This is question 3.3 from Alan Karr's Probability What is the minimum number of points a sample space must contain in order that there exist $n$ independent events none of which has probability zero or one?
I'm thinking the answer is $2^n$, but this is just from checking by hand for the values $n=2$ and $n=3$.  I thought maybe a proof by induction would be appropriate, but didn't make it terribly far with it.  
I'm merely studying probability because I want to be better at it, and this seems to be a pretty classic problem. So it seemed like a good problem to understand.  
 A: Taking into account the independence of events stated by the question and considering that by definition two subsets/events that are mutually exclusive (disjoint) cannot be independent, we can note that any pair of events $X_i$ and $X_j$ must not be disjoint. This means that the intersection between $X_i$ and $X_j$ has to contain at least one point. 
We can generalize this by noting that any intersection between the events $Y_1$, $Y_2$, $Y_3$, ... $Y_N$ has to contain at least one point, regardless of whether $Y_j$ corresponds to $X_j$ or to a "multiple" event $X_j^z$. As a result, all intersections of any combination of $Y_j$, with j = 1 to N, must include at least one point. Because the number of these combinations is $2^N$, we get that the sample space must have a minimal size of $2^N$ as well.  
Also note that this results only represents a lower bound for the size of the sample space, as stated in the question. A simple example of space with size $2^N$ over which $N$ independent events can be identified so that none has probability equal to zero or 1 is given by a sequence of $N$ coin flips. Here the sample space is characterized by $2^N$ points, and under the uniform distribution each space point has probability equal to $1/2^N$. Considering the $N$ events $Y_1$,  $Y_2$... $Y_N$, where $Y_j $ indicates the event that on the j-th flip we get (for example) head,  these are $N$ clearly independent events, each with $1/2$ probability.
A: Consider non-trivial independent events $\{A_i\}_{i=1}^n$. Now, use the fact that for any $\alpha\in\{0,1\}^n$, the family $\{B_i^{\alpha_i}\}$ is independent (see, for example, Theorem 2.5. in Klenke, Prob. Theory), where $B_i^{\alpha_i}=A_i$, if $\alpha_i=0$ and $A_i^c$, if $\alpha_i=1$.
Namely, for any $\alpha\in\{0,1\}^n$, $\bigcap_{i=1}^nB_i^{\alpha_i}$ must be non-empty (otherwise independence fails). By construction, these sets are disjoint and must contain at least one element. Hence, $|\Omega|\ge|\{0,1\}^n|=2^n$.
