$X_1, X_2, X_3,\dots$ is a sequence of random signs, does this sequence converge almost surely?
I know the answer is no and I can use Cauchy in prob to prove it. However, I am very confused about the definition of convergence a.s.
Definition should be $P(w ∈ \Omega, lim X_n(w) \to X(w))=1$; now this is very trivial but obviously the sequence of random signs converges a.s. to a random sign using this definition: $P(w_1 ∈ \Omega, lim X_n(w_1) \to X(w_1))=1/2$, $P(w_2 \in \Omega, lim X_n(w_2) \to X(w_2))=1/2$ if $X(w_1)=1/2$, $X(w_2)=-1/2$. The question didn't specify what it converges to so I can specify it converges a.s. to X defined as another random sign. And I am using the definition of a.s. convergence here, not convergence in distribution.
Please tell me what the flaw in my logic is.
Assume that $(X_n)$ is a sequence of i.i.d. random variables satisfying
$$\mathsf{P}[X_n = 1] = \mathsf{P}[X_n = -1] = \tfrac{1}{2}, \quad \forall \ n \geq 1.$$
Let $A$ be the event that $X_n$ converges. Then for each $\omega \in A$, there exists a real number $X(\omega) \in \mathbb{R}$ and a positive integer $N(\omega) \geq 1$ such that $|X_n(\omega) - X(\omega)| < \frac{1}{2}$ for all $n \geq N(\omega)$. This implies that $|X_n(\omega) - X_{N(\omega)}(\omega)| < 1$ for all $n \geq N(\omega)$, and since both $X_n$ and $X_{N(\omega)}$ take values in $\{-1, 1\}$, this forces that $X_n(\omega) = X_{N(\omega)}(\omega)$. So it follows that
$$ A \subseteq \bigcup_{N\geq 1} \{ \omega \in \Omega : X_n(\omega) = X_N(\omega) \text{ for all } n \geq N \}. $$
Splitting the RHS further depending on the value of $X_N(\omega)$, we get
\begin{align*} \mathsf{P}[A] &\leq \sum_{N\geq 1} \mathsf{P}[\{ \omega \in \Omega : X_n(\omega) = X_N(\omega) \text{ for all } n \geq N \}] \\ &\leq \sum_{N\geq 1} \mathsf{P}[\{ \omega \in \Omega : X_n(\omega) = 1 \text{ for all } n \geq N \}] \\ &\quad + \sum_{N\geq 1} \mathsf{P}[\{ \omega \in \Omega : X_n(\omega) = -1 \text{ for all } n \geq N \}] \end{align*}
But by the mutual independence, it follows that
$$ \mathsf{P}[\{ \omega \in \Omega : X_n(\omega) = 1 \text{ for all } n \geq N \}] = \prod_{n\geq N} \mathsf{P}[\{ \omega \in \Omega : X_n(\omega) = 1\}] = \prod_{n\geq N} \frac{1}{2} = 0 $$
and likewise for the second sum. So $\mathsf{P}[A] \leq 0$ and this implies that $\mathsf{P}[A] = 0$. Therefore $(X_n)$ does not converges a.s.