1.5 Working with two sets

There are finite numbers of elements in finite set. This allows us to analyze numbers of elements in different sets that results from the operations carried on them. In this module, we shall study different operations on sets in the context of practical applications. However, we shall limit ourselves to the interaction, involving two sets. The interaction, involving three sets, will be dealt in a separate module.

We use a specific notation to represent the numbers of elements in a set. For example, the numbers in set "A" is represented as "n(A)", whereas we denote numbers of elements in the union as "n( $A\cup B$ )".

Elements in the union of two sets

The area, demarcated with solid line, in the Venn’s diagram, shows the union of two sets denoted by ( $A\cup B$ ). We want to know the numbers of elements in this union in terms of numbers of elements in individual sets.

The sum of the numbers in the individual sets is generally greater than the numbers in the union. The reason is that union includes common elements only once. On the other hand, sum of the numbers of individual sets counts common elements once with each set – in total two times. Clearly, it is required that we deduct the numbers of elements, which are common to each set, from the sum of numbers of elements in individual sets. Hence,

$$n\left(A\cup B\right)=n\left(A\right)+n\left(B\right)-n\left(A\cap B\right)$$

Here, n( $A\cap B$ ) represents the numbers of elements common to two sets. As reminder only, we note that plus (+) operation is not a valid set operation. We, however, use this algebraic operation here as we are now dealing with the numbers in set - not the set.

Alternatively, we can approach this expansion in yet another way. See the representation of intersection of two sets. The union of two sets can be considered to comprise of three distinct regions. Three regions shown with different colors represent three “disjoint” sets. Clearly,

$$n\left(A\cup B\right)=n\left(A-B\right)+n\left(A\cap B\right)+n\left(B-A\right)$$

However, we observe that if we add n( $A\cup B$ ) to either of the two difference sets, then we get the complete individual set.

$$n\left(A\right)=n\left(A-B\right)+n\left(A\cap B\right)$$

$$\Rightarrow n\left(A-B\right)=n\left(A\right)-n\left(A\cap B\right)$$

and

$$n\left(B\right)=n\left(B-A\right)+n\left(A\cap B\right)$$

$$\Rightarrow n\left(B-A\right)=n\left(B\right)-n\left(A\cap B\right)$$

Substituting for the numbers of the difference set in the equation for the numbers in the union set, we have :

$$\Rightarrow n\left(A\cup B\right)=n\left(A\right)-n\left(A\cap B\right)+n\left(A\cap B\right)+n\left(B\right)-n\left(A\cap B\right)$$

$$\Rightarrow n\left(A\cup B\right)=n\left(A\right)+n\left(B\right)-n\left(A\cap B\right)$$

Numbers of elements in the union of “disjoint” sets

Since there are no common elements between two disjoint sets, the intersection between disjoint sets is an empty set. Hence,

$$n\left(A\cup B\right)=n\left(A\right)+n\left(B\right)$$

Application

Application of set theory to real situation is keyed to the interpretation of wordings and description. In order to efficiently employ the concepts of set theory to real world situations, we need to interpret description of collection appropriately.

Once collections are interpreted correctly, rest is easy. There are indeed fewer mathematical operations involved here. Most of these relate to determination of numbers of elements in a set. In this section, we shall first recapitulate or reinterpret different collections and then work with few representative situations for analysis (if we are confident then we can skip the recapitulation part).