Consider the simple symmetric random walk (SSRW) on the 1D integer lattice. If we take any finite set of integers, say, $\{-N,-N+1,\ldots,N-1,N\}$, then, with probability one, it hits every integer in this set infinitely many times. But can we say that it hits every integer infinitely many times with probability one?
Here is my attempt:
Let $\Omega$ be the sample path space of all random walk paths. And define events $$A_n=\{\text{hits all integers} -n,\ldots,n \text{ infinitely-many times}\}\subset\Omega.$$ Clearly we have $$A_1\supset A_2\supset \cdots$$ since a sample path that hits $-(n+1),\ldots,n+1$ infinitely often also hits $-n,\ldots,n$ infinitely often. So we have a decreasing sequence of events, and so by continuity of probability measure we get $$\mathbb P(\cap_{n=1}^\infty A_n)=\lim_{n\to\infty} \mathbb P(A_n)=1.$$
Now I have some unease about interpretation of $\cap_{n=1}^\infty A_n$ as a subset of the path space. It seems to be nonempty, since it must have probability one, but does $\omega\in\cap_{n=1}^\infty A_n$ really hit every integer infinitely often? I have trouble envisioning such a path. It seems kind of related to an "irrational sequence" in $\{-1,1\}^{\mathbb N}$ where we see every possible finite sequence of plus and minus 1's infinitely many times.
So we can say that the "typical" random walk path does actually hit every integer infinitely many times. I envision this as the typical path just goes up arbitrarily high and down arbitrarily low infinitely many times, and it will always go higher and lower (than any arbitrary upper and lower bounds) infinitely many times.
Am I thinking about this in a reasonable way, and is the argument correct?
You can show that $\liminf X_n=-\infty$ and $\limsup X_n=+\infty$, both with probability one. On intersection of these two events every integer must be visited infinitely often.