What is the minimum number of points a sample space must contain that there exist n independent events that wouldnt have probability of zero or one This is classic problem from most of the probability theory textbooks. What is the minimum number of points a sample space must contain in order that there exist $n$ independent events $A_1,\ldots,A_n$ none of which has probability zero or one? The answer I obtained by drawing Venn diagrams for cases $n=2,3$ was $2^n$. Cannot prove it for general case, meaning for any $n$. 
 A: If each of the events $A_1,\ldots,A_n$ has probability between zero and one, then by independence the same is true for all intersections $B_1\cap\cdots\cap B_n$ of events where $B_i$ is either $A_i$ or the complement of $A_i$.  These intersections form a collection of $2^n$ disjoint sets, and each one in the collection is nonempty (since each one has positive probability), so each one contains at least one point not in any of the others. Conclude there are at least $2^n$ points in the sample space.
You can come up with an example where the value $2^n$ is achieved.
A: Let's try proving it by induction on $n$.  Suppose for $n$ events it's $2^n$.  Then we add an $(n+1)$th event $A_{n+1}$.
Then each of the $2^n$ outcomes $\omega$ that we already had needs to be split into two separate outcomes $\omega_0$ and $\omega_1$. When the outcome is $\omega_0$, each of the $n$ events $A_1,\ldots,A_n$ either occurs or fails just as it did when the outcome was $\omega$, and the $(n+1)$th event fails. When the outcome is $\omega_1$, each of the $n$ events $A_1,\ldots,A_n$ either occurs or fails just as it did when the outcome was $\omega$, and the $(n+1)$th event occurs.
Splitting each outcome into two outcomes yields twice as many outcomes as before, thus $2^n\times 2 = 2^{n+1}$.
A: Independence among $B_1$, $B_2$,...,$B_n$ means $\prod_{i\in I}P(B_n)=P(\bigcap_{i\in I}B_n)$ for all $I\in2^{[n]}$ and $\#(I)>=2$
there are $2^n-n-1$ equations (remove the 0 and 1 $B_n$'s)
Claim: under the assumption of problem, the $2^n$ possible intersection $\bigcap_{i\in I}B_n$ ($I\in 2^{[n]}$) are all distinct. 
If not, $\bigcap_{i\in I_1}B_n=\bigcap_{i\in I_2}B_n$, then $\bigcap_{i\in I_1}B_n=(\bigcap_{i\in I_2}B_n)\cap (\bigcap_{i\in I_1}B_n)=(\bigcap_{i\in I_1 \cup I_2}B_n)$
then we have $\prod_{i\in I_1}P(B_n)=\prod_{i\in I_1 \cup I_2}P(B_n)$ => $\prod_{i\in I_2 - I_1}P(B_n)=1$, since $P(B_n)>0, \forall n$=> $P(B_n)=1, \forall n\in I_2-I_1$. it contradicts with the assumption that $P(B_n)\neq 1$. therefore, $I_2-I_1= \emptyset$. similarily, $I_1-I_2=\emptyset $, that is, $I_1=I_2$.
$B_1,..,B_n$ divides $\Omega$ into $2^n$ partitions (2 partitions ($A,A^c$) for only one event; 4 partitions ($A-B,AB,B-A,(A\cup B)^c$) for 2 events, etc), and to make all $2^n$ intersections distinct, at least one element in each partition (with positive probability).
