On page 54 of the book "Basic Stochastic Processes" of Brzezniak and Zastawniak, author proposes this example:
A coin is tossed repeatdely and you win or lose 1 pound depending on which way it lands. You start with 5 pounds and you will play until you have no money or if you have 10 pounds. If $X_{n}$ is the amount of money at $n$ toss, then the time when the game stops is: $$\tau = min\{n: X_{n} = 0, X_{n} = 10 \}$$
Prove $\tau$ is a stopping time.
Author gives the following solution: $X_{n}$ is adapted to the filtration $F_{n} = \sigma(X_{1}, X_{2}, ... X_{n})$ and $\tau$ is n stopping time because: $$\{\tau = n\} = \{0<X_{1}<10\} \cap \{0<X_{2}<10\} ... \cap \{0<X_{n-1}<10\} \cap \{X_{n} = 0\} \cap \{X_{n} = 10 \} \in F_n$$
What I don't understand is why the proof isn't like this: $$\{\tau = n\} = \{0<X_{1}<10\} \cup \{0<X_{2}<10\} ... \cup \{0<X_{n-1}<10\} \cap \{X_{n} = 0\} \cap \{X_{n} = 10 \} \in F_n$$
Because it will include all the times when I didn't have 10 or 0 pounds. $F_{n}$ is closed under intersections and unions, so my proposed solution will prove $\tau$ is a stopping time too.
With probability $1$ the process reaches one of the stop states. That is a fact about random walks that has nothing to do with filtrations.
Intersections are used in the logical formula because to stop at time $= n$ the process should not stop at time $1$ AND not stop at time $2$ AND not stop at .... time $n-1$. The purpose of the formula is to show that stopping at time $\leq t$ depends only on information in the filtration at indices $\leq t$.