User talk:Mathmensch/Archives/2018/March

Definable countable choice?
Hi, I am glad to meet an expert in set theory here. May I ask you a math question? Here it goes. Is the following true in the meta-theory of ZFC?

"For every predicate P(x) (in ZFC) such that
 * (1) existence and uniqueness of x is provable (in ZFC), and
 * (2) for this x it is provable that this x is a countable set of nonempty sets,

there exists another predicate Q(y) such that (1) holds for it too, and it is provable that this y is a choice function for that x."

Less formally, I wonder about a specific example of a sequence of nonempty sets such that there is no (or at least we do not see) a specific example of a sequence of their (chosen) elements. In other words, a specific example where countable choice really needs some choice axiom.

Thanks, Boris Tsirelson (discuss • contribs) 17:28, 6 January 2018 (UTC)


 * Hello Presumably, in your question you would replace "ZFC" by "ZF", since ZFC already includes the axiom of choice, which is pretty strong, even though there are even stronger axioms, such as the axiom of global choice, which asserts the existence of a function that is defined on all sets, and yields an element of the element to which it is applied.


 * In general, it should be noted that choice functions are not at all unique. If you take a sequence of copies of $$[n] := \{1, \ldots, n\}$$, there will be an uncountable number of choice functions for that sequence. Hence, it may also be advisable to amend the question by not requiring uniqueness for the predicate Q.


 * If these amendments are made, then we do need some sort of choice in addition to ZF. An overview over the proof may be found here.


 * A specific example of a sequence that does not have any choice function may be impossible to construct. --Mathmensch-Smalledits (discuss • contribs) 16:12, 7 February 2018 (UTC)


 * Thank you. No, I really mean ZFC. True, it proves existence of such choice function y, and of course choice functions are a lot; but generally ZFC does not give a predicate Q that "singles out" one of them. I guess that generally no such predicate exists for uncountable choice (in particular, choice from all nonempty subsets of R). But for countable choice I fail to find an example where x is singled out by some predicate P and nevertheless no one y can be singled out by a predicate Q. You write that this "may be impossible to construct". Thus, if I understand you correctly, you (like me) first, do not see such example, and second, are not sure that such example does not exist. (Assuming that by "sequence that does not have any choice function" you really mean "sequence that does not have any choice function singled out by a predicate".) Boris Tsirelson (discuss • contribs) 19:20, 7 February 2018 (UTC)
 * Oops, maybe my terminology is misleading. By predicate P(x) in ZFC I mean here a predicate definable by a formula of ZFC with only one free variable x. Boris Tsirelson (discuss • contribs) 21:02, 7 February 2018 (UTC)


 * Ah, now I see what you meant. I thought that Q was supposed to be a predicate that says: "y is a choice function of x". But it's simple to construct a Q as desired, using the axiom of choice. Just pick any choice function that you have, and define the predicate to be true on that choice function and false otherwise. However, it may be much more difficult to construct a function from all predicates satisfying (1) and (2) to predicates Q as desired, since I don't know whether all predicates form a set (if so, it can be done using the axiom of choice). --Mathmensch (discuss • contribs) 20:33, 8 March 2018 (UTC)


 * I'm not sure though whether then the formal statement that you made properly translates the written intuition that you gave. There is in any case a big difference between existence of a choice function and its concrete constructability. For instance, by the axiom of choice, there is a choice function defined on all subsets of $$\mathbb R$$, but I bet you'd have some difficulty writing down such a thing. --Mathmensch (discuss • contribs) 20:38, 8 March 2018 (UTC)


 * "Just pick any choice function that you have, and define the predicate to be true on that choice function and false otherwise"? No, look again what I mean by predicate. You cannot just insert a set (in particular, a choice function) into a formula of ZFC. More technically: an arbitrary set is not a constant of ZFC. If this set is definable (without parameters), then it is harmless to have it among constants, see Extension by definitions. But the point is just this: choice axiom gives a choice function that is (usually) not definable! In your words: not concretely constructable. In my words above: but generally ZFC does not give a predicate Q that "singles out" one of them. About definability: I am trying to write an essay on it; if you like, see User:Tsirel/sandbox. Boris Tsirelson (discuss • contribs) 21:18, 8 March 2018 (UTC)


 * From your essay, it transpires that you clearly understand computability, so I don't see any reason for additional explanations in that regard. I'll think about the remainder. --Mathmensch (discuss • contribs) 06:20, 9 March 2018 (UTC)


 * Thank you. Boris Tsirelson (discuss • contribs) 19:11, 9 March 2018 (UTC)