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!