Define the $k$-th return to $y$ as $T_y^k= \min\{n \ge T_y^{k-1} +1 : X_n = y\}$
I am trying to prove as a "left to the reader" that this is a stopping time. I understand how to argue intuitively that it is a stopping time, but writing it out formally is proving difficult.
My Attempt:
Proof by Induction: We have that $T_y = \min\{ n \ge 1 : X_n = y\}$ is a stopping time because it may be described fully by $\{T_y = n_1\} = \{X_i \ne y \text{ and } X_0, X_{n_1} = y\} \quad (1 \le i \le n_1-1)$This description depends only on the past states of the $X_i$s. With the base case established, assume that this holds for $j = k-1$ and show that this implies that $T_y^{j+1} = T_y^k$ is a stopping time.
Then $T_y^k$ may be described by $\{T_y^k = n_k\} = \{X_i \ne y \text{ and } X_{n_{k-1}}, X_{n_k} = y\} \quad (n_{k-1}+1 \le i \le n_k-1) $
Since $T_y^{k-1}$ was a stopping time, we know that $T_y^{k-1}$ can be determined by $X_0, \dots, X_{n_{k-1}}$, so we know that $X_i$ has assumed the value of $y$ $k-1$ times, and we need only consider those $X_i$ for $i \ge n_{k-1}$. We need no future information when determining $T_y^k$, and so $T_y^k$ is a stopping time.
Where are my inconsistencies? Can I make the assumptions I listed in the last paragraph?