Are there any major results provable with the Axiom of Choice that aren't provable with the Axiom of Countable Choice?
>>8171727
bamp for interest
Ramsey theory
https://en.wikipedia.org/wiki/Axiom_of_countable_choice
>>8171727
AC or an equivalent axiom is needed to prove results on uncountable sets in general.
There is no other way to attack uncountable sets, although once one result has been obtained using AC there is no need to invoke it again (for example, Heine-Borel in real analysis can give you results about uncountable closed and bounded sets without directly invoking AC, but this result is itself derived from some version of AC).
The first one that comes to mind is the fact that for every uncountable family of nonempty sets there is an element in their product. :-]
>>8171727
Every vector space has a basis, among probably many other things that haven't been proven to require it yet.
Wikipedia is your friend.
>>8173212
>Every vector space has a basis
What would this mean? If it were not for the axiom of choice then there would exist some vector space that is impossible to describe?
>>8173224
without the axiom of choice there exists an infinite-dimensional vector space with no basis
>>8173271
what's the name of it?
>>8173273
its a nonconstructive theorem. to construct a basis for an arbitrary infinite dimensional vector space you consider the set of all linearly equivalent subsets of your vector space. using zorn's lemma (equivalent to axiom of choice) lets you choose a maximal subset among all those that are linearly independent.
>>8173273
I think the statements are equivalent.