Dependent Choice (Fixed First Element)

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

Suppose that:
 * $\forall a \in S: \exists b \in S: a \mathrel{\mathcal R} b$

that is, that $\mathcal R$ is a left-total relation (specifically a serial relation).

Let $s \in S$.

Then there exists a sequence $\left\langle{x_n}\right\rangle_{n \in \N}$ in $S$ such that:
 * $x_0 = s$
 * $\forall n \in \N: x_n \mathrel{\mathcal R} x_{n+1}$

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

Let $\mathcal R'$ be the restriction of $\mathcal R$ to $S'$.

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

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

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

By the Axiom of Dependent Choice, there is a infinite sequence $\left\langle{ y_n}\right\rangle_{n \in \N}$ in $S'$ such that for each $n \in \N$, $y_n \mathrel{\mathcal R'} y_{n+1}$.

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

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

By the definition of transitive closure, there are elements $x_0, \dots, x_m$ such that $s = x_0 \mathrel{\mathcal R} x_1 \mathrel{\mathcal R} \cdots \mathrel{\mathcal R} x_m \mathrel{\mathcal R} x_m = y_0$.

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

This sequence then meets the requirements.