Dependent Choice (Fixed First Element)

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

Suppose:
 * $\forall a \in S: \exists b \in S: a \mathrel \RR b$

that is, that $\RR$ is a left-total relation (that is, a serial relation).

Let $s \in S$.

Then there exists a sequence $\sequence {x_n}_{n \mathop \in \N}$ in $S$ such that:
 * $x_0 = s$
 * $\forall n \in \N: x_n \mathrel \RR x_{n + 1}$

Proof
Let $S' = \set {y \in S: s \mathrel {\RR^+} y}$, where $\RR^+$ is the transitive closure of $\RR$.

Let $\RR'$ be the restriction of $\RR$ to $S'$.

For each $x \in S'$, there is a $y \in S$ such that $x \mathrel \RR y$. But then $s \mathrel {\RR^+} y$, so $y \in S'$, so $x \mathrel {\RR'} y$.

Thus $\RR'$ is a left-total relation on $S'$.

$S'$ is non-empty: since $\RR$ is left-total, there is a $t \in S$ such that $s \mathrel \RR t$, so $s \mathrel {\RR^+} t$, so $t \in S'$.

By Axiom of Dependent Choice, there is a infinite sequence $\sequence {y_n}_{n \mathop \in \N}$ in $S'$ such that for each $n \in \N$, $y_n \mathrel {\RR'} y_{n + 1}$.

Then by definition of restriction, $y_n \mathrel \RR y_{n + 1}$ for each $n \in \N$.

By definition of $S'$, $s \mathrel {\RR^+} y_0$.

By definition of transitive closure, there are elements $x_0, \dots, x_m$ such that $s = x_0 \mathrel \RR x_1 \mathrel \RR \cdots \mathrel \RR x_m \mathrel \RR x_m = y_0$.

Then for $n > m$, define $x_n$ as $y_{n - m}$.

This sequence then meets the requirements.