6 1. Let’s Count!
The notationA⊂Bmeans thatAis a subset ofBbut not all ofB.Inthe
chain above, we could replace the⊆signs by⊂.
If we have two sets, we can define various other sets with their help.
Theintersectionof two sets is the set consisting of those elements that are
elements of both sets. The intersection of two setsAandBis denoted by
A∩B. For example, we haveG∩D={Alice}. Two sets whose intersection
is the empty set (in other words, they have no element in common) are
calleddisjoint.
Theunionof two sets is the set consisting of those elements that are
elements of at least one of the sets. The union of two setsAandBis denoted
byA∪B. For example, we haveG∪D={Alice, Carl, Diane, Eve, Frank}.
Thedifferenceof two setsAandBis the set of elements that belong to
Abut not toB. The difference of two setsAandBis denoted byA\B.
For example, we haveG\D={Diane, Eve}.
Thesymmetric differenceof two setsAandBis the set of elements
that belong to exactly one ofA and B. The symmetric difference of
two setsAandBis denoted byA B. For example, we haveG D =
{Carl, Diane, Eve, Frank}.
Intersection, union, and the two kinds of differences are similar to addi-
tion, multiplication, and subtraction. However, they are operations onsets,
rather than operations onnumbers. Just like operations on numbers, set
operations obey many useful rules (identities). For example, for any three
setsA,B, andC,
A∩(B∪C)=(A∩B)∪(A∩C). (1.1)To see that this is so, think of an elementxthat belongs to the set on
the left-hand side. Then we havex∈Aand alsox∈B∪C. This latter
assertion is the same as saying that eitherx∈Borx∈C.Ifx∈B, then
(since we also havex∈C) we havex∈A∩B.Ifx∈C, then similarly we
getx∈A∩C. So we know thatx∈A∩Borx∈A∩C. By the definition of
the union of two sets, this is the same as saying thatx∈(A∩B)∪(A∩C).
Conversely, consider an element that belongs to the right-hand side. By
the definition of union, this means thatx∈A∩Borx∈A∩C. In the
first case, we havex∈Aand alsox∈B. In the second, we getx∈Aand
alsox∈C. So in either casex∈A, and we either havex∈Borx∈C,
which implies thatx∈B∪C. But this means thatA∩(B∪C).
This kind of argument gets a bit boring, even though there is really
nothing to it other than simple logic. One trouble with it is that it is so
lengthy that it is easy to make an error in it. There is a nice graphic way
to support such arguments. We represent the setsA,B, andCby three
overlapping circles (Figure 1.1). We imagine that the common elements of
A,B, andCare put in the common part of the three circles; those elements
ofAthat are also inBbut not inCare put in the common part of circles
AandBoutsideC, etc. This drawing is called theVenn diagramof the
three sets.