Dependent Choice for Finite Sets

Theorem
Let $\mathcal R$ be a binary relation on a non-empty set $S$.

For each $a \in S$, let $C_a = \left\{{ b \in S: a \mathrel{\mathcal R} b }\right\}$

Suppose that:


 * For all $a \in S$, $C_a$ is a non-empty finite set.

Let $s \in S$.

Then there exists a sequence $\left\langle{x_n}\right\rangle_{n \in \N}$ in $S$ such that:


 * $x_0 = s$
 * $\forall n \in \N: x_n \mathrel{\mathcal R} x_{n+1}$

Remark
This theorem follows trivially from Dependent Choice (Fixed First Element), a form of the Axiom of Dependent Choice.

The proof below shows that it follows from the weaker Axiom of Countable Choice for Finite Sets.

Proof
Define $\langle D_n \rangle$ recursively:

Let $D_0 = \left\{{s}\right\}$.

For each $n \in \N$ let $D_{n+1} = \mathcal R \left({D_n}\right)$.

Now, for each $n \in \N$ let $E_n$ be the set of all enumerations of $D_n$.

Then $E_n$ is non-empty and finite for each $n$.

By the Axiom of Countable Choice for Finite Sets, there is a sequence $\langle e_n \rangle$ such that for each $n$, $e_n \in E_n$.

Now recursively define $\left\langle{x_n}\right\rangle$:


 * Define $x_0$ as $s$.
 * Define $x_{n+1}$ as the first element in the enumeration $e_n$ which is in $\mathcal R(x_n)$.

Then $\langle x_n \rangle$ is the required sequence.

Also see

 * Axiom:Axiom of Dependent Choice which differs from this in that there is no constraint that each element is related to only a finite number of other elements.