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In "Elements of Set theory" by Enderton 54~55p

We can form something like the Cartesian product of infinitely many sets, provided that the sets are suitably indexed. More specifically, let $I$ be a set (which we will refer to as the index set) and let $H$ be a function whose domain includes $I$. Then for each $i$ in $I$ we have the set $H(i)$; we want the product of the $H(i)$'s for all $i \in I$. We define: $$X_{i\in I} H(i) = \{f\;|\; f \;\text{is a function with domain}\; I \;\text{and}\; (\forall i\in I)f(i)\in H(i)\}.$$

Axiom of Choice $\;$(second form) $\;$For any set $I$ and any function $H$ with domain $I$, if $H(i) \not =\emptyset$ for all $i$ in $I$, then $X_{i\in I} H(i) \not = \emptyset$.

The axiom has the form $\forall i\in I$, if $p(i)$ then $q(i) \iff \forall i$, if $i\in I$ then $(\text{if}\; p(i)\;\text{then}\; q(i))$. In particular, the statement is vacuously true when $I = \emptyset$. Is it necessary that we have this vacuously true part in the axiom?

op ol
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  • Well does it means for a statement that is true to be "necessary"? – Lee Mosher Feb 17 '22 at 03:06
  • @LeeMosher I meant that the axiom gives no meaning when $i\in I$. So I thought the axiom could have a restriction like "For any set $I$ which is not the empty set". For example, we don't need an axiom like "For all $i$ in $\emptyset$, if $p(i)$ then $q(i)$" – op ol Feb 17 '22 at 03:13
  • I got this question while doing this exercise "Show that the first form of the axiom of choice we can prove the second form, and conversely.". I have no idea how to handle the case for $I = \emptyset$ – op ol Feb 17 '22 at 03:16
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    The fact that the statement holds also when I is empty shows that the restriction to a not-empty index set is unnecessary. – Mauro ALLEGRANZA Feb 17 '22 at 07:19
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    It is often better to say the same thing with less. Don't be afraid of vacuous truth. Read this post for one way to understand it well. – user21820 Feb 17 '22 at 16:19

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There's no rule that every axiom in an axiomatization has to be independent of the other axioms. It's perfectly acceptable for an axiom to be provable from other axioms. (It's true that this would be redundant, so it would then be fine to remove the unnecessary axiom from the axiomatization, but there's no need to remove it.)

The same thing applies in the situation you mentioned. The axiom of choice in the form you stated is trivially true if $I = \emptyset.$ So it would be acceptable to replace the axiom

$$\text{"For any set }I\text{ and any function }H\text{ with domain }I\text{,}\\ \text{if }H(i) \not =\emptyset\text{ for all }i \in I\text{, then }\mathop{\vcenter{\huge\times}}_{i\in I} H(i) \not = \emptyset"$$

with

$$\text{"For any }\textbf{non-empty}\text{ set }I\text{ and any function }H\text{ with domain }I\text{,}\\ \text{ if }H(i) \not =\emptyset\text{ for all }i \in I\text{, then }\mathop{\vcenter{\huge\times}}_{i\in I} H(i) \not = \emptyset."$$

In fact, you could require $I$ to be infinite, since the axiom of choice for finite $I$ is provable from the axioms of ZF (without AC).

It wouldn't make any difference in terms of what you could prove.

However, there's a benefit in keeping the axioms as simple and as uniformly applicable as possible, so most people wouldn't want to restate the axiom of choice in this way.

Mitchell Spector
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    A good example of redundant axioms: replacement makes empty-set, unordered-pair and specification redundant, and yet we mention them in stating $\mathsf{ZF}$ to make obvious the contents of Zermelo set theory. – J.G. Feb 19 '22 at 08:48
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    @J.G. Absolutely. Another example: There was a thread recently about how infinitely many instances of the replacement axiom schema can be proven in ZF without replacement. – Mitchell Spector Feb 19 '22 at 08:50