How about Zorn's Lemma? The book says that under certain construction(?), Zorn's Lemma is the same as Axiom of Choice, so that's why I was confused. – jk001 Aug 31 '21 at 18:07
  • Zorn's Lemma is a theorem of ZFC (Zermelo-Fraenkel Set Theory with the Axiom of Choice). In the absence of the Axiom of Choice, one can prove that they are equivalent (assuming one allows you to prove the other as a consequence). – Arturo Magidin Aug 31 '21 at 18:08
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    Given how poorly many things are named, I think it’s fair to ask if the axiom of choice is an axiom or not. But yes, it’s an axiom. Some people choose to accept it, and they use it whenever they need it (but noting its use)— other people (more rare) choose not to accept it, and they try to develop their theory entirely without the axiom of choice. – littleO Aug 31 '21 at 18:10
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    "What are the axioms of set theory and how they are related" is too broad a question for this site. You are essentially asking for an introductory course in axiomatic Set Theory, or at the very least the equivalent of one or two full lectures in a course. If the quick intro provided by Dummit and Foote is too confusing, then I suggest picking up a book that aims to introduce formal set theory to the uninitiated. Perhaps Halmos's classic "Naive Set Theory". – Arturo Magidin Aug 31 '21 at 18:12
  • @Arturo Magidin That's fair enough and thank you for your suggestion. I would appreciate anyone's Heuristic explanation though. – jk001 Aug 31 '21 at 18:14

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Consider any set $I$. Suppose that for each $i \in I$, we specify some set $A_i$. We say that $A$ is an $I$-indexed family of sets.

We define the set $\prod\limits_{i \in I} A_i$ as $\{f \mid f : A \to \bigcup\limits_{i \in I} A_i$ and $\forall i \in I$, $f(i) \in A_i\}$.

In other words, a function $f \in \prod\limits_{i \in I} A_i$ is a function that takes as input some $i \in I$ and outputs an element of $a_i$.

This set $\prod\limits_{i \in I} A_i$ is the "cartesian product of the family $A$".

This generalises the notion of a binary Cartesian product $B \times C$. For we can define $A_0 = B$ and $A_1 = C$. Then $B \times C$ is in natural bijection with $\prod\limits_{i \in \{0, 1\}} A_i$.

The axiom of choice states that for all $I$, for every $I$-indexed family of sets $A$, if for all $i \in I$, $A_i$ has at least one element, then $\prod\limits_{i \in I} A_i$ also has at least one element.

For some reason, your text also imposes the requirement that $I$ not be empty. This requirement is just a distraction and is not necessary.

The "usual axioms of set theory" are probably the axioms of ZF.

A reasonable read for showing the equivalence of Zorn's lemma/axiom of choice/well-ordering theorem is here.

Mark Saving
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  • Although what you write is technically correct for empty index set (since the empty product does contain an element), it is likely to be confusing to someone struggling at this stage; and the statement the OP encountered specifies nonempty $I$. It might be best to add "nonempty" to "for all $I$, for every $I$-indexed..." – Arturo Magidin Aug 31 '21 at 18:22
  • @ArturoMagidin My view is exactly the opposite of yours. If a condition is not relevant in a theorem, it should be left out. Including "non-empty" causes people to needlessly think that the claim is invalid when $I$ is empty, thus increasing confusion. It also means that there is a pointless "emptiness check" which is necessary every time the axiom of choice is applied. – Mark Saving Aug 31 '21 at 18:26
  • Fair enough; as long as it was not inadvertent and is correct, that's valid. – Arturo Magidin Aug 31 '21 at 19:24