Suppose $T_A := \inf\{ n \ge 1 : X_n \in A\}$ where $A \subset \mathcal{S}$ is finite. Assume $\mathbb{P}\{T_A < \infty \; | \; X_0 = x\}= 1$ for $\forall x \in \mathcal{S}-A$. I need to show that there $\exists y \in A$ such that $y$ is recurrent.
I can intuitively argue that if we assume all $y \in A$ are transient, because $A$ is finite, this will lead to $\mathbb{P}\{T_A < \infty \; | \; X_0 = x\} < 1$ for some $x \in \mathcal{S}-A$ which is a contradiction. But are there formal results that I can use to rigorously establish this? Thanks.
If some $y\in A$ is not recurrent, then eventually it jumps to $\mathcal S-A$ (in finite time), which eventually jumps again to $A$ in view of the full probability hypothesis. Actually, the sequence keeps doing this, and it will return infinitely often to $A$ (since it is transient, by hypothesis). But since $A$ is finite, in this sequence some element $z\in A$ (not necessarily $y$) will repeat itself (infinitely often). This element $z$ will be recurrent!
So, actually all elements of $A$ are eventually recurrent, that is, for each point $y\in A$ some of the iterates will be recurrent (although not necessarily the point $y$ itself).