Build sets and follow membership
Change which elements belong to A and B. Compare the diagram, the roster and the rule for one element.
Given U = {1,2,3,4,5,6}
A={1, 2, 3} · B={3, 4, 5}
Operation
Regions show membership, not quantity. Read the elements or the counts; a larger area does not mean more elements. All complements are taken inside U. The area outside both circles is still inside U.
Neither A nor B · U∖(A∪B)
{6}
B only · B∖A
{4, 5}
A only · A∖B
{1, 2}
Both A and B · A∩B
{3} · Included
Result: A∩B = {3}Keep x only when it belongs to both A and B.
For x=3: 3∈A; 3∈B. Therefore 3∈A∩B.
A≠B in this example. Witness x=1: 1∈A; 1∉B. One such element is enough to refute equality.
Try a boundary case