As in the title, in the book 'Handbook of Analysis and its Foundations' by Schechter, the Ultrafilter principle is presented as a nonconstructive component of the Axiom of Choice, while the Axiom of Dependent Choice as a constructive component. As the author emphasizes, anyhow, UF+DC+ZF do not imply AC.
The form I know of the two are the following:
Ultrafilter Principle Any proper filter can be included in an ultrafilter. That is, if $\mathcal{F}$ is a proper filter on a set $X$, then there exists an ultrafilter $\mathcal{U}\supset\mathcal{F}$ on X.
Dependent Choice (without history)
Let $S$ be a nonempty set, let $N=\{M\subset S M\neq\emptyset\}\subset \wp(S)$ be the family of nonempty subsets of S and $f:S\rightarrow N$ a given function.
Then there exists a sequence $(x_n)\subset S$ such that $x_{n+1}\in f(x_n)$
Dependent Choice (with history) Let $S_1,S_2,S_3,…$ be nonempty subsets of $S$, $F_n=\{M\subset S_{n+1} M\neq\emptyset\}$ and let $f_n: S_1\times …\times S_n\rightarrow F_n$. Then there exists a sequence $(x_1,x_2,x_3,…)$ such that $x_{n+1}\in f_n(x_1,x_2,…,x_n)$
Anyhow, though the idea seems interesting, I can't see why in the first place the Axiom of Dependent Choice should be more constructive than the Ultrafilter Principle, nor why the two should in some sense e complementary.
Thanks for help.
Best Answer
Intuitively speaking, Dependent Choice is essentially the following principle:
Suppose that there is a process choosing sets, $X_{n,y}$, such that for every $x\in X_{n,y}$, there is a non-empty set $X_{n+1,x}$. Then there is a function $f$ with domain $\Bbb N$, such that $f(n+1)\in X_{n,f(n)}$.
It lets us have inductive constructions without having to worry about specifying exactly "the next step", and instead just prove that such a step is possible to make.
In contrast, having non-principal ultrafilters, even just on $\omega$, requires a lot more choices to be made, and there is no obvious way to understand them as an iterative construction. Even with choice, working via transfinite recursion, we can easily prove that such a construction must take more than countably many steps, starting from a free filter.