My question is really simple.
Can a set $A$ and its complement $\bar{A}$ have intersection? I cannot prove it nor find a counterexample.
EDIT
My question is general but it comes to my mind after seeing the well open problem in computational complexity, is NP=coNP?. Here NP is the set of all decision problems for which the instances where the answer is "yes" have efficiently verifiable proofs and coNP is its complement. How a set and its complement be equal?
Best Answer
Of course they have an intersection. Any two sets have an intersection.
In this case the intersection is the empty set. But that doesn't stop it from being the intersection.
As for NP and coNP, they are not complements. There are problems that are in both NP and coNP -- such as all the decidable ones -- and problems that are in neither, such as the set of true statements of first-order integer arithmetic.
CoNP is the set of decision problems for which the instances where the answer is "no" have efficiently verifiable proofs. A given set is in coNP iff its complement is in NP, but coNP itself is not the complement of NP.