My Question: Let $(\Omega,\mathcal{F},\textbf{P})$ be a probability triple such that $\Omega$ is ${countable}$. Prove that it is impossible for there to exist a sequence $A_1,A_2,\ldots \in \mathcal{F}$ which is ${independent}$ such that $\textbf{P}(A_i)=\frac{1}{2}$ for each $i$. Hint: First prove that for each $\omega\in \Omega$, and for each $n\in \mathbb{N}$ we have $P(\{\omega\})\leq \frac{1}{2^n}$. Then derive a contradiction.
My Work: Let $\omega \in \Omega$ be arbitrary. Then since $\textbf{P}(\Omega)=\textbf{P}(\bigcup \{\omega\})=1$, and $(\Omega,\mathcal{F},\textbf{P})$ is a valid probability triple, then $\textbf{P}$ is countably additive, so that $\textbf{P}(\bigcup \{\omega\})=\sum\limits_{n=1}^\infty \textbf{P}(A_n)$. Here I am not sure where to go. My problem: I am really not seeing how to go about this problem. Any help is appreciated.