On proving events have nonempty intersection if the sum of their complement is smaller than 1 Suppose for Events $A_1, A_2,\ldots,A_n$ we have that:
$$\sum\limits_{i=1}^n {\mathbb P}(A^{c}_i) < 1 $$
Does this imply:
$$\bigcap_{i=1}^{n} A_i \neq \emptyset $$
I think it does, but I couldn't manage to prove it, anybody please give some hints! Thanks a lot!

Here's a hint:
Start with n = 2. Suppose that 
$$\sum\limits_{i=1}^2 {\mathbb P}(A^{c}_i) = 1$$
We also know that 
$$P(A_1^c \cup A_2^c)=P(A_1^c)+P(A_2^c)-P(A_1^c \cap A_2^c) $$
Since probability is bounded at 1, it is clear that $P(A_1^c \cap A_2^c)$ must equal 0. 
By De Morgan's laws, we know that 
$$P(A_1^c \cap A_2^c) = P((A_1 \cup A_2)^c)$$
Since
$$P((A_1 \cup A_2)^c) = 0$$
Then
$$P(A_1 \cup A_2) = 1$$
From here, 
$$P(A_1 \cup A_2) = P(A_1) + P(A_2) - P(A_1 \cap A_2)$$
$$1 = P(A_1) + P(A_2) - P(A_1 \cap A_2)$$
$$1 = (1 - P(A_1^c)) + (1 - P(A_2^c)) - P(A_1 \cap A_2)$$
$$1 = 2 - [P(A_1^c) + P(A_2^c)] - P(A_1 \cap A_2)$$
From the last equation we can see that if, as we first conjectured, $P(A_1^c) + P(A_2^c) = 1$, then $P(A_1 \cap A_2) = 0$. Otherwise, in order for the equality to be satisfied, $P(A_1 \cap A_2)$ must be greater than 0.
Now try to extend this to n > 2 and your proof will be complete.
