Bomb#20
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Whoa![/keanureeves]The prima facie hole in the argument is here: "During their consultations the night before, prisoners have agreed on a CHOICE function: For each equivalence class, they have assigned a particular representative sequence."
The Axiom of Choice guarantees that such a choice function exists; it doesn't guarantee that prisoners can all agree on one. In general there will be infinitely many choice functions for a prisoner to choose from among. By what mechanism will the prisoners be able to pick out one from all the possibilities and ensure that they're all going to use the same choice function when it comes to deciding what color to say their hats are?
Is this objection to the mathematical abstraction, or to "real-world" limitations? For example, instead of an infinite number of prisoners, could we postulate a Hydra-like creature with a single brain to come up with the choice function, but that had to use his infinitely-many fire-walled brainlets during the guessing phase?
Maybe that plugs the hole in your proof; I'm not sure. Either way, full marks for creativity!
Elusiveness I'd say, not flaw. The choice functions you can hold in your hands and write down, you don't need an extra axiom for. The whole point of the axiom is to make the elusive ones available.Or maybe you've hit on the fundamental flaw (or elusiveness) in the Axiom of Choice: It's one thing to say "There exists a choice function" and another thing to actually hold that function in your hands, or be able to write it down.
It's clear that each equivalence class is countable and that there are uncountably many of them. But I don't think that means you're only using a weak form of AC. There's a known weak form called the Axiom of countable choice, but that asserts the existence of choice functions on countable sets of sets, not on uncountable sets of countable sets. I haven't heard that limiting the size of the member sets to countable infinities makes the axiom any weaker.The rationals are proven to be countable if they can be mapped one-to-one to a subset of the natural numbers. An inelegant brute-force way is to map each a/b to (2a3b). (Multiply this by 5 to represent -a/b.) Insist that a/b be in lowest terms to avoid duplication.
I think that a similar approach can be used to show that each of the equivalence classes I introduced upthread is countable (though this demonstration uses AC at the outset). ...
If this is correct — and by now I'm giving myself at best a 30% chance — then it means we are using only a weak form of the Axiom of Choice in the proof upthread: We are imposing a choice function on an (uncountable) collection of COUNTABLE sets.