Aperiodic irreducible chain

For each $x\in E$, $R_x$ is nonempty. If $E=\{x\}$, [F1] gives $p(x,x)=1$, so $p^{(1)}(x,x)>0$. If $E$ has at least two states, fix an arbitrary $x$ and take $y\mathrm{\ne }x$; by [F3]–[F4], there are $r,s\in {\mathbb{N}}_0$ with $p^{(r)}(x,y)>0$ and $p^{(s)}(y,x)>0$. Since $x\mathrm{\ne }y$, [F2] forces $r,s\ge 1$, and [F6] gives $p^{(r+s)}(x,x)\ge p^{(r)}(x,y)p^{(s)}(y,x)>0$. Thus $d(x)$ is a positive integer by [F5] in either case.

For any $x,y\in E$, irreducibility [F3] makes them communicate, so [F7] gives $d(x)=d(y)$. Step 1.1 shows this common value is positive. It is therefore independent of the chosen state and defines $\mathrm{per}(p)$; declaring aperiodicity by the condition $\mathrm{per}(p)=1$ is well-defined.

The empty state space is excluded in the definition because no state period could be chosen as a common value. For a one-state chain, step 1.1 gives period one and hence aperiodicity. On a deterministic cycle of length $k\ge 1$, label the states $x_0,\dots ,x_{k-1}$ and let the transition from $x_j$ go to $x_{(j+1)<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>k}$ with probability one. Iteration gives $p^{(n)}(x_j,x_j)=1$ when $k$ divides $n$ and zero otherwise, so positive return times are precisely the positive multiples of $k$ and the period is $k$; a one-state absorbing chain is the case $k=1$. The return set begins at $n=1$, so $p^{(0)}(x,x)=1$ does not make every chain aperiodic. The proof uses only fixed-pair routes and finite row sums, with no choice function or AC. This is a definition, not an iff theorem.