4 ms·

The OP's argument seems to be just that you cannot always construct a choice function. But the entire point of the axiom of choice is to say "that's fine; you'

• rini17 4y ago
  How is that different from claiming, say, there's an integer betwen three and four, oh we can't construct it but that's not a problem, gonna use it in the proofs anyway?
  • jbnicolai 4y ago
    That's essentially what axioms are. Wether it's reasonable or not depends on your intuition of both axiom & implications. This is where the axiom of choice is interesting; many people feel that it seems intuitively reasonable, but the implications are counterintuitive (e.g. banach-tarski)
  • contravariant 4y ago
    Well an integer between three and four is going to cause some contradictions, depending on your definition of 'integer'. You can assume there exists a number between 3 and 4 though, no problems there.
  • jacobmartin 4y ago
    This is an interesting question that gets to the heart of the constructivism debate, which I won't touch on here. As for the integer example, the integers are all easily (finitely) constructed (successors to 0 and their additive opposites), so it isn't a consistent concept to say "This is an unconstructable integer." Sets, on the other hand, are not necessarily always constructable, unless you only take the constructivist axioms.
  • xigoi 4y ago
    An axiom has to be consistent with other axioms (that is, adding it doesn't introduce a contradiction).
  • xen0 4y ago
    That would let me derive your mother's phone number. https://xkcd.com/704/ https://xkcd.com/704/ It's different because there's nothing that says such a choice function _cannot_ exist (short of an axiom that says explicitly so).
• lisper 4y ago
  > The OP's argument seems to be just that you cannot always construct a choice function. No, the argument is that if you cannot describe the choice then you have a problem. My argument does not assume anything about the choice function, it's merely an informal observation about its output.
  • xen0 4y ago
    So the statement "let x be an element of y" is already problematic? If so, your problem is not with the axiom of choice. Edit: "Not being able to describe the choice a choice function makes" and "not being able to construct it" are really the same thing.
    • lisper 4y ago
      > So the statement "let x be an element of y" is already problematic? Hm, good point. No, that's not the problem because that's just the same as saying "let x be an undefinable number". The problematic statement is (I think -- I need to think about this some more) "let f be a choice function on (infinite) sets of undefinable numbers". If x=f(S) that is not the same as "let x be an undefinable number" unless S is the set of ALL undefinable numbers, which it need not be.
      • xen0 4y ago
        You're right that there's a problem with 'Let f be a choice function on...', but that's the problem the axiom of choice solves. Sure, it's non-constructive; you don't know much about f, just that it exists and f(s) is in s (for all sets s that are subsets of S). The question is specifically what weird undefinable numbers have to do with anything; why those are specifically a problem. Let S be the set of all non-finitely defined real numbers. Let (Sn) be the sequence of sets such that Sn = {s in S : n-1 < s < n} Am I allowed to say "let s1 be an element of S1"? Or "s2 an element of S2"? (Note that we don't need the axiom of choice to get a choice function for this sequence; I can easily construct one; does that change anything?).
        • deleted 4y ago