I have found the following theorem in a book:
Let $s \in S$ be any state of an irreducible Markov chain on state space $S=\{0,1,2,...\}$. The chain is recurrent if there exists a solution $\{ y_j : j \not= s \}$ to the inequalities $$ y_i \geq \sum_{j: j \not=s} p_{ij} y_j, \quad i \not=s,$$ such that $y_i \rightarrow \infty$ as $i \rightarrow \infty$.
Now my professor said that I can't use it if $S$ is finite. I have to give an additional condition in that case, because $i$ is in the sense of a state and not an integer number. Is he right?
I think he is wrong, since one can see in the following example I figured out. Am I right?
Example: Let $S=\{0,1,2 \}$ with transition matrix $$P=\begin{pmatrix} 2/3 & 1/3 & 0 \\ 2/3 & 0 & 1/3 \\ 0 & 2/3 & 1/3 \end{pmatrix}$$.
I take $s=0$ and claim that $\{ y_j=j : j > 0 \}$ is a solution to the inequalities above. Indeed: $$i=1 : \ y_1=1 \geq p_{11}y_1 + p_{12}y_2=2/3.$$ $$i=2 : \ y_2=2 \geq p_{21}y_1 + p_{22}y_2=4/3.$$ And for $i>2$, we have that $i \geq 0$ always. Further $y_i \rightarrow \infty$ as $i \rightarrow \infty$.
As stated, the theorem explicitly requires S to be an infinite set.
However, it is also true for finite sets (trivially): Every finite irreducible Markov chain is recurrent. This follows from the pigeon hole principle: if there are finitely many states, then over infinitely many time steps, some state must be visited infinitely many times. Then using irreducibility, every state must be visited infinitely many times.