Given the number of results that are independent of ZF. It seems that once you've found a proof of a theorem that uses the axiom of choice, the odds are that it will be independent of ZF. So my question is:


-Is there any result that has a solution in ZFC which relies on AC, but has another proof that can be done only in ZF?


I'm especially interested in those results which people thought that were going to be independent of ZF, because only a proof relying on AC was known. And then someone found a proof in ZF.


When I say axiom of choice I'm also including weaker versions like countable choice or DC.



If a result is apparently provable with AC, is actually independent of ZF?
Construction of the Haar integral for locally compact Hausdorff group $G$ ... as a linear functional on $C\_{00}(G)$ ... is often done using the Axiom of Choice. In the Hewitt & Ross textbook ABSTRACT HARMONIC ANALYSIS this is Theorem (15.5), and they do it without AC. They do then have an exercise (15.25) where they outline the much shorter proof using Tihonov's Theorem.


