Question:
Given a simple random walk $S_n$ with starting point $S_0 = a$ such that $0 \leq a \leq N$,
Prove that $P(S_n \in \{0, N\}$ for some $n \in \mathbb{N}) = 1$
I think this is a pretty simple question in simple random walks, but my textbook doesn't seem to give a proof for it. I rewrote the above as:
$P(S_n \in \{0, N\}$ for some $n \in \mathbb{N}) = P((S_n = 0 \text{ for some n}) \ \cup \ (S_n = N \text{ for some n}))$,
and then I realised that $(S_n = 0 \text{ for some n})$ and $(S_n = N \text{ for some n})$ are not disjoint events, so I can't separate these two events.
Another approach that was suggested to me was to partition this probability into disjoint runs of length k, and then to consider the event that one of these runs contain only steps to the right. However I got no where close to the probability 1 with this method, so I'm not too sure what this approach is supposed to lead to.
How does one prove that this simple random walk will be absorbed with probability 1?