Can a set and its complement have intersection?

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?

• How do you define the complement? – Lee Mosher Feb 8 '17 at 2:03
• Um... by definition.... If foozlenucks are precisely everything that is not a duck, then is it possible for there to be a duck that is a foozlenuck? – fleablood Feb 8 '17 at 2:13
• Take an element $x$ from the intersection, is it in A? – dave Feb 8 '17 at 2:14
• A decision-problem is in coNP iff its complement (as a problem not as a set) is in NP. So it's prefectly possible to be in both NP and coNP, this just depends on the problem at hand. – Henno Brandsma Feb 8 '17 at 5:11