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\label{prop:orbit-countable}
Each orbit ray $O_g$ (for $g \in \{\alpha, \pi, \gamma, \eta\}$)
is countably infinite.
More precisely, the map
$\varphi_g : \mathbb{N} \to O_g$ defined by
$\varphi_g(n) = \rho^n(g)$ is a bijection.

\emph{Well-definedness.}
$\varphi_g(n) = \rho^n(g) \in O_g$
by $\KAxiom{3}$ (orbit closure under $\rho$).

\emph{Surjectivity.}
By definition, $O_g = \{\rho^n(g) : n \geq 0\}$,
so every element of $O_g$ is in the image of $\varphi_g$.

\emph{Injectivity.}
Suppose $\varphi_g(n) = \varphi_g(m)$,
i.e., $\rho^n(g) = \rho^m(g)$.
Without loss of generality, assume $n \leq m$.
If $n = m$, we are done.
If $n < m$, write $m = n + k$ with $k \geq 1$.
Then $\rho^n(g) = \rho^{n+k}(g) = \rho^k(\rho^n(g))$.
This says $\rho^n(g)$ is periodic under $\rho$ with period~$k$.
But by $\KAxiom{4}$ (no-jump), $\rho$ strictly advances the depth:
$\rho(\rho^j(g)) = \rho^{j+1}(g)$, and the depth increases
from $j$ to $j+1$ at each step.
Therefore $\rho^k(\rho^n(g))$ has depth $n+k > n$,
while $\rho^n(g)$ has depth $n$.
Since depth uniquely identifies an orbit element
(by $\KAxiom{4}$), they cannot be equal.
Contradiction.
Hence $n = m$, and $\varphi_g$ is injective.