- #1
abiyo
- 43
- 0
So if we have a finite collection of disjoint non-empty sets, one can show using ZF only(with no need of AC) there is a choice function. I understand the reason for this. My confusion is when one goes to non-finite collection of sets. For example if the index set is the Natural numbers, why
do we need AC? Why will this case be different from the finite case?
do we need AC? Why will this case be different from the finite case?