Different classes can have different recurrence types

The entries of $p$ are nonnegative and its two row sums are $1+0=1$ and $\frac{1}{2}+\frac{1}{2}=1$, so the displayed table is a stochastic matrix.

The row from $0$ is concentrated at $0$. By induction using [F3], $p^{(n)}(0,0)=1$ and $p^{(n)}(0,1)=0$ for every $n\ge 0$. Meanwhile $p^{(1)}(1,0)=\frac{1}{2}>0$, so $1\to 0$ but $0\to ̸1$. By [F1] and the zero-step identity [F2], the communication relation on these two states is equality; hence its two communicating classes are $\{0\}$ and $\{1\}$.

From state $0$, the chain stays at $0$ at every step because $p(0,0)=1$. Thus $T_0^{+}=1$ almost surely and ${\mathbb{P}}_0(T_0^{+}<\mathrm{\infty })=1$; state $0$ is recurrent by [F5].

From state $1$, $T_1^{+}=1$ on the first-step transition to $1$, which has probability $\frac{1}{2}$. On the other first-step transition, $X_1=0$ and the chain then stays at $0$, so $T_1^{+}=\mathrm{\infty }$. Hence ${\mathbb{P}}_1(T_1^{+}<\mathrm{\infty })=\frac{1}{2}<1$, and state $1$ is transient by [F4, F5, given]. This verifies the claimed difference in recurrence type across the two distinct classes.

The witness has two states, so the empty-space and one-state cases cannot arise here. The zero transition $p(0,1)=0$ is essential to the class separation, and the absorbing row at $0$ supplies the no-return branch from $1$. The return time starts at $n=1$, so the initial visit at time zero is not counted. The computation uses only the explicit finite transition table and no choice function; it proves a one-way counterexample, not an iff statement.