Markov chain periodicity: finite vs general state space The definitions pertaining to Markov chains are generally different for finite state space and general state space Markov chains. This is mainly due to the fact that one cannot talk about moving to a state $x$ for general state spaces, and one cannot write down the transition "matrix". However, in my experience, most definitions for general state space Markov chains like irreducibility, recurrence etc are a general extension from the finite state space.
I recently realized that this is not true for the definition of periodicity. For general state space $\mathcal{X}$ the definition of aperiodicity is (you can find a reference here):

A Markov chain with stationary distribution $ \pi(\cdot)$ is aperiodic
  if there does not exists $d  \geq 2$ and disjoint sets
  $\mathcal{X}_1, \dots, \mathcal{X}_d \subseteq \mathcal{X}$ with $P(x,
 \mathcal{X}_{i+1}) = 1$ for all $x$  in $\mathcal{X}_i$ ($1 \leq i\leq
 d-1$) and $P(x, \mathcal{X}_1) = 1$ for all $x$ in $\mathcal{X}_d$
  such that $\pi(\mathcal{X}_1) > 0$. Otherwise the chain is periodic
  with period $d$.

This is the definition that makes more sense to me since it essentially means that the Markov chain cannot get stuck in a loop indefinitely.
For finite state space Markov chains, the definition is (can be found here).

For a state $x$, define its period $d_x$ as  $$d_x = \gcd \{n\geq 1:
 P^n(x,x) > 0\} \,.$$ If the Markov chain is irreducible, then all
  states share the same period $d$. If $d = 1$, then the Markov chain is
  called aperiodic.

This definition does not necessarily say that the Markov chain will get stuck in a loop. So my question is, is there anyway that the general state space definition is a generalization from the finite state space definition?
 A: The following argument shows how the "gcd" criterion leads to "stuck in a loop."

For an irreducible Markov chain with state space $\cal S$,
suppose  that for some $i\in{\cal S}$ and $d\geq 1$, we have $(n\geq0: p_{ii}(n)>0)\subset d\mathbb{Z}$.
By irreducibility, for any state $j$ we can find non-negative integers $m,m^\prime$
so that $p_{ij}(m)>0$ and $p_{ji}(m^\prime)>0$. Then
$0< p_{ij}(m)p_{ji}(m^\prime)\leq p_{ii}(m+m^\prime)$ so that
$m+m^\prime \equiv 0 \pmod{d}$.
In particular, if $p_{ij}(m)>0$
and $p_{ij}(m^{\prime\prime})>0$, then $m\equiv m^{\prime\prime}\pmod{d}$.
So for $0\leq r<d$ we can partition $\cal S$ as follows:
$$C_r=\left\{j\in{\cal S}: p_{ij}(n)>0\Rightarrow n\equiv r\pmod{d}\right\}.$$
We also let $C_d=C_0$.
If $j\in C_r$ and $k\in C_s$, for any $n,n^\prime,n^{\prime\prime}$
with $0<p_{ij}(n)p_{jk}(n^\prime)p_{ki}(n^{\prime\prime})\leq p_{ii}(n+n^\prime+n^{\prime\prime})$,
we get $n+n^\prime+n^{\prime\prime}  \equiv 0 \pmod{d}$, or $n^\prime\equiv s-r \pmod{d}$.
Taking $n^\prime=1$ for example, we see that  $p_{jk}>0$ implies $s\equiv r+1\pmod{d}$ so
$$\sum_{k\in C_{r+1}}p_{jk}=1 \mbox{ for any }j\in C_r.\tag 1$$
Note that $i\in C_0$ and therefore by (1) and induction,
$C_r$ is non-empty for all $0\leq r<d$.

Added: What I mean by "stuck in a loop" is that the Markov chain will cycle through $d$ subsets of states. Here is a picture of what this looks like when $d=5$:

Notation: 


*

*The notation $\sum_{k\in C_{r+1}}$ means that you sum over all states $k$ that lie in the subset $C_{r+1}.$

*The notation $\mathbb{Z}=\{\dots, -2,-1,0,1,2,\dots\}$ means the set of
integers, and $d \mathbb{Z}=\{dz: z\in\mathbb{Z}\}=\{\dots, -2d,-d,0,d,2d,\dots\}$ is the subset that consists of all multiples of $d$. 
