3 ms·

So how does anyone form an intuition for even defining the Cartesian product of infinite sets in general, without assuming the AC in the first place?

xamuel 9y ago
Consider something like the Cartesian product of countably infinitely many copies of the naturals: this is nothing more than the set of all countably-infinitary tuples (n1,n2,...). It's obviously nonempty, e.g. because it contains (0,0,...). It seems intuitively obvious that similar reasoning should make any arbitrary Cartesian product of nonempty sets be empty.
- danharaj 9y ago
  That reasoning applies only to products of an arbitrary family of the same set, provided you know an element of that set. That is an exceptionally simple special case, don't you think?
- cuspycode 9y ago
  Yes, that particular example is intuitive. But I don't see how a similar reasoning should apply to arbitrary infinite sets however. That's a huge intuitive leap in my opinion, and it's one that I can't follow.
  - xamuel 9y ago
    Ok, you're presenting to an audience consisting of lots of tables, each with at least one person at it. You say: "Will every table please have a representative stand up." To question the AC is to suggest that for some configuration of (sufficiently many) tables, it's impossible for the audience to fulfill your request, on account of there not existing a function f(table)=representative. This sounds absurd, because obviously each table can independently choose a representative, and it isn't like they need to coordinate with the other tables in order to fulfill your request.
    - cuspycode 9y ago
      "Lots of tables" is the finite case, for which the AC is a theorem. An infinite set of tables is qualitatively different. An arbitrary f(table) function will have to contain an infinite amount of information, which is beyond anything my intuition can grasp. For f(table) functions that contain finite information, the intuition is easy and the AC is reduced to a theorem.

xamuel 9y ago

Consider something like the Cartesian product of countably infinitely many copies of the naturals: this is nothing more than the set of all countably-infinitary tuples (n1,n2,...). It's obviously nonempty, e.g. because it contains (0,0,...). It seems intuitively obvious that similar reasoning should make any arbitrary Cartesian product of nonempty sets be empty.

danharaj 9y ago
That reasoning applies only to products of an arbitrary family of the same set, provided you know an element of that set. That is an exceptionally simple special case, don't you think?
cuspycode 9y ago
Yes, that particular example is intuitive. But I don't see how a similar reasoning should apply to arbitrary infinite sets however. That's a huge intuitive leap in my opinion, and it's one that I can't follow.
- xamuel 9y ago
  Ok, you're presenting to an audience consisting of lots of tables, each with at least one person at it. You say: "Will every table please have a representative stand up." To question the AC is to suggest that for some configuration of (sufficiently many) tables, it's impossible for the audience to fulfill your request, on account of there not existing a function f(table)=representative. This sounds absurd, because obviously each table can independently choose a representative, and it isn't like they need to coordinate with the other tables in order to fulfill your request.
  - cuspycode 9y ago
    "Lots of tables" is the finite case, for which the AC is a theorem. An infinite set of tables is qualitatively different. An arbitrary f(table) function will have to contain an infinite amount of information, which is beyond anything my intuition can grasp. For f(table) functions that contain finite information, the intuition is easy and the AC is reduced to a theorem.