Questions about Cartesian product

I have some questions about this:
1.- The Axiom of Choice guarantees a choice function from a collection
of sets, but it would seem that the existence of the Cartesian product
possibly requires more than one choice function on the same sets. How
can this be?

2.- It is intuitively obvious that the product of sets A and B is the
empty set if, say, A is empty. How can this be proved?

3.- In Royden's book he defines the product as a collection of sets.
Others define it as a set of j-tuples(e.g. (x_1, x_2,...,x_j)), which,
as I understand them, are not considered sets. Which is the correct
definition?

Any and all help is appreciated.

.

Relevant Pages

  • Re: Dik T. Winter says: Definition: sum{i in N} i = 0
    ... > Kardinalzahlen sich nach ihrer Größe ordnen lassen und in dieser ... it is not given but its existence is stated. ... Axiom VI eine reines Existenz-Axiom nicht mehr und nicht weniger. ... that if a choice function exists there is a well-ordering. ...
    (sci.math)
  • Re: A poosible additional axiom to ZFC?
    ... >>element of S is a singleton) then the choice function is ... The point of the new axiom, though, is that these choice functions do _not_ ... exist solely on the basis of AC, and therefore they're not that unique ... and we neen AC for the existence. ...
    (sci.logic)
  • Re: Set Theory Problem
    ... in Zermelo Fraenkel set theory without axiom of choice? ... It's consistent with ZF that the set of functions from ... existence of a choice function ...
    (sci.math)
  • Re: Dont get Axiom of Choice?
    ... you are defining no particular choice function. ... Constructivists/definitionists require that a mathematical object be ... constructed/defined if we are to admit its existence. ... axiom is controversial. ...
    (sci.math)
  • Re: Questions about Cartesian product
    ... but it would seem that the existence of the Cartesian product ... |possibly requires more than one choice function on the same sets. ... the axiom of choice implies that for any function f:x->y and subset s ... s is empty, but the general case follows easily as a consequence of ...
    (sci.math)
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