My name is Annatala. I'm a graduate student in computer science, and in my spare time I study Set Theory for fun (I'm weird).

I've been slooooowly going through Jech's wonderfully dense Set Theory book, with occasional Wolfram and/or Wikipedia reference. I have a question about AC that I'm not quite certain about, despite how simple it is. (I insist I have a fairly good understanding of basic set theory despite how stupid I'm about to sound!)

I'm curious why AC is not trivially true. I'm not asking for Cohen's forcing proof of ZF+ $\displaystyle \neg$AC; I'm asking for something much simpler.

Let's say we have a countably infinite family of nonempty setsS= {A,B,C, ...} and want to show a choice functionfexists wheref= { (A,a), (B,b), ... }, withabeing an element ofA, etc. Clearly, iffexists,f$\displaystyle \subset$S$\displaystyle \times$ $\displaystyle \bigcup$S. Since this Cartesian product must include every possible combination ofSand the elements of the elements ofS, it seems "intuitively true" that a satisfactoryfmust be one of these combinations. But we cannot deduce its existence whenSis not "well-orderable", even if all setsXinShave cardinality 2 (unless, that is, we already have a constructive way to select from eachX).

Even weirder, I hear tell that AC is equivalent to the statement: "the Cartesian product of any family of nonempty sets is nonempty". This is rather mind-boggling as the Cartesian product is well-defined with a simple formula, and the only way (again, intuitively) you can get an empty product is if one of the sets is empty.

My best guess about all this is that perhaps it is because ZF does not admit as sets collections which cannot be constructed via separation. Is this the right track? And wouldn't this make the Universe rather...well, tiny?

If this is the case, it certainly makes downward Lowenheim-Skolem straightforward to grasp!

Anyway; any advice, pearls of wisdom, or Pinkie Pie references are greatly appreciated. Thanks!

Anna