As most of us, I struggled a lot when I first heard about Axiom of Choice (AC) and its consequences. Some things that can be derived from AC don't agree with my intuition. Consider for instance Zermelo's theorem applied to $\mathbb{R}.$
For many years I wished to abandon this evil axiom, but I was powerless. I mistakenly thought that assuming $\neg AC$ implies that all classical results using AC are then lost. Equivalence of Cauchy and Haine continuity for example. I wasn't aware that there are alternatives to AC.
Finally when dealing with the characterization of Noetherian rings I came across Axiom of Dependent Choice (DC). It sounded so right to me. I loved it since I read it for the very first time.
Since that day I try to re-examine all results that use AC to find out whether DC is enough.
Question. What results follows from DC and what results require full AC?
I am interested in the results form all mathematics. Set theory, topology, algebra, logic, etc.
Obviously all the results that are equivalent to AC fall to the latter group.
Kurt Godel in the 1930's showed that Con(ZF) implies Con(ZFC). Paul Cohen in the 1960's showed that Con(ZF) implies Con(ZF$+\;\neg$AC). So you have your choice, so to speak.
DC implies CC (Countable Choice). CC is: If $f$ is a set-valued function with dom($f)= \omega,$ and $f(n)\ne \phi$ for all $n,$ then $\prod_{n\in \omega}f(n)\ne \phi.$
CC implies that a countable union of countable sets is countable, and hence that $\omega_1$ is a regular cardinal, which matters in the theory of Borel sets, among other things.
In ZF the words finite and infinite must be used with caution. A Tarski-finite set is a bijective image (or any functional image) of a member of $\omega.$ A set $S$ is Dedekind-infinite iff there is an injection from $S$ to a proper subset of S. It is an elementary exercise to show that $S$ is Dedekind-infinite iff $S$ has a countably infinite subset.
It has been shown that it is equi-consistent with ZF that there is a set which is neither Tarski-finite nor Dedekind-infinite.
We can use DC to prove that a set $S$ that is not Tarski-finite is Dedekind infinite, as follows: For $n\in \omega$ let $F_n$ be the set of injective functions from $n$ into $S.$ Let $F=\cup_{n\in \omega}F_n.$ For $f,g \in F,$ let $f<^*g$ iff dom($f)\subsetneqq$ dom($g)$ and $g|_{\text {dom}(f)}=f.$
Then $<^*$ is a binary relation on $F$ and dom$(<^*)=F.$ Now $\phi$ (the empty function) belongs to $F.$ So DC implies there is a sequence $(f_n)_{n\in \omega}$ in $F$ with $f_0=\phi$ and $f_n<^*f_{n+1}$ for all $n\in \omega.$ Then $\cup_{n\in \omega}f_n$ is an injection from $\omega$ into $S.$
I dk whether that can be proved using only CC.