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Working with three sets

Module by: Sunil Kumar Singh. E-mail the author

Working with three sets is similar as working with two sets. The underlying characteristics of set operations are same. The union and intersection of sets are carried out with three sets - one after another. Notably, union and intersection operations are commutative. This allows us to extend these set operations to third set in any sequence. Venn’s diagram enables us to visualize resulting set. In this module, we shall first formulate expression for the numbers in the union of three sets. Subsequently, we shall apply the formulation to real time analysis - using both graphical (Venn diagram) and analytical methods.

The union, involving three sets, can be considered in terms of union of a set with "union of other two sets". In that sense, union of three sets represent elements which belong to either of three sets. Here, we want to find the expression of numbers of elements in the union set, which is represented as :

n A B C n A B C

Before, we work on the expansion of this term, let us first find out what does the term " A B C A B C " represent on Venn’s diagram? The figure below shows the representation of this term :

Figure 1: The union set represent elements of three sets combined together.
Union of three sets
 Union of three sets  (th1.gif)

The set shown in the figure above consists of following class of elements :

  1. The elements, which are exclusive of sets “A”, “B” and “C” respectively.
  2. The elements, which are common to a pair of two sets at a time.
  3. The elements, which are common to all three sets.

In the figure below, the common areas between a pair of two sets are marked “1”, “2” and “3”. The common area among all three sets is marked “4”.

Figure 2: The union consists of disjointed regions.
Union of three sets
 Union of three sets  (th2.gif)

Union of three sets

As discussed earlier, the sum of numbers of individual sets is greater than the number of elements in “ A B C A B C ”, unless the sets are disjoint sets. It is imperative that we account for the repetition of common elements. Proceeding as in the case of union of two sets, we deduct the intersections between each pair of sets as :

n A B C = n A + n B + n C - n A B - n A C - n B C n A B C = n A + n B + n C - n A B - n A C - n B C

In this manner, we account for common elements between two sets. However, we have deducted elements "common to all three sets" in this process – three times. On the other hand, the elements "common to all three sets" are present in the numbers of each of the individual sets - in total three times as there are three sets. Ultimately, we find that we have not counted the elements common to all sets at all. It means that we need to account for the elements common to all three sets. In order to add this number, we first need to know – what does this common area (marked 4) represent symbolically?

In the earlier module, we have seen that the area marked “4” is represented by “ A B C A B C ”. Hence, the correct expansion for the numbers of elements in the union, involving three set, is :

n A B C = n A + n B + n C - n A B - n A C - n B C + n A B C n A B C = n A + n B + n C - n A B - n A C - n B C + n A B C

Note:

This result is an important result as the same is used while studying probability.

Union of three sets (Analytical method)

We can achieve this result analytically as well. Here, we consider “A” as one set and “ B C B C “ as other set. Then, we apply the relation, which has been obtained for the numbers in the union of two sets as :

n A B C = n A + n B C n [ A B C ] n A B C = n A + n B C n [ A B C ]

Applying result for the union of two sets for “ n B C n B C ”, we have :

n B C = n B + n C n B C n B C = n B + n C n B C

Putting in the expression for “ n A B C n A B C ”,

n A B C = n A + n B + n C n B C n [ A B C ] n A B C = n A + n B + n C n B C n [ A B C ]

At this stage, our task is to evaluate “ n [ A B C ] n [ A B C ] ”, Recall that we have worked with the distributive property of “intersection operator over union operator”. Following distributive property,

n [ A B C ] = n [ A B A C ] n [ A B C ] = n [ A B A C ]

We can treat each of the terms in the small bracket on the right hand side of the above equation as a set. Applying relation obtained for the numbers in the union of two sets again, we have :

n [ A B C ] = n A B + n A C n [ A B A C ] n [ A B C ] = n A B + n A C n [ A B A C ]

The last term in above equation is :

[ A B A C ] = A B C [ A B A C ] = A B C

Hence,

n [ A B C ] = n A B + n A C n [ A B C ] n [ A B C ] = n A B + n A C n [ A B C ]

Now, putting this expression in the expression of the numbers in the union involving three sets and rearranging terms, we have :

n A B C = n A + n B + n C - n A B - n A C - n B C + n A B C n A B C = n A + n B + n C - n A B - n A C - n B C + n A B C

In the nutshell, we find that numbers of elements in the union, here, is equal to the sum of numbers in the individual sets, minus elements common to two sets taken at a time, plus elements common to all three sets.

Illustration

In this section, we shall work with an example, which is quite intuitive of the analysis, involving three sets. We shall see that analysis of set operations in terms of Venn’s diagram is very direct and simple. As such, we shall first attempt analyze situation with Venn’s diagram.

However, we need to emphasize that extension of set concepts to calculus, probability and other branches of mathematics require that we develop analytical skill with respect to set operations. Keeping this aspect in mind, we shall also work the solution, using analytical method.

Problem : In a town, a total 100000 people read newspaper. Out of these, 40 % read newspaper “A”, 30 % read newspaper “B”, 10 % read newspaper “C”. It is found that 5% read both “A” and “B”; 4% read both “A” and “C”; and 3% read both “B” and “C”. Also, 2% of the people read all three newspapers. Find numbers (i) who read only “A” (ii) who read only “B” (iii) who read neither of three newspapers.

We define three sets “A”, “B” and “C”, corresponding to people reading newspapers “A”, “B” and “C” respectively. From question, we have :

n A = 0.4 X 100000 = 40000 n A = 0.4 X 100000 = 40000

n B = 0.3 X 100000 = 30000 n B = 0.3 X 100000 = 30000

n C = 0.1 X 100000 = 10000 n C = 0.1 X 100000 = 10000

n A B = 0.05 X 100000 = 5000 n A B = 0.05 X 100000 = 5000

n A C = 0.04 X 100000 = 4000 n A C = 0.04 X 100000 = 4000

n B C = 0.03 X 100000 = 3000 n B C = 0.03 X 100000 = 3000

n A B C = 0.2 X 100000 = 2000 n A B C = 0.2 X 100000 = 2000

Venn's diagram method

We observe that sum of the individual sets is less than total numbers of people , reading newspaper, in the town. Hence, total reading population represents universal set, “U”. The representations of these sets are shown on Venn’s diagram. Note that we have split the elements common to a pair of two sets in two parts (a) elements exclusive to intersection of two sets and (b) elements common to all three sets.

Figure 3: The regions common to sets.
Union of three sets
 Union of three sets  (th3.gif)

From the diagram,

(i) The required set is the region of “A” not common to “B” and “C”. This region represents elements, which are exclusive to set “A”. Thus, numbers of people reading only “A” is :

n 1 = 40000 3000 + 2000 + 2000 = 33000 n 1 = 40000 3000 + 2000 + 2000 = 33000

(ii) The required set is the region of “B” not common to “A” and “C”. This region represents elements, which are exclusive to set “B”. Thus, numbers of people reading only “B” is :

n 2 = 30000 3000 + 1000 + 2000 = 24000 n 2 = 30000 3000 + 1000 + 2000 = 24000

(iii) The required set is the remaining region of universal set “U” i.e. complement of the union of three sets. Now, proceeding as before, people who read newspaper “C” only is (see Venn’s diagram above):

n 3 = 100000 2000 + 1000 + 2000 = 5000 n 3 = 100000 2000 + 1000 + 2000 = 5000

Hence, the required number is :

Figure 4: The regions representing people, who read neither of three newspapers.
Union of three sets
 Union of three sets  (th4.gif)

n 4 = 100000 33000 + 24000 + 5000 + 3000 + 2000 + 1000 + 2000 n 4 = 100000 33000 + 24000 + 5000 + 3000 + 2000 + 1000 + 2000

n 4 = 30000 n 4 = 30000

Analytical method

(i) Here we are required to find the numbers of people reading only “A”. It is clear that this set is part of the people, who do not read newspapers “B” and “C”. As discussed in the case of two sets, the numbers of people who read neither “B” and “C” is given by De-morgan’s first equation,

B C = B C B C = B C

Figure 5: The region representing people, who read neither of two newspapers.
Intersection of two complement sets
 Intersection of two complement sets  (th5.gif)

For clarity, we have shown this region in the Venn’s diagram. We should realize that this intersection of two sets also includes people who read newspaper “A”. However, we are required to know numbers of people, who read newspaper “A” only. The exclusion people reading other newspaper as well are not part of our required set.

The remaining set (refer Venn’s diagram) represent area consisting of people who exclusively read newspaper “A” or who do not read any of three newspapers. Now, we need the intersection of set A with the remaining region to obtain the numbers, who read only newspaper “A”. Hence, required number is :

n 1 = n A B C n 1 = n A B C

Using De-morgan's equation,

n 1 = n A B C = n [ A B C ] n 1 = n A B C = n [ A B C ]

n 1 = n [ A [ U B C ] ] = n [ A U A B C ] ] n 1 = n [ A [ U B C ] ] = n [ A U A B C ] ]

n 1 = n [ A A B C ] = n A n [ A B C ] n 1 = n [ A A B C ] = n A n [ A B C ]

Using distributive property of intersection over union,

n 1 = n A n [ A B A C ] n 1 = n A n [ A B A C ]

Using formula of expansion of union of two sets,

n 1 = n A [ n A B + n A C n [ A B A C ] ] n 1 = n A [ n A B + n A C n [ A B A C ] ]

n 1 = n A [ n A B + n A C n A B C ] n 1 = n A [ n A B + n A C n A B C ]

We see that values of each term on the right hand side are given. Putting these values,

n 1 = 40000 [ 5000 + 4000 2000 ] = 33000 n 1 = 40000 [ 5000 + 4000 2000 ] = 33000

(ii) Here we are required to find the numbers of people reading only “B”. Proceeding as before,

n 2 = n B [ n B A + n B C n B A C ] n 2 = n B [ n B A + n B C n B A C ]

n 2 = n B [ n A B + n B C n A B C ] n 2 = n B [ n A B + n B C n A B C ]

Putting values,

n 2 = 30000 [ 5000 + 3000 2000 ] = 24000 n 2 = 30000 [ 5000 + 3000 2000 ] = 24000

(iii) In order to find the numbers of people, who read neither of three newspapers, we first find union of three sets. The union represents people who read either of these newspapers – one, two or all three. Clearly, people, who do not read either of these papers constitute a complement of this union.

n 3 = A B C = U A B C n 3 = A B C = U A B C

Using expansion for the numbers in the union of three sets,

n 3 = U [ n A + n B + n C n A B n A C n B C + n A B C ] n 3 = U [ n A + n B + n C n A B n A C n B C + n A B C ]

Putting values, we have :

n 3 = 100000 [ 40000 + 30000 + 10000 5000 4000 3000 + 2000 ] = 30000 n 3 = 100000 [ 40000 + 30000 + 10000 5000 4000 3000 + 2000 ] = 30000

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