Timeline for Finite axiom of choice: how do you prove it from just ZF?
Current License: CC BY-SA 2.5
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| Jul 20, 2010 at 0:04 | comment | added | Carl Mummert | @Asaf Karagila: You can only write that formula in the case when the product is externally finite. As the other answers indicate, it is enough that the product is internally finite, in which case an induction argument shows the choice function exists. Moreover, the standard reading of "a finite number of sets" is the internal one. Andreas Blass's comment is safe because he only refers to choosing from a single set. | |
| Jul 20, 2010 at 0:03 | comment | added | user7758 | Thank you, this is very helpful. As I indicated in my comments to Thomas Scanlon's answer, my confusion arose because I was looking for an "explicit choice function". | |
| Jul 19, 2010 at 23:47 | comment | added | Asaf Karagila♦ | I was in the process of writing this myself, I'll just add one more thing to clarify: As Andreas said, when dealing with a finite number of sets you can explicitly write the function in the form of a formula. However, in the infinite case you cannot write this sort of formula (it would require infinitely many quantifiers to mention which element is chosen) and therefore you're in need of a stronger tool that would guarantee the existence of such function. | |
| Jul 19, 2010 at 23:40 | history | answered | Andreas Blass | CC BY-SA 2.5 |