# Confusion over Axiom of Choice.

1. Jan 8, 2014

### QIsReluctant

Hello,

What "choice" does the Axiom of Choice permit us to make? I've searched high and low and not found a satisfactory answer. To me it seems to add no new information to its hypotheses: Given an arbitrary collection of non-empty sets, isn't it true, without assuming anything and just by the fact that each set has at least one element, that you can map each set to an element within itself? Or is the rub that the statement is not provable or disprovable from the axioms of ZF (which I wish I were more familiar with)?

One author suggests that you need a hard-and-fast rule (e.g., if I'm using the domain of positive-length real intevals let f(interval) = the midpoint), but then says that we don't need AC for any finite interval because we are able, by induction, to pick set-by-set within the domain. I guess I need a more precise definition ...

2. Jan 8, 2014

### pwsnafu

ZF can do induction, hence it can do finite choice. Induction doesn't let you do countable choice.

Uncountable choice leads to Banach-Tarski, which is counter intuitive.

3. Jan 8, 2014

### mathman

The axiom of choice is needed when the number in the collection is non-countable. In that case the construction of the choice set can't be made, since you have to make an uncountable number of choices.

4. Jan 8, 2014

### jgens

It is actually worse than this. A (weak) form of the axiom of choice is needed in the countable case as well.

5. Jan 8, 2014

### economicsnerd

But who wants to live in a world without countable choice?

6. Jan 8, 2014

### jgens

Those who subscribe to (ultra)finitism perhaps? :tongue:

7. Jan 9, 2014

### pwsnafu

To make this more precise. ACω is not provable in ZF. This means:
ZF is strickly weaker than ZF+ACω, which is strictly weaker than ZF+DC, which is strictly weaker than ZF+AC.

AC is Axiom of Choice, ACω is Axiom of Countable Choice and DC is Axiom of Dependent Choice, in case anybody was asking.