Question about the necessary condition for disjoint set I ran into a problem asking about the necessary condition for disjoint sets and am quite puzzled by the solution to it. It is given as follows: 
Let $A$ and $B$ be subsets of a universal set. Which of the following is a necessary condition for $A$ and $B$ to be
disjoint?
a) either $A = \emptyset$ or $B = \emptyset$
b) whenever $x \not\in A$, it must occur that $x \in B$
c) whenever $x \not\in A$, it must occur that $x \not \in B$
d) whenever $x \in A$, it must occur that $x \in B$
e) whenever $x \in A$, it must occur that $x \not\in B$
I know for sure that a, c, and d are not a necessary condition for disjoint sets. I can see why (e) is true but why is (b) not true ? I kind of some how have an intuition that it is due to the fact that the empty set is also disjoint from itself and is it because choice (b) violates the fact that the empty set can be disjoint from itself ? (this is just a guess and would like to confirm for sure). 
All help would be appreciated.
Thank you
 A: Condition (b) says that anything not in $A$, is in $B$. That is, together $A$ and $B$ contain everything. This has nothing to do with disjointness: two sets are disjoint exactly when they have no elements in common. They can be "small" - that is, their union can be not all of the universal set. As you point out, we could take $A, B$ to be the emptyset - but maybe a more intuitive example would be to have our universal set be $\mathbb{N}$ and set $A=\{1\}$, $B=\{2\}$. Then condition (b) fails (e.g. $x=3$) but $A$ and $B$ are disjoint. Does this help?
A: When $x \not\in A$ then it is possible that $x \not\in B$. It's not must that $x \in B$ (however it is possible too). If $x \not\in A$, then $x$ can also be the part of universal set and not of the set $B$.
A: To see why (b) is not true Assume that  $B\subsetneq A^c$.
 In other words Assume that $A\cap B = \emptyset$ and $A\cup B $ is not universal set ($A\cup B \subsetneq U$).

 In this way you can find an $x$ that $x \not\in A$ and $x \not\in B.$

