Let $P$ be a probability measure on $\Omega$ with $|\Omega| \le\aleph_0$. Then there is no sequence $A_i$ consisting of independent events with $P(A_i)=p \; \forall i$.
$0<p<1$
Proof by contradiction:
Let $A_i$ be a sequence of independent events with $P(A_i)=p \; \forall i$, whereas $0<p<1$.
$P(\bigcup A_i)=\sum_{i=1}^{n=\aleph_0}P(A_i)=np=\infty$.
Now, because $\Omega = \bigcup A_i$, it should $P(\Omega)=P(\bigcup A_i)$ but $P(\Omega)=1 \neq P(\bigcup A_i)=\infty$
So there exists no such sequence $A_i$
The key point: consider a point $\omega \in \Omega$ and let $\psi_{\omega}$ be the associated probability. for each $i$, $\omega$ is either in $A_i$ or its complement. so, if our sequence existed, we get $\psi_{\omega}$ less than or equal to an infinite product of $p's$ and $(1-p)'s$. Hence $\psi_{\omega}=0\;\forall \omega\in \Omega$. But by countable additivity that is impossible.