I have a problem for a homework that says A and B are events. It then requires me to decide if A and Ω (the sample space) are independent.
My initial guess is to say yes, because if the sample space occurs, that should not influence the probability that A will occur on the next trial.
My proof for this is as follows:
P(Ω & A) = P(A) * P(Ω|A) where P(Ω|A) = P(A)
so P(Ω & A) = P(A) * P(A)
but if they were independent then P(Ω & A) = P(A) * P(Ω) and since P(Ω) = 1,
P(Ω & A) = P(A) and not P(A) * P(A) meaning they are not independent.
So is my proof correct, or my intuition correct? Very confused right now and would appreciate some guidance.
Your intuition is not quite right. To say that $A$ and $B$ are independent events does not say anything about the probability that $B$ will occur on the next trial. Independence of events involves only one trial. I make one observation. Given that $A$ occurred on that trial, what is the probability that $B$ occurred on the same trial?
When $B=\Omega$, knowing that $\Omega$ occurred does not change the probability that $A$ happened since knowing $\Omega$ occurred gives you no information at all (it always occurs). Intuitively, this is why $\Omega$ and $A$ are independent.
Your proof is not correct. $P(\Omega\mid A)$ is not equal to $P(A)$ in general. It's also worth noting that it's not quite correct to just say "we get $P(A)^2$ instead of $P(A)$ therefore this is wrong", since it could be that $P(A)^2=P(A)$, for your reasoning to be complete you would need to mention that $x^2=x$ is only true when $x$ is $1$ or $0$.
A correct proof is to write down the definition of independence for $\Omega$ and $A$:
$$P(\Omega\cap A)=P(\Omega)P(A)$$
and note that $\Omega\cap A=A$ as sets and $P(\Omega)=1$, so that the equation reduces to $P(A)=P(A)$.