How does Axiom of Choice imply Axiom of Dependent Choice?

Question

Axiom of Choice:

Given a collection $A$ of nonempty sets, there exists a function $$c: A \to \bigcup_{A_{i} \in A}A_{i}$$ such that $c(A_{i})\in A_{i}$ for all $A_{i} \in A$.

Axiom of Dependent Choice:

Given a nonempty set $A$ and a binary relation $\mathcal{R}$ on $A$ such that for all $a\in A$, there exists $b\in A$ such that $a\mathcal{R}b$. There exists a sequence $$(a_{n})_{n\in \mathbb{N}}$$ such that $a_{n}\mathcal{R}a_{n+1}$ for all $n \in \mathbb{N}$.

Here is my incomplete proof that Axiom of Choice implies Axiom of Dependent Choice:

For $a\in A$, let $R(a)=\{b\in A\mid a\mathcal{R}b\}\implies R(a)\neq\varnothing$ for all $a\in A$.

Using Axiom of Choice for the indexed family of sets $(R(a))_{a\in A}$, there exists a mapping

$$f:A\to A$$ such that $$\forall a\in A:f(a)\in R(a)$$

$\text{That is }\forall a\in A:a\mathcal{R}f(a)$. Let $B=\{(a,f(a))\mid a\in A\}$

I don't know how to proceed to prove the existence of the required sequence $(a_{n})_{n\in \mathbb{N}}$ from set $B$.

Please help me complete my proof! Many thanks for you!

Answer 1

You're basically done already: pick some $a\in A$ and consider the sequence $$a, f(a),f(f(a)), f(f(f(a))), ...$$ The $f$ that choice provides us is exactly the "strategy" we use to build the desired sequence.

Answer 2

Once you have $f$, use the assumption that $A$ is nonempty to choose $x \in A$. Now, you can define a sequence recursively by $a_0 = x$, $a_{n+1} = f(a_n)$. (So for example, $a_1 = f(x)$, $a_2 = f(f(x))$, $a_3 = f(f(f(x)))$, etc.)

Answer 3

I think you're almost there: now define $$ a_0 := a, a_n := f^n(a)$$ for some $a \in A$.