Basic Set Theory - inside-economics.cominside-economics.com/s/Mathematics For Economists/1....

www.inside-economics.com

Inside ECONOMICS

Mathematics for Economists

Basic Set Theory

Introduction to Sets

Set theory is an important topic in mathematics and applied economics. Sets are much more extensively used than people think. In applied economics we often work with sets of objects or data points which are contained within a set of possible values. Essentially a set is a collection of objects. These objects are known as elements. The idea of sets is broad and as such a set can be made up of almost anything conceivable. For instance it could be a set of numbers, books, alphabetical letters, individuals, firms, households, cars, Elvis Presley or students. An element is an individual member of the set. For example let set 𝐴 contain three elements 1,2 and 3. We can write the set A formally as 𝐴 = {1,2,3}. Additionally we can write 2 as an element of A as, 2 ∈ 𝐴. If an element does not belong to a set we use the sign ∉. There we can write −3∉𝐴 which means that −3 is not an element in the set 𝐴. Example 1: Thinking about sets in applied economics

In microeconomics we are often studying an income as a function of education, experience and other factors. When we define income we know that it must be a positive number as no one, for a given quantity of labour, can have a negative income. In other words it is reasonable to assume that no one goes to work and pays their manager for the privilege of providing their labour. Therefore we define income as a set 𝐼: [0, +∞), where income has a lower bound of 0 and an upper bound of positive infinity which although very unlikely, is possible. In addition if labour is paid per hour and measured in hours per day spent at work, the set of possible hours at work for a given day is the set 𝐿: {0,1,2 … ,24}. The minimum hours in a day that someone can work is zero and the maximum labour input per day is 24 hours. It is important to note that when we present sets the order in which we write the element is inconsequential and does not matter. For instance if we want to show the set of integer numbers we can denote the set in various ways. The set of integer number ℤ can be expressed in set notation in many ways such as:

ℤ: {0,1,−1,2,−2,3,−3, … } or, ℤ: {…− 3,−2,−1,0,1,2,3, … } or, ℤ: {… ,−1,0,1, … } Sets of Sets: It is possible for elements of a set to be sets. For example the 𝐴 has three elements of which one element {1,2,3} is actually a set.

𝐴 = {𝑋, {1,2,3},𝑍}

It is important to note that the set and, not the objects contained within the set, is an element of 𝐴 . For instance {1,2,3} is an element 𝐴, however the object 1 is not an element 𝐴.


Inside ECONOMICS

Finite Sets: If all the elements of a set can be listed then the set is called a finite set. The sets A and B are examples of finite sets. Set A has 3 elements, set B 5 elements and set C has 6 elements.

𝐴 = {𝑋,𝑌,𝑍}

𝐵 = {1,2,3,4,5}

𝐶 = {1000,2001,3003,4004,5999,6555}

Infinite Sets: An infinite set is a set where we cannot possibly list all the numbers. Therefore we denote an infinite set as follows.

𝐷 = {1,2,3,4, … } or 𝐷 = {1,2,3,4, …∞}

For instance we can write the sets of natural numbers N and the set of real numbers ℝ in the following way.

𝑁 = {1,2,3,4, …∞}

ℝ = {1,2,3,4, …∞}

Intervals: If we have two number 𝑎 and 𝑏 then the set of all the numbers between 𝑎 and 𝑏 and is called an interval of the set. An open interval is where the end points 𝑎 and 𝑏 are excluded and is denoted as:

(𝑎, 𝑏) = {𝑥 ∈ ℝ1:𝑎 < 𝑥 < 𝑏}

A closed interval is where the end points 𝑎 and 𝑏 are included and is denoted as:

[𝑎, 𝑏] = {𝑥 ∈ ℝ1:𝑎 ≤ 𝑥 ≤ 𝑏}

If only one of the two end points is included in the interval then is called a half closed or half open set which we denote as (𝑎, 𝑏] or [𝑎, 𝑏). Given the above definitions we know that there are types of infinite intervals where we have the number 𝑎 and ±∞ :

(−∞,𝑎] = {𝑥 ∈ ℝ1: 𝑥 ≤ 𝑎}

(−∞,𝑎) = {𝑥 ∈ ℝ1: 𝑥 < 𝑎}

[𝑎, +∞) = {𝑥 ∈ ℝ1: 𝑥 ≥ 𝑎}

(𝑎, +∞) = {𝑥 ∈ ℝ1: 𝑥 > 𝑎}

(−∞, +∞) = ℝ1

The Empty Set: The empty set or null set contains no elements. Every subset contains the empty set, including itself. So we may not always expressly show the empty set when we write out a set but it is important to note that the empty set will be a subset of any set. We denote the empty set as the following:

𝐸 = { } or ∅

The empty set is the only set in which every property holds. Please note that the empty set is a subset of every set


Inside ECONOMICS

Equal Sets: Equal sets are two sets which contain the same elements but the elements are in different order. Although the sets A and B, below have different order they are equal because they contain the same elements. Remember that the order of a set is irrelevant. If 𝐴 = {5,3,2,4,1} and 𝐵 = {1,2,3,4,5}, then the sets are equal (𝐴 = 𝐵)

Subsets: A subset is a set within a set. The set 𝐴 is a subset of 𝐵 if and only if every element of 𝐴 is an element of 𝐵 which is denoted as 𝐴 ⊂ 𝐵. A subset is defined as,

𝐴 ⊂ 𝐵 𝑖𝑓 𝑎𝑛𝑑 𝑜𝑛𝑙𝑦 𝑖𝑓 ∀𝑥 ∈ 𝐴 , 𝑥 ∈ 𝐵 Also because these sets are equal it holds that if 𝐴 ⊂ 𝐵 and 𝐵 ⊂ 𝐴 then 𝐴 = 𝐵

If 𝐴 = {1,2,4,7} and 𝐵 = {𝟏,𝟐, 3,𝟒} then 𝐴 ⊂ 𝐵 because every element of 𝐴 is an element in 𝐵.

Proper Subset: A proper subset is a subset where not every element in 𝐵 is an element in 𝐴. This means that 𝐵 is a larger set that contains the smaller set 𝐴.We denote the proper subset as 𝐴 ⊊ 𝐵. Formally:

𝐴 ⊊ 𝐵 𝑖𝑓 𝑎𝑛𝑑 𝑜𝑛𝑙𝑦 𝑖𝑓 (𝑥 ∈ 𝐴 , 𝑥 ∈ 𝐵 𝑎𝑛𝑑 𝑡ℎ𝑒𝑟𝑒 𝑒𝑥𝑖𝑠𝑡𝑠 𝑧 ∈ 𝐵 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝑧 ∉ 𝐴)

Essentially the proper set means that 𝐴 is the subset of 𝐵 but 𝐵 is larger than 𝐴. So there is at least one more element in 𝐵 than in 𝐴.

If 𝐴 = {1,2,4,7} and 𝐵 = {𝟏,𝟐, 3,𝟒, 5,6,𝟕, 8,9,10} then 𝐴 ⊊ 𝐵 because every element of 𝐴 is an element in 𝐵, and 𝐵 has at least one more element than 𝐴. In fact in this case 𝐵 has 6 more elements than 𝐴.

The domain of a function defined as a subset: Some functions are based only on subsets of ℝ. Given the function 𝑓 the set of numbers 𝑥 which the function 𝑓(𝑥) is defined is called the domain of 𝑓. There are reasons why the domain of a function may be limited. The most common reasons for restricted the domain is that we cannot divide by zero, and cannot take the square root or logarithm of a negative number. For example 𝑓(𝑥) = 1

𝑥2 the domain is

all 𝑥 except for −1 or +1. The domain of a function may also be restricted by the economic application. In economic applications we mostly deal with sets with non negative numbers and therefore the domain is restricted. If the function 𝑓(𝑥) is the cost to produce 100 computers then we can denote the domain of 𝑓 as:

ℝ+ ≡ {𝑥 ∈ ℝ1:𝑥 ≥ 0}

Example 2: Subsets

In the following diagram 𝐷 is a proper subset of the set 𝐴 because every element in 𝐷 is also an element in 𝐴.

𝐷 is a subset of 𝐴 if and only if every element of 𝐷 is an element of 𝐴, 𝐴 ⊊ 𝐵

𝐴

𝐷

𝐷 is a subset of 𝐴 because if we pick an arbitrary element of 𝐷 say 𝑥1 then this element is also in the set 𝐴 𝑥1


Inside ECONOMICS

Example 3: Subsets

In the diagram below set 𝐷 is not a subset of 𝐴 because, although some elements are, not all elements in 𝐷 are elements in the set 𝐴. To be a subset every element of 𝐷 must be an element in 𝐴.

The Universe::The universe is the space of all possible elements. For example, when working with real numbers, then ℝ could be the universe. Intersection of two sets: The set containing the elements that are common to both sets is called the intersection and is denoted by ∩. The intersection of two sets is denoted as: 𝐴 ∩ 𝐵 = {𝑥 ∈ 𝐸|𝑥 ∈ 𝐴 𝑎𝑛𝑑 𝑥 ∈ 𝐵} The intersection of many sets: Given a collection of many sets 𝐴𝑛, where the index values 𝑛 that belong to some other set called 𝑁, then:

�𝐴𝑛𝑛∈𝑁

= {𝑎| ∃ 𝑎 ∈ 𝐴𝑛 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑛 ∈ 𝑁}

Example 4: Intersection

Suppose that we have the following two sets 𝑄 = {𝟐, 4,𝟔, 8} and 𝑅 = {1,𝟐, 5,𝟔, 7,9,11}. The intersection of 𝑄 and 𝑅 is 𝑄 ∩ 𝑅 = {2,6}, where 2,6 ∈ 𝑄 ∩ 𝑅. The intersection will include all the values that are elements in both sets.

𝑄 ∩ 𝑅 = {2,6}, where 2,6 ∈ 𝑄 ∩ 𝑅

𝑅

𝑄

𝑄 is not a subset of 𝑅 because not every element in 𝑄 is an element in the set. The intersection of the two sets contains the two elements common to both sets.

2 6

𝐷 is a subset of 𝐴 if and only if every element of 𝐷 is an element of 𝐴

𝐴

𝐷

𝐷 is not a subset of 𝐴 because if we pick an arbitrary element of 𝐷 say 𝑥1 then this element is not an element in the set 𝐴. Remember that every element in the subset must be an element in the set.

𝑥1 𝑥2


Inside ECONOMICS

Example 5: The intersection between 3 sets Suppose we have 3 sets 𝐴: {1,2,3,4}, 𝐵: {2,3,4,5,6,7} and 𝐶: {3,4,8,9,10} the intersection of 𝐴,𝐵 and 𝐶 will be the set of elements that are common to all three sets. That is 𝐴 ∩ 𝐵 ∩ 𝐶 = {3,4}

Example 6: Intersection

Find the intersection between the following sets, 𝐴: {1,2,3} and 𝐵: {4,5,6} ?

In this case there is not intersection between the sets. This is known as a disjoint.

Unions of two sets The union of two sets is the set which contains all members of the two sets and is denoted by the symbol ∪. The union of two sets is denoted as:

𝐴 ∪ 𝐵 = {𝑥 ∈ 𝐸|𝑥 ∈ 𝐴 𝑜𝑟 𝑥 ∈ 𝐵} The unions of many sets: Given a collection of many sets 𝐴𝑛, where the index values 𝑛 that belong to some other set called 𝑁, then:

�𝐴𝑛𝑛∈𝑁

= {𝑎| ∃ 𝑛 ∈ 𝑁 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝑎 ∈ 𝐴𝑛}

𝐵 𝐴 As you can see there is no overlap in between these sets and therefore There are no elements common to both sets and therefore no intersection.

𝐵

𝐴 The intersection between the three sets is the set containing 3 and 4. It is easy to notice that 𝐴 ∩ 𝐵 = {2,3,4} and that 𝐴 ∩ 𝐶 = {3,4} and 𝐵 ∩ 𝐶 = {3,4}

𝐶

3 4

2

𝐴 ∩ 𝐵 ∩ 𝐶 = {3,4} where 3, 4 ∈ 𝐴 ∩ 𝐵 ∩ 𝐶

𝐴 ∩ 𝐴 = 𝐴

𝐴 ∩ 𝐵 = 𝐵 ∩ 𝐴

(𝐴 ∩ 𝐵) ∩ 𝐷 = (𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐷)

For any sets 𝐴,𝐵 and 𝐷, the following statements hold:


Inside ECONOMICS

Example 7: The Union of two sets

If we have the sets 𝑄 = {2,4,8} and 𝑅 = {1,2,5,6,7,9,11} then the union will be all the elements of both sets. This means that every element of both sets is part of the union. Therefore the union between the sets 𝑄 and 𝑅 is 𝑄 ∪ 𝑅 = {1,2,4,5,6,7,8,9,11}.

Example 8: The union between 3 Sets

Again, we have 3 sets 𝑆: {1,2,3,23}, 𝑇: {1,2,6,7,15} and 𝑍: {−3,−2,1,9,10} the union of 𝐴,𝐵 and 𝐶 will be the set of all elements of three sets. That is 𝑆 ∪ 𝑇 ∪ 𝑍 = {−3,−2,1,2,3,6,7,9,10,15,23}

Disjoint The sets 𝑋 and 𝑌 are defined as a disjoint if there are no common elements between the two sets. In other words each set is completely different to the other set. We have already seen a disjoin in example 6. Example 9: The Disjoint

If 𝑋: {2,4,6,8} and 𝑌: {1,3,5,7,9,11} then we call this a disjoint, as there are no elements common to both sets.

𝐵

𝐴 The union between the three sets is the set containing all the elements of all the sets. It includes all the elements in the combined space of the three sets in the diagram.

𝐶

𝑄 ∪ 𝑅 = {1,2,4,5,6,7,8,9,11}

𝑅

𝑄

The union of the two sets contain all the elements of both sets. No element in either set is left out.

2

11 8 4

1 5 6 7 9

𝑨 ∪ 𝑨 = 𝑨

𝑨 ∪ 𝑩 = 𝑩 ∪ 𝑨

(𝑨 ∪ 𝑩) ∪ 𝑫 = 𝑨 ∪ (𝑩 ∪ 𝑫)

For any sets 𝐴,𝐵 and 𝐷, the following statements hold:


Inside ECONOMICS

The Compliment: Let 𝐴 be a set in space 𝑅 then we define the compliment as 𝐴𝑐 = {𝑥 ∈ 𝑅|𝑥 ∉ 𝐴} .The compliment of A and B is a all the elements in the sample space that are not elements of A or B. This is shown in general form in the following diagram below. Where 𝑈 represents the union between the sets 𝐴 and 𝐵 we define the compliment of the union as 𝑈′ or 𝑈𝑐.

𝑈′ = {𝑥: 𝑥 ∈ 𝑈 𝑎𝑛𝑑 𝑥 ∉ 𝐴 ∪ 𝐵}

Example 10: Complement of Set

Suppose we have the set 𝐴: {𝑎, 𝑏, 𝑐,𝑑} in the space 𝑅: {𝑎, 𝑏, 𝑐,𝑑, 𝑒,𝑓,𝑔,ℎ} then the compliment of the set 𝐴 is 𝐴𝑐 = {𝑥 ∈ 𝑅|𝑥 ∉ 𝐴} = {𝑒,𝑓,𝑔,ℎ}

Example 11: Compliment of a Set

Suppose we have the set 𝐴: {1,2,3,4} in the space 𝑅: {1,2,3,4,5, … ,100} then the compliment of the set 𝐴 is 𝐴𝑐 = {𝑥 ∈ 𝑅|𝑥 ∉ 𝐴} = {5,6,7, … 100}

𝐴𝑐 𝐴

The area 𝐴𝑐 comprises all the elements that are not in the set 𝐴. 𝐸 represents the entire space including both 𝐴 and 𝐴𝑐

𝐸

𝐴𝑐 𝐴

The area 𝐴𝑐 comprises all the elements that are not in the set 𝐴.

𝐵 𝐴 As you can see there is no overlap in

between these sets and therefore There are no elements common to both sets and there is a disjoint.


Inside ECONOMICS

The difference between two sets: If we have two sets 𝐴 and 𝐵 and we subtract 𝐵 from 𝐴 then we have the difference between the two sets. If we define the space of these sets as 𝑆 then we can write this as:

𝐴 − 𝐵 = {𝑥 ∈ 𝑆|𝑥 ∈ 𝐴 𝑎𝑛𝑑 𝑥 ∉ 𝐵}

This operation will remove all the elements of 𝐴 that are also elements of 𝐵. Essentially the intersection between 𝐴 and 𝐵 is excluded from 𝐴.

Example 12: Difference between sets

Suppose we have the sets 𝑍: {−1,−2,−3,4} and 𝑉: {1,2,3,4,5} , in the space 𝑆 then the difference between the sets will be 𝑍 − 𝑉 = {𝑥 ∈ 𝑆|𝑥 ∈ 𝐴 𝑎𝑛𝑑 𝑥 ∉ 𝑉} .

𝑍 − 𝑉 = {𝑥 ∈ 𝑆|𝑥 ∈ 𝐴 𝑎𝑛𝑑 𝑥 ∉ 𝑉} = {−1,−2,−3}

The Symmetric Difference: The symmetric difference between two sets is the operation where the intersection of two sets is subtracted from the union of two sets. If we have two sets 𝐴 and 𝐵 then the symmetric difference is:

𝐴∆𝐵 = {𝑥 ∈ 𝑆|[𝑥 ∈ 𝐴 𝑎𝑛𝑑 𝑥 ∉ 𝐵] 𝑜𝑟 [𝑥 ∈ 𝐵 𝑎𝑛𝑑 𝑥 ∉ 𝐴]} This is the same as

𝐴∆𝐵: (𝐴 − 𝐵) ∪ (𝐵 − 𝐴) 𝑜𝑟,

𝐴∆𝐵: (𝐴 ∪ 𝐵) − (𝐵 ∩ 𝐴)

Example 13: Symmetric Difference

Find 𝑋∆𝑌, where 𝑋: {−3,−2,1,2,4} and 𝑌: {4,5}

𝑋∆𝑌: (𝑋 − 𝑌) ∪ (𝑌 − 𝑋)

𝑋∆𝑌: {−3,−2,1,2} ∪ {5} = {−3,−2,1,2,5} Example 14: Symmetric Difference

Find [𝐴∆(𝐵 ∩ 𝐶)] ∩ 𝐴 ,where 𝐴: {3,4,9,13}, 𝐵: {−1,2,4,7,12} and 𝐶: {−1,4,7,9,12,13}

𝑉

𝑍

The intersection of 𝐴 and 𝐵, which is 4, is now taken out of the set 𝑍 − 𝑉. In this diagram we have explicitly shown the space the two sets operate in.

𝑍 − 𝑉 = {−1,−2,−3}

4 -1 -2

1 2 3 5 -3

𝑆

(𝑨𝒄)𝒄 = 𝑨

𝑨 ∪ 𝑨𝒄 = 𝑬

𝑨 ∩ 𝑨𝒄 = ∅

For Compliments the following formulas hold:


Inside ECONOMICS

𝐴∆(𝐵 ∩ 𝐶) = [𝐴 − (𝐵 ∩ 𝐶)] ∪ [(𝐵 ∩ 𝐶) − 𝐴]

As (𝐵 ∩ 𝐶) = {−1,4,7,12}, therefore

[𝐴 − (𝐵 ∩ 𝐶)] ∪ [(𝐵 ∩ 𝐶) − 𝐴] = {3,9} ∪ {−1,7,12} = {−1,3,7,9,12}

Therefore [𝐴∆(𝐵 ∩ 𝐶)] ∩ 𝐴 is

[𝐴∆(𝐵 ∩ 𝐶)] ∩ 𝐴 = {−1,3,7,9,12} ∩ {3,4,9,13} = {3,9}

[𝐴∆(𝐵 ∩ 𝐶)] ∩ 𝐴 = {3,9}

Example 15: The Symmetric Difference

Suppose we have two sets 𝑋: {1,2,3,4,5,6} and 𝑌: {−1,1,2,3} then the symmetric difference between the two sets is:

𝑋∆𝑌: (𝑋 − 𝑌) ∪ (𝑌 − 𝑋) = {4,5,6} ∪ {−1} = {−1,4,5,6}

We could also use the following equation to derive the same result.

𝑋∆𝑌: (𝑋 ∪ 𝑌) − (𝑌 ∩ 𝑋) = {−1,1,2,3,4,5,6, } − {1,2,3} = {−1,4,5,6}

The Cartesian Product

The Cartesian Product is the pairs of elements resulting from the multiplication of two sets. So if we have two sets 𝑉 and 𝑋 then the Cartesian Product of these two sets is the set made up of all the ordered pairs, (𝑣, 𝑥).

𝑉 × 𝑋 = {(𝑣, 𝑥)|𝑣 ∈ 𝑉, 𝑥 ∈ 𝑋}

For sets 𝐴,𝐵, … ,𝑍 the Cartesian product is denoted formally as:

𝐴 × 𝐵 × … × 𝑍 = {(𝑎, 𝑏, … , 𝑧)|𝑎 ∈ 𝐴, 𝑏 ∈ 𝐵, … , 𝑧 ∈ 𝑍}

Example 16: The Cartesian Product

If 𝐴 = {2,4,6,8} and 𝐵: {1,3}, then the Cartesian Product is

𝐴 × 𝐵 = {(2,1), (2,3), (4,1), (4,3), (6,1), (6,3), (8,1)(8,3)} Example 17: The Cartesian Product

If 𝐴 = {𝑥, 𝑧} and 𝐵: {2, 𝑐, 𝑥, 320}, then the Cartesian Product is

𝑋 𝑌 The intersection of 𝑋 and 𝑌 is 𝑌 ∩ 𝑋= (1,2,3) The union of 𝑋 and 𝑌 is 𝑋 ∪ 𝑌 =(-1,1,2,3,4,5,6) This means that we get all the elements of both sets and then subtract the elements of the intersection to end up with {−1,4,5,6}

4 -1 1 2 3 5

6

𝑋 𝑌 The 𝑋 − 𝑌 = {4,5,6, } The 𝑌 − 𝑋 = {−1, } The union of 𝑋 − 𝑌 and 𝑌 − 𝑋 is (𝑋 − 𝑌) ∪ (𝑌 − 𝑋) = {−1,4,5,6}

4 -1 1 2 3 5

6


Inside ECONOMICS

𝐴 × 𝐵 = {(𝑥, 2), (𝑥, 𝑐), (𝑥, 𝑥), (𝑥, 320), (𝑧, 2), (𝑧, 𝑐), (𝑧, 𝑥)(𝑧, 320)} Example 18: The Cartesian Product

If 𝐴 = {3, 𝑧} and 𝐵: {2, 𝑏} and 𝐶: {1,7}, then the Cartesian Product is

𝐴 × 𝐵 × 𝐶 = {(3,2), (3, 𝑏), (𝑧, 2), (𝑧, 𝑏), (3,1), (3,7), (𝑧, 1)(𝑧, 7)(2,1)(2,7), (𝑏, 1), (𝑏, 7)}

Proof Techniques for Statements About Sets

In this text we do not explicitly deal with mathematical proofs as it is not essential in gaining intuition of basic mathematical applications. However this being said, having the ability and technique to construct proofs can be extremely important when working with pure mathematics and abstract problems solving. Subsequently as sets are a fundamental part of mathematics, and proofs of sets are relatively simplistic compared to various other mathematical proofs, it is probably a good starting point. If you wish to skip this section the next section 3.3 describes the properties of set operations.

To make proofs about sets we usually pick an arbitrary element from one set and prove that it either belongs or does not belong to the other set or sets. For instance we may want to prove that the set 𝐶 is equal to set 𝐷, that is 𝐶 = 𝐷. To prove these sets are equal we prove that 𝐶 ⊂ 𝐷 and that 𝐷 ⊂ 𝐶. To prove that 𝐶 is contained in the set 𝐷, 𝐶 ⊂ 𝐷 we simply pick a random element from 𝐶 such as 𝑐 and show that it is contained within 𝐷 that is 𝑐 ∈ 𝐷. Then to prove that 𝐷 ⊂ 𝐶 we just follow the same logic by choosing a random element of 𝐷, say 𝑑 and show that 𝑑 ∈ 𝐶. Basically there are two parts to the proof but the same technique is used. It is important to note that the element chosen should not have any special characteristics and should just be a generic element of the set.

If we want to prove that 𝐶 ⊄ 𝐷 then we have to find an element 𝑐 that does not belong 𝐷. In this case the arbitrary 𝑐 can have a special characteristic and being generic is not important.

Please note that we do not have to follow this methodology strictly. Once you become familiar with the properties of set operations, you can often use other properties to construct your argument. Also another way to construct proofs about sets is to construct a truth table, which will be covered further in this text.

Properties of Set Operations

There are certain properties which always hold for any sets 𝐴,𝐵 and 𝐷. To prove theses properties is quite simple and you can just use the proof techniques above for given sets to show the following hold.

𝑨 ∪ 𝑨 = 𝑨

𝑨 ∩ 𝑨 = 𝑨

𝑨 ∪ 𝑩 = 𝑩 ∪ 𝑨

𝑨 ∩ 𝑩 = 𝑩 ∩ 𝑨

(𝑨 ∪ 𝑩) ∪ 𝑫 = 𝑨 ∪ (𝑩 ∪𝑫)


Inside ECONOMICS

(𝑨 ∩ 𝑩) ∩ 𝑫 = (𝑨 ∩ 𝑩) ∪ (𝑨 ∩𝑫)

The last two properties are proved in the following examples.

Complements: For Compliments the following formulas hold: (𝑨𝒄)𝒄 = 𝑨


𝑨 ∩ 𝑨𝒄 = ∅

(𝑨 ∪ 𝑩)𝒄 = 𝑨 ∪ 𝑩

De Morgan's Law: A well known pair of results known as De Morgan's Law holds for every sets 𝐴 and 𝐵.

𝑨𝒄 ∪ 𝑩𝒄 = (𝑨 ∩ 𝑩)𝒄 which is also expressed as 𝑨 ∪ 𝑩 = (𝑨𝒄 ∩ 𝑩𝒄)𝒄

𝑨𝒄 ∩ 𝑩𝒄 = (𝑨 ∪ 𝑩)𝒄 which is also expressed as 𝑨 ∩ 𝑩 = (𝑨𝒄 ∪ 𝑩𝒄)𝒄

More Results,

𝑰𝒇 𝑨 ⊂ 𝑩 𝒕𝒉𝒆𝒏 𝑨− 𝑩 = ∅

𝑨 × (𝑩 ∪𝑫) = (𝑨 × 𝑩) ∪ (𝑨 × 𝑫)

(𝑨 × 𝑩) ∪ (𝑪 ×𝑫) = (𝑨 ∩ 𝑪) × (𝑩 ∩ 𝑫)

Example 19: Set Proof Prove that 𝐴 ∪ (𝐵 ∩ 𝐷) = (𝐴 ∪ 𝐵) ∩ (𝐴 ∪ 𝐷).

This is equivalent to claiming that:

𝐴 ∪ (𝐵 ∩ 𝐷) ⊂ (𝐴 ∪ 𝐵) ∩ (𝐴 ∪ 𝐷) and (𝐴 ∪ 𝐵) ∩ (𝐴 ∪ 𝐷) ⊂ 𝐴 ∪ (𝐵 ∩ 𝐷).

We need both inclusions to be true to prove the equality holds.

Part 1: Show that 𝑨𝑼(𝑩∩ 𝑫) ⊂ (𝑨 ∪ 𝑩) ∩ (𝑨 ∪ 𝑫)

Now we want to pick an arbitrary element 𝑥 in the first term 𝐴𝑈(𝐵 ∩ 𝐷), that is 𝑥 ∈ 𝐴𝑈(𝐵 ∩ 𝐷). As this is a union between 𝐴 and the intersection of 𝐵 and 𝐷, then we know that 𝑥 can belong to either 𝐴 or 𝐵 ∩ 𝐷. This means that 𝑥 ∈ 𝐴 or 𝑥 ∈ 𝐵 ∩ 𝐷, and we must examine both cases.

1. If 𝑥 ∈ 𝐴 then we know that 𝑥 must belong to the second term, because 𝑥 ∈ 𝐴 ∪ 𝐵 and 𝑥 ∈ 𝐴 ∪ 𝐷. Therefore it must be the case that 𝑥 ∈ (𝐴 ∪ 𝐵) ∩ (𝐴 ∪ 𝐷).

2. If 𝑥 ∈ 𝐵 ∩ 𝐷 then it must be the case that 𝑥 belongs to both 𝐵 and 𝐷. Therefore it must be the case that 𝑥 belongs to both 𝐴 ∪ 𝐵 and 𝐴 ∪ 𝐷, which we express as 𝑥 ∈ 𝐵 ⊂ 𝐴 ∪ 𝐵 and 𝑥 ∈ 𝐷 ⊂ 𝐴 ∪ 𝐷. This implies that 𝑥 ∈ (𝐴 ∪ 𝐵) ∩ (𝐴 ∪ 𝐷).

In both cases we always have 𝑥 ∈ (𝐴 ∪ 𝐵) ∩ (𝐴 ∪ 𝐷).


Inside ECONOMICS

Part 2: Show that (𝑨 ∪ 𝑩) ∩ (𝑨 ∪ 𝑫) ⊂ 𝑨 ∪ (𝑩 ∩ 𝑫)

Let 𝑧 be an arbitrary element and 𝑧 ∈ (𝐴 ∪ 𝐵) ∩ (𝐴 ∪ 𝐷). This means that 𝑧 must belong to both 𝐴 ∪ 𝐵 and 𝐴 ∪ 𝐷, that is 𝑧 ∈ (𝐴 ∪ 𝐵) and 𝑧 ∈ (𝐴 ∪ 𝐷). This means that 𝑧 is either an element in 𝐴 or 𝐵 and 𝐷. This means that 𝑧 ∈ 𝐴 or 𝑍 ∈ 𝐵 ∩ 𝐷. Again we have two arguments.

1. If 𝑧 ∈ 𝐴, then it must be the case that 𝑧 ∈ 𝐴 ⊂ 𝐴 ∪ (𝐵 ∩ 𝐷)

2. If 𝑧 ∈ 𝐵 ∩ 𝐷 then it must be the case that 𝑧 ∈ 𝐵 ∩ 𝐷 ⊂ 𝐴 ∪ (𝐵 ∩ 𝐷)

In both cases we have the same result 𝑧 ∈ 𝐴 ∪ (𝐵 ∩ 𝐷)

As 𝐴 ∪ (𝐵 ∩ 𝐷) ⊂ (𝐴 ∪ 𝐵) ∩ (𝐴 ∪ 𝐷) and (𝐴 ∪ 𝐵) ∩ (𝐴 ∪ 𝐷) ⊂ 𝐴 ∪ (𝐵 ∩ 𝐷), it must be the case that 𝐴 ∪ (𝐵 ∩ 𝐷) = (𝐴 ∪ 𝐵) ∩ (𝐴 ∪ 𝐷).

Proof Concluded. Example 20: Set Proofs Prove that 𝐴 ∩ (𝐵 ∪ 𝐷) = (𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐷)


𝐴 ∩ (𝐵 ∪ 𝐷) ⊂ (𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐷) and (𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐷) ⊂ 𝐴 ∩ (𝐵 ∪ 𝐷)

We need both inclusions to be true to prove the equality holds.

Part 1: Show that 𝑨 ∩ (𝑩 ∪ 𝑫) ⊂ (𝑨 ∩ 𝑩) ∪ (𝑨 ∩ 𝑫)

Let 𝑥 be a randomly chosen element in 𝐴 ∩ (𝐵 ∪ 𝐷). Therefore 𝑥 ∈ 𝐴 and 𝑥 ∈ 𝐵 ∪ 𝐷 and both sets must be examined.

1. If 𝑥 ∈ 𝐴 then it must be the case that 𝑥 ∈ 𝐴 ∩ 𝐵 and 𝑥 ∈ 𝐴 ∩ 𝐷. If 𝑥 belongs to both 𝐴 ∩ 𝐵 and 𝐴 ∩ 𝐷, then 𝑥 must belong to their union. This means that 𝑥 ∈ 𝐴 ⊂ (𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐷).

2. If 𝑥 ∈ 𝐵 ∪ 𝐷 then 𝑥 can belong to either 𝐵 and or 𝐷. If 𝑥 ∈ 𝐵 then 𝑥 will belong to the set (𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐷), 𝑥 ∈ 𝐵 ⊂ (𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐷). However if 𝑥 does not belong to 𝐵, that is 𝑥 ∉ 𝐵 then the element 𝑥 must belong to the set 𝐷 as 𝑥 ∈ 𝐵 ∪ 𝐷. If 𝑥 belongs to the set 𝐷 it must be the case that 𝑥 ∈ 𝐷 ⊂ (𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐷). Therefore it must be that 𝑥 ∈ 𝐵 ∪ 𝐷 ⊂(𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐷).

Therefore 𝐴 ∩ (𝐵 ∪ 𝐷) ⊂ (𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐷)

Part 2: Show that (𝑨 ∩ 𝑩) ∪ (𝑨 ∩ 𝑫) ⊂ 𝑨 ∩ (𝑩 ∪ 𝑫)

Let 𝑞 be an arbitrarily chosen element from (𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐷), therefore 𝑞 ∈ (𝐴 ∩ 𝐵) ∪ (𝐴 ∩𝐷). This means that 𝑞 belongs to (𝐴 ∩ 𝐵) and (𝐴 ∩ 𝐷). Both these sets need to be examined.

1. If 𝑞 ∈ (𝐴 ∩ 𝐵) then 𝑞 must belong to both 𝐴 and 𝐵. This means that 𝑞 ∈ 𝐴 and 𝑞 ∈ 𝐵, and it must be the case that 𝑞 ∈ 𝐴 ⊂ 𝐴 ∩ (𝐵 ∪ 𝐷) and 𝑞 ∈ 𝐵 ⊂ 𝐴 ∩ (𝐵 ∪ 𝐷).


Inside ECONOMICS

2. If 𝑞 ∈ (𝐴 ∩ 𝐷) then 𝑞 must belong to both 𝐴 and 𝐷. Therefore 𝑞 ∈ 𝐴 ⊂ 𝐴 ∩ (𝐵 ∪ 𝐷) and 𝑞 ∈ 𝐷 ⊂ 𝐴 ∩ (𝐵 ∪ 𝐷). Hence (𝐴 ∩ 𝐷) ⊂ 𝐴 ∩ (𝐵 ∪ 𝐷).

Therefore (𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐷) ⊂ 𝐴 ∩ (𝐵 ∪ 𝐷)

As 𝐴 ∩ (𝐵 ∪ 𝐷) ⊂ (𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐷) and (𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐷) ⊂ 𝐴 ∩ (𝐵 ∪ 𝐷) it must be the case that 𝐴 ∩ (𝐵 ∪ 𝐷) = (𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐷).

Proof Concluded. Example 21: Proof of Sets Prove that (𝑋 × 𝑌) ∩ (𝑊 × 𝑍) = (𝑋 ∩𝑊) × (𝑌 ∩ 𝑍)


(𝑋 × 𝑌) ∩ (𝑊 × 𝑍) ⊂ (𝑋 ∩𝑊) × (𝑌 ∩ 𝑍) and (𝑋 ∩𝑊) × (𝑌 × 𝑍) ⊂ (𝑋 × 𝑌) ∩ (𝑊 ∩ 𝑍)

Again we need both inclusions to be true to prove the equality holds.

Part 1: Show that (𝑿 × 𝒀) ∩ (𝑾 × 𝒁) ⊂ (𝑿 ∩𝑾) × (𝒀 ∩ 𝒁)

Let 𝑎 ∈ (𝑋 × 𝑌) ∩ (𝑊 × 𝑍) be arbitrary. Suppose that 𝑎 = (𝑏, 𝑐) and therefore 𝑏 ∈ 𝑋 and 𝑏 ∈ 𝑊 which means that 𝑏 ∈ 𝑋 ∩𝑊. In addition 𝑐 ∈ 𝑌 and 𝑐 ∈ 𝑍 and therefore 𝑐 ∈ 𝑌 ∩𝑊. Therefore because a randomly chose element 𝑎 = (𝑏, 𝑐) is contained in both (𝑋 ∩𝑊) and (𝑌 ∩ 𝑍) it must be the case that 𝑎 = (𝑏, 𝑐) ∈ (𝑋 ∩𝑊) × (𝑌 ∩ 𝑍). This means we have shown that a randomly chosen element 𝑎 from the first term (𝑋 × 𝑌) ∩ (𝑊 × 𝑍) is an element of the second term (𝑋 ∩𝑊) × (𝑌 ∩ 𝑍) . Now we must be able to show the reverse.

Part 2: Show that (𝑿 ∩𝑾) × (𝒀 ∩ 𝒁) ⊂ (𝑿 × 𝒀) ∩ (𝑾 × 𝒁)

Suppose we pick a random element in 𝑑 contained in (𝑋 ∩𝑊) × (𝑌 ∩ 𝑍). This is 𝑑 ∈ (𝑋 ∩𝑊) × (𝑌 × 𝑍) is arbitrary. Also suppose that 𝑑 = (𝑒,𝑓) ∈ (𝑋 ∩𝑊) × (𝑌 ∩ 𝑍) where 𝑒 ∈ (𝑋 ∩𝑊) and 𝑦 ∈ (𝑌 ∩ 𝑍). This implies that 𝑒 ∈ 𝑋, 𝑒 ∈ 𝑊, 𝑓 ∈ 𝑌, and 𝑓 ∈ 𝑍. Thus 𝑑 = (𝑒,𝑓) ∈ (𝑋 × 𝑌) and 𝑑 = (𝑒,𝑓) ∈ (𝑊 × 𝑍). Therefore 𝑑 ∈ (𝑋 × 𝑌) ∩ (𝑊 × 𝑍). This shows that a randomly chosen element 𝑑 belonging to the first term (𝑋 ∩𝑊) × (𝑌 ∩ 𝑍) is also an element contained in the second term (𝑋 × 𝑌) ∩ (𝑊 × 𝑍).

As (𝑋 × 𝑌) ∩ (𝑊 × 𝑍) ⊂ (𝑋 ∩𝑊) × (𝑌 ∩ 𝑍) and (𝑋 ∩𝑊) × (𝑌 ∩ 𝑍) ⊂ (𝑋 × 𝑌) ∩ (𝑊 × 𝑍), have been proved we have also proved that (𝑋 × 𝑌) ∩ (𝑊 × 𝑍) = (𝑋 ∩𝑊) × (𝑌 ∩ 𝑍).

Proof Concluded.

De Morgan's Law: A well known pair of results known as De Morgan's Law holds for every sets 𝐴 and 𝐵.



Firstly we display these equalities in the following diagrams and the proceed with a formal proof.


Inside ECONOMICS

Example 22: De Morgan's Laws

Prove the following equalities hold. 𝑨𝒄 ∪ 𝑩𝒄 = (𝑨 ∩ 𝑩)𝒄

𝑨𝒄 ∩ 𝑩𝒄 = (𝑨 ∪ 𝑩)𝒄

First we will prove 𝐴𝑐 ∪ 𝐵𝑐 = (𝐴 ∩ 𝐵)𝑐 by showing 𝐴𝑐 ∪ 𝐵𝑐 ⊂ (𝐴 ∩ 𝐵)𝑐 and (𝐴 ∩ 𝐵)𝑐 ⊂ 𝐴𝑐 ∪𝐵𝑐 1. Prove 𝑨𝒄 ∪ 𝑩𝒄 = (𝑨 ∩ 𝑩)𝒄 Part 1: Show that 𝑨𝒄 ∪ 𝑩𝒄 ⊂ (𝑨 ∩ 𝑩)𝒄

(𝐴 ∪ 𝐵)𝑐is represented by the total grey area surround the union of the sets. This is equal to 𝐴𝑐 ∩ 𝐵𝑐 as we can see by putting the above two diagrams together then counting or observing the grey area. So this picture is the amalgamation of the two above diagrams for 𝐴𝑐 and 𝐵𝑐.

𝑨𝒄 ∩ 𝑩𝒄 = (𝑨 ∪ 𝑩)𝒄 𝑺

𝑩 𝑨

𝑨𝒄 ∩ 𝑩𝒄 = (𝑨 ∪ 𝑩)𝒄

If we have the intersection 𝐴 ∩ 𝐵 then the compliment of this set will be the space surrounding the intersection. This is represented by the grey area labelled (𝐴 ∩ 𝐵)𝑐, which surrounds 𝐴 ∩ 𝐵.

(𝑨 ∩ 𝑩)𝒄 = 𝑨𝒄 ∪ 𝑩𝒄 𝑺 𝑨

(𝑨 ∩ 𝑩)𝒄 = 𝑨𝒄 ∪ 𝑩𝒄

The compliment 𝐵𝑐 is represented by the total grey area surround the set 𝐵.

The compliment is actually the total space subtract the space of set 𝐵. This means that 𝐵𝑐 = 𝑆 − 𝐵.

𝑩𝒄

𝑺 𝑩

𝑩𝒄

The compliment 𝐴𝑐 is represented by the total grey area surround the set 𝐴. The compliment is actually the total space subtract the space of set 𝐴. This means that 𝐴𝑐 = 𝑆 − 𝐴.

𝑨𝒄

𝑺

𝑨

𝑨𝒄

𝑨 𝑩

The union of 𝐴 and 𝐵 includes all the elements and represented in the diagram as the space of 𝐴, 𝐴 ∩ 𝐵 and 𝐵.

The intersection of the two sets is simply 𝐴 ∩ 𝐵.

𝑆 represent the space in which these sets occur.

𝑨 ∩ 𝑩

𝑺


Inside ECONOMICS

Let 𝑥 be a randomly chosen point in 𝐴𝑐 ∪ 𝐵𝑐. This means that 𝑥 can be in either 𝐴𝑐 or 𝐵𝑐, that is 𝑥 ∈ 𝐴𝑐 or 𝑥 ∈ 𝐵𝑐. If 𝑥 ∈ 𝐴𝑐 then it must be the case that 𝑥 ∉ 𝐴 and because 𝐴 is a subset of 𝐴 ∩ 𝐵 it must be that 𝑥 ∉ 𝐴 ∩ 𝐵. (If 𝑥 is not an element in 𝐴 then it can never be an element in the intersection of 𝐴 and any set). Subsequently 𝑥 ∈ (𝐴 ∩ 𝐵)𝑐. Also if 𝑥 ∈ 𝐵𝑐 then it must be the case that 𝑥 ∉ 𝐵 and because 𝐵 is a subset of 𝐴 ∩ 𝐵 it must be that 𝑥 ∉ 𝐴 ∩ 𝐵. Hence 𝑥 ∈ (𝐴 ∩ 𝐵)𝑐 .So either way we always have 𝑥 ∈ (𝐴 ∩ 𝐵)𝑐. This means that 𝐴𝑐 ∪ 𝐵𝑐 ⊂ (𝐴 ∩ 𝐵)𝑐 holds.

Part 2: Show that (𝑨 ∩ 𝑩)𝒄 ⊂ 𝑨𝒄 ∪ 𝑩𝒄

Let 𝑥 be an arbitrary point of (𝐴 ∩ 𝐵)𝑐. Then 𝑥 ∉ 𝐴 ∩ 𝐵 which means that 𝑥 is not an element in either set, 𝑥 ∉ 𝐴 and 𝑥 ∉ 𝐵. Therefore it must be the case that 𝑥 ∈ 𝐴𝑐 or 𝑥 ∈ 𝐵𝑐. Hence 𝑥 ∈ 𝐴𝑐 ∪ 𝐵𝑐. This means that (𝐴 ∩ 𝐵)𝑐 ⊂ 𝐴𝑐 ∪ 𝐵𝑐 holds.

Therefore 𝐴𝑐 ∪ 𝐵𝑐 = (𝐴 ∩ 𝐵)𝑐 holds. Proof Concluded. 2. Prove 𝑨𝒄 ∩ 𝑩𝒄 = (𝑨 ∪ 𝑩)𝒄

For the second equality 𝐴𝑐 ∩ 𝐵𝑐 = (𝐴 ∪ 𝐵)𝑐, we need to show that 𝐴𝑐 ∩ 𝐵𝑐 ⊂ (𝐴 ∪ 𝐵)𝑐 and that , (𝐴 ∪ 𝐵)𝑐 ⊂ 𝐴𝑐 ∩ 𝐵𝑐. We can use the same method above to prove this. However we can instead use the first equality 𝐴𝑐 ∪ 𝐵𝑐 = (𝐴 ∩ 𝐵)𝑐 (as we have proved this holds) and the properties of the set operations, in particular (𝐴𝑐)𝑐 = 𝐴.

Let 𝐴 = 𝑊𝑐 and let 𝐵 = 𝑍𝑐. We can then by applying the property (𝐴𝑐)𝑐 = 𝐴 to 𝐴 = 𝑊𝑐 and 𝐵 = 𝑍𝑐. This is 𝑊 = 𝐴𝑐 and 𝑍 = 𝐵𝑐. Now we just substitute our values into the first equation which we know is true because we have just proved it above.

Using 𝐴𝑐 ∪ 𝐵𝑐 = (𝐴 ∩ 𝐵)𝑐 and the fact that (𝐴𝑐)𝑐 = 𝐴

𝑊 ∪ 𝑍 = 𝐴𝑐 ∪ 𝐵𝑐

(𝐴 ∩ 𝐵)𝑐 = (𝑊𝑐 ∩ 𝑍𝑐)𝑐

Therefore,

𝑊 ∪ 𝑍 = (𝑊𝑐 ∩ 𝑍𝑐)𝑐

𝐴𝑐 ∪ 𝐵𝑐 = (𝐴 ∩ 𝐵)𝑐

We know that these two equations are equal. To get them into the correct form we just take the compliment. Just for clarification (𝐴𝑐)𝑐 = 𝐴 but ((𝐴𝑐)𝑐)𝑐 = 𝐴𝑐, so when we take the compliment of (𝑊𝑐 ∩ 𝑍𝑐)𝑐 we are actually doing ((𝑊𝑐 ∩ 𝑍𝑐)𝑐)𝑐 = 𝑊𝑐 ∩ 𝑍𝑐.

(𝑊 ∪ 𝑍)𝑐 = 𝑊𝑐 ∩ 𝑍𝑐

This is the equality we want to prove with a distinct notation for the sets and is equivalent to:

(𝐴 ∪ 𝐵)𝑐 = 𝐴𝑐 ∩ 𝐵𝑐


Inside ECONOMICS

Therefore we have shown that,

𝐴𝑐 ∩ 𝐵𝑐 = 𝑊𝑐 ∩ 𝑍𝑐 = (𝑊 ∪𝑍)𝑐 = (𝐴 ∪ 𝐵)𝑐

Although it is unnecessary, If we reverse the equality we have exactly what we wanted to prove 𝐴𝑐 ∩ 𝐵𝑐 = (𝐴 ∪ 𝐵)𝑐.

Proof Concluded.

Please note that you can also express De Morgan's Laws as:

𝐴 ∪ 𝐵 = (𝐴𝑐 ∩ 𝐵𝑐)𝑐

𝐴 ∩ 𝐵 = (𝐴𝑐 ∪ 𝐵𝑐)𝑐

Example 23: more proofs of set operations

Prove that for any pair of sets 𝑸 and 𝑹,

(𝑸𝒄) × (𝑹𝒄) ⊂ (𝑸 × 𝑹)𝒄

Here we are just trying to see whether an element of (𝑄𝑐) × (𝑅𝑐) is contained within the set (𝑄 × 𝑅)𝑐

Firstly because it is a multiplication proof we need to use an arbitrary element 𝑥 = (𝑞. 𝑟). This arbitrary element is obviously the product of 𝑞 from 𝑄 and 𝑟 from 𝑅. Therefore if 𝑥 ∈ 𝑄𝑐 × 𝑅𝑐, then 𝑞 ∉ 𝑄 and 𝑟 ∉ 𝑅. This means that 𝑥 ∉ 𝑄 × 𝑅 and therefore it must be the case that 𝑥 ∈ (𝑄 × 𝑅)𝑐.

Therefore (𝑄𝑐) × (𝑅𝑐) ⊂ (𝑄 × 𝑅)𝑐.

Proof Concluded. Example 24: Proofs of Set Operations

Prove whether the following is true or false.

For any two sets 𝑪 and 𝑫

𝑪∆𝑫 = (𝑪 ∪ 𝑫𝒄)∆(𝑫∪ 𝑪𝒄)

First, write out the equation using the difference, 𝐴∆𝐵: (𝐴 − 𝐵) ∪ (𝐵 − 𝐴)

𝐶∆𝐷 = (𝐶 ∪ 𝐷𝑐)∆(𝐷 ∪ 𝐶𝑐) = [(𝐶 ∪ 𝐷𝑐) ∪ (𝐷 ∪ 𝐶𝑐)] − [(𝐶 ∪ 𝐷𝑐) ∩ (𝐷 ∪ 𝐶𝑐)]

The first term is the union of the two unions, and contains all points in the universe. This means that if we subtract something from this, the remaining set is the complement of the second term.(If X is a set and S is the universe of points then, S − X = Xc) In this case all that remains is the complement of the second term.

𝐶∆𝐷 = (𝐶 ∪ 𝐷𝑐)∆(𝐷 ∪ 𝐶𝑐) = [(𝐶 ∪ 𝐷𝑐) ∩ (𝐷 ∪ 𝐶𝑐)]𝑐

This term is equivalent to (𝐶 ∪ 𝐷𝑐)𝑐 ∪ (𝐷 ∪ 𝐶𝑐)𝑐


Inside ECONOMICS

(Hint: whenever you have term that is a complement of a complement then you should always think back to De Morgan's Laws and see if you can use the laws in your proof)

𝑨𝒄 ∪ 𝑩𝒄 = (𝑨 ∩ 𝑩)𝒄

𝑨𝒄 ∩ 𝑩𝒄 = (𝑨 ∪ 𝑩)𝒄

[(C ∪ Dc) ∩ (D ∪ Cc)]c = (C ∪ Dc)c ∪ (D ∪ Cc)c = �CC ∩ D� ∪ (Dc ∩ C)

However we know that Cc ∩ D = D − C and that Dc ∩ C = C − D.

This means that �CC ∩ D� ∪ (Dc ∩ C) = (D − C) ∪ (C − D) = C∆D

Therefore we have shown that,

C∆D = (C ∪ Dc)∆(D ∪ Cc) = [(C ∪ Dc) ∩ (D ∪ Cc)]c = (C ∪ Dc)c ∪ (D ∪ Cc)c =

= �CC ∩ D� ∪ (Dc ∩ C) = (D − C) ∪ (C − D) = C∆D

𝐂∆𝐃 = (𝐂 ∪ 𝐃𝐜)∆(𝐃 ∪ 𝐂𝐜) is true.

Proof Concluded.

Logic and Truth Tables for Sets

Truth tables can be applied to set operations. To use truth tables we simply put a number 1 into the box if it is true and zero if it is false. This is a simple way to sort out confusing set information in an ordered way.

If 𝐴 then 𝐵 is denoted 𝐴 ⟹ 𝐵. The logic equivalence of " 𝐴 ⟹ 𝐵" is "˧𝐵 ⟹ ˧𝐴" , which means if not 𝐵 then not 𝐴. Also note that there is logical equivalence between the statements "˧(𝐴 𝑜𝑟 𝐵)" and "(˧𝐴) and (˧𝐵)". In truth tables we use the rules of logic and subsequently the order or statements is significant. Consider the following truth tables

𝐴 𝐵 𝐴 and 𝐵 𝐴 or 𝐵 ˧(𝐴 𝑜𝑟 𝐵) 𝐴 ⟹ 𝐵 0 0 0 0 1 1 0 1 0 1 1 1 1 0 0 1 0 0 1 1 1 1 0 1

𝐴 𝐵 ˧𝐴 ˧𝐵 (˧𝐴) and(˧𝐵) "˧𝐵 ⟹ ˧𝐴 0 0 1 1 1 1 0 1 1 0 0 1 1 0 0 1 0 0 1 1 0 0 0 1


Inside ECONOMICS

Example 29: Truth Tables and Sets

We can write this as, if 𝐴 then 𝐵, 𝐴 ⟹ 𝐵, to prove that 𝐴 is a proper subset of 𝐵, we can use truth tables.

Therefore if 𝑥 ∈ 𝐴 and 𝑥 ∉ 𝐵 then 𝐴 is not a subset of 𝐵. However an element can belong to 𝐵 and not be contained within 𝐴, which is true because 𝐴 is a proper subset of 𝐵. Therefore to provide evidence that proves that 𝐴 is not a subset of 𝐵 all we have to do is pick an arbitrary element and show that the third row is true. From the diagram we can see that 𝐴 ⟹ 𝐵 and therefore 𝐴 ⊊ 𝐵.

We can apply these rules to sets using more general notation.

The first row says that if the element is not in 𝐴 then it is in 𝐴𝑐 . If the element is not in 𝐵 then the element is in the compliment 𝐵𝑐. By using the truth table we can see where an element belongs based on the information. For instance in the last row the element is in both 𝐴 and 𝐵 and therefore must be in the intersection of these two sets.


Show that 𝐴 ∩ 𝐵 = (𝐴𝑐 ∪ 𝐵𝑐)𝑐

𝐴 is a subset of 𝐵 if and only if every element of 𝐵 is an element of 𝐴, 𝐴 ⊊ 𝐵

𝐵

𝐴

𝐴 is a subset of 𝐵 because if we pick an arbitrary element of 𝐴 say 𝑥1 then this element is also in the set 𝐵 𝑥

𝐴𝑐 ∪ 𝐵𝑐 (𝐴𝑐 ∪ 𝐵𝑐)𝑐 𝐴 ∩ 𝐵 1 0 0 1 0 0 0 0 0 0 1 1

𝐴 𝐵 𝐴𝑐 𝐵𝑐 0 0 1 1 0 1 1 0 1 0 0 1

𝐴 𝐵 ˧𝐴 ˧𝐵 𝐴 ⟹ 𝐵 "˧𝐵 ⟹ ˧𝐴 0 0 1 1 1 1 0 1 1 0 1 1 1 0 0 1 0 0 1 1 0 0 1 1


Inside ECONOMICS

From the truth table we have constructed we can see that if an element 𝑥 belongs to (𝐴𝑐 ∪ 𝐵𝑐)𝑐, that is 𝑥 ∈ (𝐴𝑐 ∪ 𝐵𝑐)𝑐 then it must be the case that 𝑥 ∈ 𝐴 ∩ 𝐵.


Show that (𝐴 ∩ 𝐵) ∩ 𝐷 = (𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐷)

The truth table above contains all the possible scenarios and from this we can see that if 𝑥 ∈ (𝐴 ∩ 𝐵) ∩ 𝐷 then 𝑥 ∈ (𝐴 ∩ 𝐵) and 𝑥 ∈ (𝐴 ∩ 𝐷) which means that 𝑥 ∈ (𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐷). Therefore (𝐴 ∩ 𝐵) ∩ 𝐷 = (𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐷).

Power Sets

The power set is denoted 2𝐴 and is the set of all subsets of the set 𝐴. A power set is a set where each element of the set is also a set. If there are 2 elements in contained in the set then the power set will contain 4 elements where each of these elements is itself a set. This means that if a set has exactly 𝑘 elements then it's power set will have 2𝑘 elements. This concept is best explained through demonstration.

If set 𝐴 = {1,2,3} then the power set is 2𝐴 = {∅, {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3}, } .We can also write this power set as 2𝐴 = {∅, {1}, {2}, {3}, {1,2}, {1,3}, {2,3},𝐴}.

As the power set is a set, we can compute the power set of a power set. The power set of a power set is denoted as 22𝐴. For set 𝐴 above the power set of the power set would have 28 elements, which are sets.

If 𝑍 = {𝑎, 𝑏} then, the power set is

2𝑍 = �∅, {𝑎}, {𝑏}, {𝑎, 𝑏}�

and the power set of the power set is

22𝑍 = �2𝑍,∅, {∅}, �{𝑎}�, �{𝑏}�,𝑍, �∅, {𝑎}�, �∅, {𝑏}�, {∅,𝑍}, �{𝑎}, {𝑏}�, �{𝑎},𝐴�,

�{𝑏},𝐴�, �∅, {𝑎}, {𝑏}�, {∅, {𝑎},𝐴}, {∅, {𝑏},𝐴}, �{𝑎}, {𝑏},𝐴��

The attributes and mechanics of power sets are further demonstrated in the following examples.

Example 25: Power Sets

𝐴 𝐵 𝐷 (𝐴 ∩ 𝐵) ∩ 𝐷 (𝐴 ∩ 𝐵) (𝐴 ∩ 𝐷) 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 1 1 0 0 1 0 1 0 1 0 0 1 1 1 1 1 1 1


Inside ECONOMICS

Suppose we have two sets 𝐴 = {𝑠, 𝑡, 𝑥} and 𝐵 = {𝑥,𝑦}.

Calculate 2𝐴 − 2𝐵 and 2𝐴−𝐵

First lets define the 2𝐴 and 2𝐵.

2𝐴 = {∅, {𝑠}, {𝑡}, {𝑥}, {𝑠, 𝑡}, {𝑠, 𝑥}, {𝑡, 𝑥},𝐴}

2𝐵 = {∅, {𝑥}, {𝑦},𝐵}

Therefore,

2𝐴 − 2𝐵 = {∅, {𝑠}, {𝑡}, {𝑠, 𝑡}, {𝑠, 𝑥}, {𝑡, 𝑥},𝐴}

2𝐴−𝐵 = �∅, {𝑠}, {𝑡}, {𝑠, 𝑡}�


For the sets 𝑋 = {1,5} and 𝑌 = {1,3}, then calculate 2𝑋 ∩ 2𝑌

2𝑌 = �∅, {1}, {3}, {1,3}�

Therefore

2𝑋 ∩ 2𝑌 = �∅, {1}�


For the sets 𝑋 = {𝑎, 𝑏} and 𝑌 = {𝑎, 𝑏, 𝑐}, then calculate 2𝑋 ∩ 2𝑌

2𝑥 = {∅, {𝑎}, {𝑏},𝑋}

2𝑌 = {∅, {𝑎}, {𝑏}, {𝑐}, {𝑎, 𝑏}, {𝑎, 𝑐}, {𝑏, 𝑐},𝑌}

Therefore

2𝑋 ∩ 2𝑌 = �∅, {𝑎}, {𝑏}, {𝑎, 𝑏}�


Prove that for any sets 𝐴 and 𝐵:

2𝐴 ∩ 2𝐵 = 2𝐴∩𝐵

First we prove that 2𝐴 ∩ 2𝐵 ⊂ 2𝐴∩𝐵. Let 𝑥 be arbitrary point contained in 2𝐴 ∩ 2𝐵, that is 𝑥 ∈ 2𝐴 ∩ 2𝐵. This means that 𝑥 ∈ 2𝐴 and 𝑥 ∈ 2𝐵. As 𝐴 ⊂ 2𝐴 and 𝐵 ⊂ 2𝐵 this implies that 𝑥 ∈ 𝐴 and that 𝑥 ∈ 𝐵. Therefore it must be the case that 𝑥 ∈ 𝐴 ∩ 𝐵. This implies that 𝑥 is an element in the power set of the intersection 2𝐴∩𝐵 because 𝐴 ∩ 𝐵 ⊂ 2𝐴∩𝐵. This proves that 2𝐴 ∩ 2𝐵 ⊂ 2𝐴∩𝐵.

Now we prove that 2𝐴∩𝐵 ⊂ 2𝐴 ∩ 2𝐵. Suppose that 𝑥 ∈ 2𝐴∩𝐵, where 𝑥 is arbitrary. The power set 2𝐴∩𝐵 contains only elements in the intersection of 𝐴 and 𝐵 and therefore 𝐴 ∩ 𝐵 ⊂ 2𝐴∩𝐵


Inside ECONOMICS

and 𝑥 ∈ 𝐴 ∩ 𝐵. As 𝑥 ∈ 𝐴 ∩ 𝐵 then 𝑥 ∈ 𝐴 and 𝑥 ∈ 𝐵, and 𝐴 ⊂ 2𝐴 and 𝐵 ⊂ 2𝐵 it must be the case that 𝑥 ∈ 2𝐴 ∩ 2𝐵. This proves that 2𝐴∩𝐵 ⊂ 2𝐴 ∩ 2𝐵.

As we have proved 2𝐴 ∩ 2𝐵 ⊂ 2𝐴∩𝐵 and 2𝐴∩𝐵 ⊂ 2𝐴 ∩ 2𝐵 it must be the case that the equation holds, that is 2𝐴 ∩ 2𝐵 = 2𝐴∩𝐵.

𝑨 ∪ 𝑨 = 𝑨

𝑨 ∩ 𝑨 = 𝑨

𝑨 ∪ 𝑩 = 𝑩 ∪ 𝑨

𝑨 ∩ 𝑩 = 𝑩 ∩ 𝑨

(𝑨 ∪ 𝑩) ∪ 𝑫 = 𝑨 ∪ (𝑩 ∪ 𝑫)

(𝑨 ∩ 𝑩) ∩ 𝑫 = (𝑨 ∩ 𝑩) ∪ (𝑨 ∩ 𝑫)

(𝑨𝒄)𝒄 = 𝑨


𝑨 ∩ 𝑨𝒄 = ∅

(𝑨 ∪ 𝑩)𝒄 = 𝑨 ∪ 𝑩

𝑰𝒇 𝑨 ⊂ 𝑩 𝒕𝒉𝒆𝒏 𝑨 − 𝑩 = ∅

𝑨 × (𝑩 ∪ 𝑫) = (𝑨 × 𝑩) ∪ (𝑨 × 𝑫)

(𝑨 × 𝑩) ∩ (𝑪 × 𝑫) = (𝑨 ∩ 𝑪) × (𝑩 ∩ 𝑫)

Summary of the Properties of Set Operations Basic Properties

Complements

De Morgan's Law



Other Properties


Inside ECONOMICS

Mathematics For Economists

Partitions of a Set

Introduction to Partitions

A finite partition of set 𝑅 is a finite collection of all the non empty subsets contained within 𝑅, based on a given division of set. In other words the original set is divided or separated into a collection of non empty subsets. The concept of partitions is widely used in game theory and information based applications.

Consider a non empty set 𝑅. A finite partition of 𝑅 is a finite collection of non empty subsets 𝐵1,𝐵2, … ,𝐵𝑘 ⊂ 𝑅 for some 𝑘 ∈ {1,2,3, … } such that 𝐵𝑖 ∩ 𝐵𝑗 = ∅ 𝑖𝑓 𝑖 ≠ 𝑗 and,

�𝐵𝑖 = 𝑅𝑘

𝑖=1

We can also say that the set 𝐵 = {𝐵1,𝐵2, … ,𝐵𝑘} is a partition of 𝑅.

Example 29: Economic Application of Partitions

It is common for economists to forecast the state of the future economy or model uncertainty. Suppose that a set includes four different states of the world where each state is a decomposition of economic variables. This means that every state of the world consists of variables such as inflation, interest rates and gross domestic product. Also let 𝑗 be an agent and 𝜑𝑗 is the agents knowledge partition. The knowledge partition is the agents decomposition of the space in several subsets. This means that it is the agents choice of the variables inflation, interest rates and gross domestic product. Although there are several or many possible states of the economy, only one is actually going to happen. We define this state as the true state 𝑤∗.

The agents knowledge partition is:

𝜑𝑗 = �𝜑1𝑗,𝜑2

𝑗, … ,𝜑𝑘𝑗� , where 𝜑𝑖

𝑗 ∩ 𝜑𝑗𝑗 = ∅ 𝑖𝑓 𝑖 ≠ 𝑗 and

⋃ 𝜑𝑖𝑗 = Ω𝑘

𝑖=1 where Ω represents the total state space

If the agents forecasts that the future state is in the partition 𝜑𝑗 = �𝜑1𝑗,𝜑2

𝑗 ,𝜑3𝑗,𝜑4

𝑗� where each partition 𝜑𝑛

𝑗 is a set of possible future states 𝑤. This means that all the possible future states are contained in the space 𝛀 which is divided into these four partitions. Therefore the true state 𝑤∗ must belong to one of these four sets.


Inside ECONOMICS

Eventually agent 𝑗 learns that the true state 𝑤∗ is in 𝜑1𝑗, that is 𝑤∗ ∈ 𝜑1

𝑗

Refinements

Consider the two partitions 𝐵 = {𝐵1,𝐵2, … ,𝐵𝑘} and 𝑉 = {𝑉1,𝑉2, … ,𝑉𝑘} of a non empty set 𝑆. The partition 𝐵 refines 𝐶 if and only if that for every element of 𝐶𝑖 ∈ 𝐶 there are elements 𝐵𝑗1, … ,𝐵𝑗𝑠 ∈ 𝐵 such that:

�𝐵𝑗𝑟 = 𝐶𝑖

𝑘

𝑟=1

This means that 𝐵 is a refinement of 𝐶 if each element 𝐶𝑖 of partition 𝐶 can be written as the union of some elements 𝐵𝑗𝑟 of the partition 𝐵.

Any partition of 𝐵 of 𝑆 refines itself.

Example 30: Refinements

Let 𝐴 = {2,4,6,8}. Suppose we have the following partitions of the set 𝐴: 𝑅 = �{2,4}, {6,8}� and 𝑊 = �{2,6}, {4,8}�. List all the refinement of 𝑅 and all the refinements of 𝑊.

Refinements of 𝑅:

𝑅1 = 𝑅 = �{2,4}, {6,8}�

𝑅2 = �{2}, {4}, {6,8}�

𝑅3 = �{2,4}, {6}, {8}�

𝑅4 = �{2}, {4}, {6}, {8}�

Refinements of 𝑊:

𝑊1 = 𝑊 = �{2,6}, {4,8}�

𝑊2 = �{2}, {6}, {4,8}�

𝑊3 = �{2,6}, {4}, {8}�

𝑊4 = �{2}, {4}, {6}, {8}�

𝜑1𝑗

𝛀 Each area is a partition 𝜑𝑗 for agent 𝑗 and 𝑤 represents the possible states of the world within that partition. The 𝑤∗ is the true state of the world. In this example 𝑤∗ ∈ 𝜑1𝑖 .

𝑤∗

𝜑3𝑗

w w

w

w

w

w w

𝜑4𝑗

𝜑2𝑗

𝛀= {𝑤1,𝑤2, …𝑤𝑧} and 𝜑𝑗 = �𝜑1𝑗 ,𝜑2

𝑗 ,𝜑3𝑗 ,𝜑4

𝑗�


Inside ECONOMICS

Common Refinement: For a non empty set 𝑆 the Partition 𝐵 is a common refinement of partitions 𝐶 and 𝐷, if 𝐵 refines both partitions 𝐶 and 𝐷. (in Example 30 above the only common refinement is 𝑅4 = 𝑊4)

Proper Refinement: For a non empty set 𝑆 the partition 𝐵 is a proper refinement of partitions 𝐶 if partition 𝐵 refines partition 𝐶 and they are different partitions.

Coarsening: If partition 𝐵 refines partition 𝐶, then we describe the partition 𝐶 as a coarsening of partition 𝐵.


Consider the following partitions, 𝑋 = �{1}, {2}, {3,4}�, 𝑌 = �{1,2,3}{4}�, 𝑍 = �{1,23}, {4}� and 𝑆 = {1,2,3,4,5}.

The partition 𝑋 is a proper refinement of the partitions 𝑌 and 𝑍. This means that 𝑌 and 𝑍is are proper coarsenings of 𝑋. The partitions 𝑌 and 𝑍 are equal and this means that partition 𝑌 is a refinement of partition 𝑍 and vice versa. However these refinements are not proper refinements. Likewise partition 𝑍 is a coarsening of partition 𝑌 and vice versa, but they are mot proper coarsenings. The partition 𝑋 is a common refinement of both 𝑍 and 𝑌. Partition 𝑌 is a common coarsening of 𝑋 and 𝑍. Also partition 𝑍 is common coarsening of 𝑋 and 𝑌. For simplification refer to the table below.


Suppose a set 𝑆 has exactly 100 elements. Consider a partition of this set, 𝑄 which contains exactly 99 elements. That is 𝑄 = {𝑄1,𝑄2, …𝑄99}, where all 𝑄𝑖 ≠ ∅, 𝑆 = 𝑄1 ∪ 𝑄2 ∪ …∪ 𝑄99 and 𝑄𝑖 ∪ 𝑄𝑗 = ∅, if 𝑖 ≠ 𝑗. How many distinct refinements does the partition 𝑄 have?

As the partition 𝑄 has exactly 99 elements and there is 100 elements in the set 𝑆. It must be the case that exactly one 𝑄𝑖 contains two elements. All the other 𝑄𝑗 are sets having just one element. By using generic numbers we can show this.

If 𝑆 = {1,2, … ,100} then let:

𝑄 = {𝑄1,𝑄2, …𝑄99} = �{1}, {2}, … {99,100}�

Now find the refinements of this partition. Please note that the indexes 1 and 2 below are just to label the refinement number and do not correspond to the sets in the partition 𝑄 above.

𝑄1 = 𝑄 = �{1}, {2}, … {99,100}�

Refinement of Coarsening ofCommon

Refinements ofProper

Refinements ofCommon

Coarsenings ofProper

Coarsenings ofX Z,Y Y,ZY Z Z X X,Z XZ Y Y X X,Y X


Inside ECONOMICS

𝑄2 = �{1}, {2}, … {99}, {100}�

Therefore there are two distinct refinements.


The set 𝑆 has exactly 100 elements. Consider two element partitions 𝑄 of the set 𝑆. That is 𝑄 = {𝑄1,𝑄2}, where 𝑄1 ≠ ∅, 𝑄2 ≠ ∅, 𝑄1 ∩ 𝑄2 ≠ ∅ and 𝑆 = 𝑄1 ∪ 𝑄2.

(a)How many different 2 element partitions does 𝑆 have?

(b)How many different 2 element partitions does 𝑆 have if we require that #𝑄1 = 1 and #𝑄2 = 99?

For any 2 element partition 𝑄 = {𝑄1,𝑄2} of 𝑆 it must be the case that once we fix one of the elements say 𝑄1 then 𝑄2 = 𝑆 − 𝑄1. Therefore to find the different 2 element partitions we just need to know the number of non empty proper subsets of 𝑆 and divide by 2. We divided by two because we know that if there are 100 elements then there will be 50 2 element partitions. This is demonstrated in the following hypothetical partitions:

𝑄 = �{1}, {2, … ,100}�,

𝑄 = �{2}, {1, … ,100}�, …

𝑄 = �{1,2}, {3, … ,100}�…

𝑄 = �{1,2, … 50}, {51, … ,100}�…

𝑄 = �{1,2, … 99}, {100}�

To find the number of non empty proper subsets of 𝑆 we use the power set of 𝐴. Remember that the power set is the set of all subsets of the set 𝐴. A power set is a set where each element of the set is also a set. As there are 100 elements the number of non empty proper subsets will be 2100. That is:

2𝑄{∅, {1}, {2}, … , {100}, {1,2}, {1,3}, … , {1, . . ,50}, … , {1, … ,99},𝑄}

However we want the number of non empty proper subsets and the power set includes both the empty set and the set 𝑄, which is not a proper subset. Therefore we just subtract two away from the power set to get the number of non empty proper subsets.

2𝑄 − 2

Now we just divide the number of non empty proper subsets by 2 because we want to know the number of 2 element partitions. Therefore:

2𝑄 − 22

=2100 − 2

2= 2100 − 1

(a)There are 2100 − 1 different 2 element partitions.


Inside ECONOMICS

(b) If #𝑄1 = 1 then 𝑄2 = 𝑆 − 𝑄1, meaning there is no freedom to choose the elements of 𝑄2. In fact the elements of 𝑄2 is characterized completely by 𝑄1. Every time we choose an element of 𝑄1 we have a different 2 element partition. We can choose 100 possible elements for 𝑄1. Therefore there are 100 elements if #𝑄1 = 1 and #𝑄2 = 99.


Inside ECONOMICS


Joins and Meets

Introduction to Joins and Meets

The meet and join are binary operation that occur over the set of all partitions of a set. The meet and join of two partitions allows us to organise the information common to both sets in order to find the finest common refinement and finest common coarsening. Joins and Meets have useful applications in information, game theory and knowledge models.

Meet: The meet of 𝑃 and 𝑄 is the finest common coarsening of 𝑃 and 𝑄. The meet of 𝑃 and 𝑄 is denoted as 𝑃⋀𝑄.

Join: The join of 𝑃 and 𝑄 is the finest common refinement of 𝑃 and 𝑄. The join of 𝑃 and 𝑄 is denoted as 𝑃⋁𝑄

This concept is easily explained through an example.

Example 33: Joins and Meets

Let 𝑠 = {2,4,6,5,7} and suppose that we have the following partitions:

𝑋 = �{2,4}, {6,5,7}�

𝑌 = �{2,6}, {4,5,7}�

In this case the meet and join are, respectively

𝑃⋀𝑄 = �{2,4,5,6,7}�

𝑃⋁𝑄 = �{2}, {4}, {6}, {5,7}�

Although the partition �{2}, {4}, {5}, {6}, {7}� is also a common refinement, there is a coarser common refinement and the join is the coarsest common refinement of 𝑃 and 𝑄.

Another Method to Calculate Meets and Joins

For each 𝑠 ∈ 𝑆 let 𝑃(𝑠) denote the unique 𝑃𝑖 ∈ 𝑃 such that 𝑠 ∈ 𝑃𝑖. Given two partitions 𝑃 and 𝑄 of the same set 𝑆, we can define their meet and their join as follows. Suppose that we represent the meet and join as.

𝑃⋀𝑄 = 𝑀1 ∪𝑀2 ∪ …∪𝑀𝑚

𝑃⋁𝑄 = 𝐽1 ∪ 𝐽2 ∪ …∪ 𝐽𝑛

for some positive 𝑛 and 𝑚. Then the elements 𝑠1 and 𝑠2 are in the same element 𝐽𝑖 of the join if and only if they are both in the same element in the partition 𝑃 and they are also in the same element in 𝑄.

An Alternative Definitions for the Join


Inside ECONOMICS

For every 𝑠 ∈ 𝑆, the join is defined as 𝐽(𝑠) by:

𝐽(𝑠) = 𝑃(𝑠) ∩ 𝑄(𝑠),∀𝑠 ∈ 𝑆

This means that the elements of the join are subsets of the elements of each partition. Therefore the join 𝐽 is a refinement of 𝑃 and also a refinement of 𝑄.

In the above example the elements of the join are {2}, {4}, {6}, and , {5,7}, and the elements of the partition 𝑋 are {2,4} and {6,5,7}. Therefore every element of the join is a subset of the elements in the partition. For instance the element {2} in join is a subset of the element {2,4} from the partition 𝑋. The same logic applies to the partition 𝑌 .Thus the join must be a refinement of the partition 𝑋 and the partition 𝑌.

An Alternative Definition for the Meet

For any 𝐴 ⊂ 𝑆 let, 𝑃(𝐴) = ⋃ 𝑃(𝑠)𝑠∈𝐴 and 𝑄(𝐴) = ⋃ 𝑄(𝑠)𝑠∈𝐴 , then we define the elements of the meet as:

For every 𝑠 ∈ 𝑆:

𝑀(𝑠) = 𝑃(𝑠) ∪ 𝑄(𝑠) ∪ 𝑃�𝑄(𝑠)� ∪ 𝑄�𝑃(𝑠)� ∪ 𝑃 �𝑄�𝑃(𝑠)�� ∪ …

The above statement requires calculating the union of infinitely many sets, but since 𝑠 is finite, every 𝑀(𝑠) must also be a finite set, and, at some finite point, there is nothing more to add to the infinite union.

Example 34: Join and Meets

Let 𝑠 = {1,2,3,4,5,6,7,9,12} and suppose that we have the following partitions:

𝑃 = �{1,2,9,12}, {3,4}, {6,7}, {5}�

𝑄 = �{1,2}, {3,4,6}, {7}, {5,9,12}�

In this case the join and meet are, respectively

𝑃⋁𝑄 = �{1,2}, {3,4}, {5}, {6}, {7}, {9,12}�

𝑃⋀𝑄 = �{1,2,5,9,12}, {3,4,6,7}�

Note that any two points that are in the same element of the join are also together in both partitions 𝑃 and 𝑄. To demonstrate what this actually means we can draw arrows connecting the two point elements of the join and elements of partitions.

𝑃 = �{1,2,9,12}, {3,4}, {6,7}, {5}�

𝑃⋁𝑄 = �{1,2}, {3,4}, {5}, {6}, {7}, {9,12}�

Note that for the meet every pair of points that are in the same element of the meet, for instance 1 and 5, there is a path linking them. The path is mathematically represented as


Inside ECONOMICS

𝑥0 ∈ 𝑃(𝑥1) and 𝑥1 ∈ 𝑄(𝑥2). In this particular example the path linking 1 and 5 is 9. That is for 𝑥0 ∈ 𝑃(𝑥1) we have 1 ∈ 𝑃(9) and for 𝑥1 ∈ 𝑄(𝑥2) is 9 ∈ 𝑄(5).

Example 35: Joins and Meets

Let 𝑠 = {1,2,3,4,5,6,7,9,12} and suppose that we have the following partitions:

𝑃 = �{1,2,9,12}, {3,4}, {6,7}, {5}�

𝑄 = �{1,2}, {3,4,6}, {7}, {5,9,12}�

𝑆 = �{1}, {2,3,5}, {4,6,7}, {9,12}�

The join and meet for partition 𝑃 and partition 𝑄, are respectively

𝑃⋁𝑄 = �{1,2}, {3,4}, {5}, {6}, {7}, {9,12}�

𝑃⋀𝑄 = �{1,2,5,9,12}, {3,4,6,7}�

The join and meet for partition 𝑃 and partition 𝑆, are respectively

𝑃⋁𝑆 = �{1}, {2}, {3}, {4}, {5}, {6,7}, {9,12}�

𝑃⋀𝑆 = {{1,2,3,4,6,7,9,12}{5}}

We can also compute (𝑃⋁𝑆)⋀𝑄

(𝑃⋁𝑆)⋀𝑄 = �{1,2,3,4,6,7,9,12}, {5}�

and (𝑃⋀𝑆)⋁𝑄

(𝑃⋀𝑆)⋁𝑄 = �{1,2}, {3,4,6}, {5}, {7}, {9,12}�

also 𝑄⋁𝑆

𝑄⋁𝑆 = �{1}, {2}, {3}, {4,6}, {5}, {7}, {9,12}�

and (𝑄⋁𝑆)⋀𝑃

(𝑄⋁𝑆)⋀𝑃 = �{1,2,9,12}, {3,4,6,7}, {5}�

Diagrammatic Method to Calculate the Join and Meet

Often it is useful to use matrix diagrams to calculate joins and meets. If partition 𝐴 has five elements and partition 𝐵 has four elements we construct a 4 × 5 matrix. Partition 𝐵 is represented in at the top for the columns and partition 𝐴 the rows. Once we setup the matrix we can simply use it by ensuring that each state is written in the same row that contains this state in the partition 𝐴 and in the same column it appears in partition 𝐵.

To calculate the connected components we choose any state and draw all the vertical and horizontal lines from the this state to other non empty cells. Upon reaching another state we simply repeat the process of drawing horizontal and vertical lines. We continue this until


Inside ECONOMICS

there are no new non empty cell that can be reached. Essentially we just follow the connected points of each state until we cannot find any more connected points. It is important to note that some states are directly reachable through a single connection but other states are indirectly reachable. All the reachable states, directly or indirectly, are included in the connected component state. The meet is the collection of all the sets of connected components and the join is obtained by looking at each non empty cell and taking the partition.

Example 36: Diagrammatic Calculation of Joins and Meets

Let 𝑆 = {1,2,3,4,5,6,7,8, } and consider the partition 𝐴 = �{1,2}, {3,4}, {6,7}, {5}, {8}� and 𝐵 = �{1,6}, {4,5}, {2,7,8}, {3}�. The we can represent these partitions in the diagram below. As partition 𝐴 has five elements and partition 𝐵 has four elements we construct a 4 × 5 matrix.

We start with the state 𝑤 = 1 and draw a line to every to every other state that is connected directly to this state. The states directly reachable from 𝑤 = 1 are 𝑤 = 2 and 𝑤 = 6. The states that are indirectly reachable from 𝑤 = 1 are those that are reachable from the directly reachable states. Therefore we draw lines from state 𝑤 = 2 to 𝑤 = 7 and 𝑤 = 8, and then draw lines from state 𝑤 = 6 to 𝑤 = 1 and 𝑤 = 7. Now check if there are any reachable states from 𝑤 = 8. As there are not any reachable states from 𝑤 = 8 we have exhausted all the reachable states. The states 𝑤 = 1, = 𝑤 = 2,𝑤 = 6,𝑤 = 7 and 𝑤 = 8 are the elements that are in the first element of the meet. That is {1,2,6,7,8}.

Now we must go to the next state that is not in part of the element above. Hence we start at 𝑤 = 4 and draw a line to the reachable states 𝑤 = 3 and 𝑤 = 5. Then we find all the reachable sets from 𝑤 = 3 and 𝑤 = 5. In this case there are no reachable sets from 𝑤 = 3 and 𝑤 = 5. Therefore the states 𝑤 = 3,𝑤 = 4 and 𝑤 = 5 are elements in the an element of the meet. That is {3,4,5}

Now we simply combine the two elements to calculate the meet:

𝐵⋀𝐴 = �{1,2,6,7,8}, {3,4,5}�

The join is given by the partition of all the non empty in the cells above. This means to calculate the join we just write the set of all the connecting states in the non empty cells.

𝐵⋁𝐴 = �{1}, {2}, {3}, {4}, {5}, {6}, {7}, {8}�

{1,2} {3,4} {6,7} {5} {8}

{1,6} 1 6

{4,5} 4 5

{2,7,8} 2 7 8

{3} 3

A

B


Inside ECONOMICS


Consider the space 𝑆 = {1,2,3,4,5,6,7,8} and the partitions 𝑄 = �{1,5,8}, {4,2}, {6,7}, {3}� and 𝑃 = �{1}, {3,4,6}, {7}, {2,5}, {8}�. Then we have the following 4 × 5 matrix.

Start at 𝑤 = 1 and draw the lines that directly reachable. The only line directly reachable in this case is 𝑤 = 5 and 𝑤 = 8. We then draw lines the directly reachable from states 𝑤 = 5 and 𝑤 = 8. In this case the only reachable state is 𝑤 = 2. We continue this process and draw lines to all the directly reachable states from 𝑤 = 2. This takes us to the state 𝑤 = 4. Again we continue the process 𝑤 = 4 to 𝑤 = 6 and 𝑤 = 6 to 𝑤 = 3 and 𝑤 = 7. All the states 𝑤 = 2,𝑤 = 4,𝑤 = 6,𝑤 = 7,𝑤 = 3 are all indirectly reachable from state 𝑤 = 1. Therefore the meet and join are, respectively:

𝑄⋀𝑃 = �{1,2,3,4,5,6,7,8}�

𝑄⋁𝑃 = �{1}, {2}, {3}, {4}, {5}, {6}, {7}, {8}�


Consider the space 𝑆 = {𝑎, 𝑏, 𝑐,𝑑, 𝑒,𝑓,𝑔, ℎ} and the partitions 𝑋 = �{𝑎,𝑑}, {𝑏, 𝑐}, {𝑒,ℎ}, {𝑓,𝑔, 𝑖}� and 𝑌 = �{𝑎, 𝑏}, {𝑐,𝑑}, {𝑒,𝑓,𝑔}, {ℎ}, {𝑖}�. Then we have the following 4 × 5 matrix.

Start at 𝑤 = 𝑎 then draw lines to the directly reachable states 𝑤 = 𝑏 and 𝑤 = 𝑑. Then draw the reachable states from 𝑤 = 𝑏 and 𝑤 = 𝑑 and so on until we exhaust all the directly and indirectly reachable states of 𝑤 = 𝑎. The collection of the state is an element in the meet of 𝑋 and 𝑌. That is {𝑎, 𝑏, 𝑐,𝑑}. Now we go to the next state in 𝑋, 𝑤 = 𝑒 and follow the same process as that used for 𝑤 = 𝑎. This is represented by the directly reachable states from 𝑤 = 𝑒 to 𝑤 = ℎ and 𝑤 = 𝑓,𝑔 ,and the indirectly reachable states from 𝑤 = 𝑓,𝑔 to 𝑤 = 𝑖. Therefore the meet and join are, respectively:

{1} {3,4,6} {7} {2,5} {8}

{1,5,8} 1 5 8

{4,2} 4 2

{6,7} 6 7

{3} 3

P

Q

{a,b} {c,d} {e,f,g} {h} {i}

{a,d} a d

{b,c} b c

{e,h} e h

{f,g,i} f,g i

Y

X


Inside ECONOMICS

𝑄⋀𝑃 = �{𝑎, 𝑏, 𝑐,𝑑}, {𝑒,𝑓,𝑔,ℎ, 𝑖}�

𝑄⋁𝑃 = �{𝑎}, {𝑏}, {𝑐}, {𝑑}, {𝑒}, {𝑓,𝑔}, {ℎ}, {𝑖}�


Suppose that two firms are bidding on a patent for a new technology. There are two companies that wish to purchase the patent of this technology from the inventor. However both firms have done different analysis on the possible outcomes this technology will have once it is commercialised and ready for the market. The possible states of the world are 𝑠 = {𝑎, 𝑏, 𝑐,𝑑, 𝑒,𝑓}. The firms will base their bids upon what they believe the possible state will be. However neither firm has perfect knowledge of the likely outcome. However each players refinement represents their knowledge of the potential states of the world. A refinement means that the firm has potentially better knowledge. For instance {𝑎,𝑑} means that the future state could be either 𝑎 or 𝑑. Once further knowledge or information is gained we may be able to refine this to {a} and {d}. Therefore in this case the join represents direct communication between the firms and the meet represents common knowledge.

Firm 1's partition of knowledge is 𝐹𝑖𝑟𝑚1 = �{𝑎,𝑑}, {𝑏, 𝑒}, {𝑐}, {𝑓}�

Firm 2's partition of knowledge is 𝐹𝑖𝑟𝑚2 = �{𝑎, 𝑏}, {𝑐}, {𝑑, 𝑒,𝑓}�

Therefore the meet and join are, respectively:

𝐹𝑖𝑟𝑚1⋀𝐹𝑖𝑟𝑚2 = �{𝑎, 𝑏,𝑑, 𝑒, 𝑓}, {𝑐}�

𝐹𝑖𝑟𝑚1⋁𝐹𝑖𝑟𝑚2 = �{𝑎}, {𝑏}, {𝑐}, {𝑑}, {𝑒}, {𝑓}�

The knowledge common to both firms is that the state of world is in either the first or second element of the meet. Therefore if we believe the true state of the world once the product is released will be in the second element we know that the true state must be 𝑐. However if a firm believes that the true state lies in the first element of the meet then the firms know it can be either 𝑎, 𝑏,𝑑, 𝑒 and 𝑓. This means that the finest common coarsening is a representation of general knowledge. Alternatively we could say the meet is a refinement of the space 𝑆 and is therefore information, but because the partition has few elements it is not specific knowledge, and because it is the coarsest common refinement it is the most general knowledge a firm can have.

The join is the coarsest common refinement and represents the knowledge or information as resulting from direct communication between the firms. So once they share information they have a much better understanding of the possible states. Therefore if they believe the true state, upon commercialisation of the technology, lies in the first element of the join they

{a,d} {b,e} {c} {f}{a,b} a b{c} c

{d,e,f} d e fFirm 2

Firm 1


Inside ECONOMICS

know that the state must be 𝑎 and cannot be any other state. Each refinement represents an increase in information or knowledge and therefore the firm can be more specific about the possible state.


Suppose that the national secret service is seeking specific information about possible threats and opens a dialogue with other nations intelligence services. Let the set of all possible states be 𝑆 = {𝑎, 𝑏, 𝑐,𝑑, 𝑒,𝑓,𝑔,ℎ} and there are four nations with the following partitions of knowledge on the matter:

Nation 1: 𝑁1: �{𝑎, 𝑏,𝑔}, {𝑐,ℎ}, {𝑑, 𝑒,𝑓}�

Nation 2: 𝑁2: �{𝑎, 𝑏, 𝑐,𝑑, }, {𝑒,𝑓,𝑔, ℎ}�

Nation 3: 𝑁3: �{𝑎, 𝑏, 𝑐}, {𝑑}, {𝑒, 𝑓,𝑔}�

Nation 4: 𝑁4: �{𝑎}, {𝑏}, {𝑐,𝑑}, {𝑒, ℎ}, {𝑔}, {𝑓}�

Suppose that Nation 1 is the nation seeking information regarding possible threats. Nation 1 firstly shares information with Nation 2. This is depicted in the following diagram.

Now Nation 1 has refined its knowledge partition as the join between its knowledge partition 𝑁1 and Nation 2's knowledge partition 𝑁2.

𝑁1⋁𝑁2 = �{𝑎, 𝑏}, {𝑐}, {𝑑}, {𝑒,𝑓}, {𝑔}, {ℎ}�

Now Nation 1 uses the join as its own knowledge partition when it opens information sharing with Nation 3.

Again the meet is the common knowledge and the join is the refinement of knowledge base on sharing information. Therefore when Nation 1 uses 𝑁1⋁𝑁2 to refine its knowledge and uncover the possible threats.

(𝑁1⋁𝑁2)⋁𝑁3 = �{𝑎, 𝑏}, {𝑐}, {𝑑}, {𝑒,𝑓}, {𝑔}, {ℎ}�

{a,b,g} {c,h} {d,e,f}

{a,b,c,d} ab c d

{e,f,g,h} g h efNat

ion

2

Nation 1

{a,b} {c} {d} {e,f} {g} {h}

{a,b,c} ab c

{d} d

{e,f,g,h} e,f g h

Nation 1

Nat

ion

3


Inside ECONOMICS

The join has not refined the partition and therefore has not revealed more information on the possible threats to Nation 1. However Nation 1 decides to open discussions with Nation 4.

Nation 4 more knowledge than the other nations. This is evident as the matrix has a larger dimension than the previous matrices. We can see that the join of these partitions reveals more information about the situation.

(𝑁1⋁𝑁2)⋁𝑁4 = �{𝑎}, {𝑏}, {𝑐}, {𝑑}, {𝑒}, {𝑓}, {𝑔}, {ℎ}�

In fact this partition is the finest refinement and allows Nation 1 the best possible information about the possible threats. It is also important to note that the meet is also becoming more refined as we have finer partitions. This means that with more information sharing the level of common knowledge is becoming more refined.

{a,b} {c} {d} {e,f} {g} {h}{a} a{b} b

{c,d} c d{e,h} e h{g} g{f} f

Nation 1N

atio

n 4

A Note on the Concept of Common Knowledge

The meet represents the Common Knowledge of players. Generally the notion of common knowledge requires an infinite chain of statements. Obviously, if the state space is finite, after some point statements become abundant. For example once all agents know a state is a possible state, and all agents know that all agents know that all agents know that a state is a possible state and this goes on and on indefinitely. Nevertheless, conceptually speaking, common knowledge requires infinitely many checks.


Inside ECONOMICS


The Inclusion-Exclusion Principle

Inclusion Exclusion Principle

The Inclusion-Exclusion Principle is a useful tool to calculate the number of elements in the union and has wide application in probability calculations.

The Inclusion Exclusion for Two Sets

The number of elements in a set 𝐴 ∪ 𝐵 is the number of elements in A plus the elements in B, minus the number of elements in 𝐴 ∩ 𝐵 as this number was counted twice in n(A)+n(B) therefore.

#(𝐴 ∪ 𝐵) = #𝐴 + #𝐵 − #(𝐴 ∩ 𝐵)

It is simple to see why #(𝐴 ∪ 𝐵) = #𝐴 + #𝐵 − #(𝐴 ∩ 𝐵). Let 𝑥 = #(𝐴 − 𝐵), 𝑦 = #(𝐵 − 𝐴) and 𝑧 = #(𝐴 ∩ 𝐵). The number of elements in 𝐴 ∪ 𝐵 is exactly 𝑥 + 𝑦 + 𝑧, and the number of elements in 𝐴 is 𝑥 + 𝑧. The number of elements in 𝐵 is 𝑦 + 𝑧 and thus:

#(𝐴 ∪ 𝐵) = 𝑥 + 𝑦 + 𝑧 = (𝑥 + 𝑧) + (𝑦 + 𝑧) − 𝑧 = #𝐴 + #𝐵 − #(𝐴 ∩ 𝐵)

The inclusion exclusion principle for Probability: If 𝐴 and 𝐵 are two events then the probability that one event or the other will occur, 𝑃𝑟(𝐴 𝑜𝑟 𝐵), is given by;

𝑃𝑟(𝐴 𝑜𝑟 𝐵) = 𝑃𝑟(𝐴 ∪ 𝐵) = 𝑃𝑟(𝐴) + 𝑃𝑟(𝐵) − 𝑃𝑟 (𝐴 ∩ 𝐵)

Example 41: The Inclusion-Exclusion Principle for Two Sets

Suppose that 60 people were interviewed for a position at a bank. 26 had an economics degree, 40 had and finance degree and 9 had both an economics and finance degree.

(a) What is the probability that a person selected has both a finance and economics degree?

𝑃𝑟(𝐴 ∩ 𝐵) =9

60

(b) What is the probability that a person has a finance degree or an economics degree? Hint use the inclusion exclusion principle

𝑨 𝑩

The union of 𝐴 and 𝐵 includes all the elements and represented in the diagram as the space of 𝐴, 𝐴 ∩ 𝐵 and 𝐵.

The intersection of the two sets is simply 𝐴 ∩ 𝐵.

𝑆 represent the space in which these sets occur. This is sometimes referred to as the

𝑨 ∩ 𝑩

𝑺


Inside ECONOMICS

𝑃𝑟(𝐸𝑐𝑜 ∪ 𝐹𝑖𝑛) =2660

+4060

−9

60=

5760


A firm has 14 employees: 5 are women and 9 are men. Three of the women and 5 of the men are older than 40. If a person is chosen at random from this firm, what is the probability that the person is male or over 40 years old?

𝑃𝑟(𝑚𝑎𝑛 ∪ 𝑜𝑣𝑒𝑟 40) = 𝑃𝑟(𝑚𝑎𝑛) + 𝑃𝑟(𝑜𝑣𝑒𝑟 40) − 𝑃𝑟 (𝑚𝑎𝑛 ∩ 𝑜𝑣𝑒𝑟 40

𝑃𝑟(𝑚𝑎𝑛 ∪ 𝑜𝑣𝑒𝑟 40) =9

14+

814

−5

14=

1214

=67

Therefore the probability that person is a male or over 40 year old is 67.


The board of directors consist of 9 liberals, 5 are female and 4 male, and 9 labour where 5 are male and 4 are female. If a director is picked at random to attend a meeting on behalf of the company what is the probability that person will be a labour voter or female?

𝑃𝑟(𝑙𝑎𝑏𝑜𝑢𝑟 ∪ 𝑓𝑒𝑚𝑎𝑙𝑒) = 𝑃𝑟(𝑙𝑎𝑏𝑜𝑢𝑟) + 𝑃𝑟(𝑓𝑒𝑚𝑎𝑙𝑒)− 𝑃𝑟 (𝑙𝑎𝑏𝑜𝑢𝑟 ∩ 𝑓𝑒𝑚𝑎𝑙𝑒 )

𝑃𝑟(𝑙𝑎𝑏𝑜𝑢𝑟 ∪ 𝑓𝑒𝑚𝑎𝑙𝑒) =9

18+

918

−4

18=

1418

=79

Therefore the probability that a director is a labour voter or female is 79.


How many number between 1 and 100 are multipliers of 𝑛 where 𝑛 = {5,7,35}. The quantity of multipliers of 5 and 7 is equal to:

𝑎5 = 20

𝑎7 = 14

𝑎35 = 2

As 35 and its multiples are multiples of both 5 and 7 they represent the intersection. This means that we will double count these numbers and therefore they must be excluded. Therefore we can now use the inclusion exclusion principle to work out the number of multiples of 5 and 7 between 1 and 100. By the inclusion-exclusion principle:

𝑎5 + 𝑎7 − 𝑎35 = 20 + 14 − 2 = 32

There are 32 number between 1 and 100 that are multipliers of 5 and 7.

The Inclusion-Exclusion Principle for Three Sets

Following the logic for two sets we can also use the inclusion-exclusion principle for three sets. We state the inclusion-exclusion principle for three sets as:


Inside ECONOMICS

#(𝐴 ∪ 𝐵 ∪ 𝐶) = #𝐴 + #𝐵 + #𝐶 − #(𝐴 ∩ 𝐵) − #(𝐴 ∩ 𝐶) − #(𝐵 ∩ 𝐶) + #(𝐴 ∩ 𝐵 ∩ 𝐶)

To see why this is true we need to use the property (𝐴 ∪ 𝐵) ∩ 𝐶 = (𝐴 ∩ 𝐶) ∪ (𝐵 ∩ 𝐶).

#(𝐴 ∪ 𝐵 ∪ 𝐶) = #�(𝐴 ∪ 𝐵) ∪ 𝐶� = #(𝐴 ∪ 𝐵) + #𝐶 − #�(𝐴 ∪ 𝐵) ∩ 𝐶� =

= #𝐴 + #𝐵 − #(𝐴 ∩ 𝐵) + #𝐶 − #�(𝐴 ∩ 𝐶) ∪ (𝐵 ∩ 𝐶)� =

= #𝐴 + #𝐵 + #𝐶 − #(𝐴 ∩ 𝐵) − �#(𝐴 ∩ 𝐶) + #(𝐵 ∩ 𝐶) − #�(𝐴 ∩ 𝐶) ∩ (𝐵 ∩ 𝐶)�� =

= #𝐴 + #𝐵 + #𝐶 − #(𝐴 ∩ 𝐵) − #(𝐴 ∩ 𝐶) − #(𝐵 ∩ 𝐶) + #(𝐴 ∩ 𝐵 ∩ 𝐶)

In the final step we have used the property (𝐴 ∪ 𝐵) ∩ 𝐶 = (𝐴 ∩ 𝐶) ∪ (𝐵 ∩ 𝐶).

Example 45: The Inclusion-Exclusion Principle for Three Sets

Suppose that it is Sports Day at a local school, and there are three major events, Shot-Put, 100m Sprint and High Jump. There are 300 students at sports day who have entered the competition. We know that 60 contestants did the Shot-Put, 80 competed in the 100m and 50 the High Jump. We know that 30 students competed in both the Shot-Put and the 100m Sprint, 25 competed in both the Shot-Put and High Jump and 15 competed in both the 100m Sprint and the High Jump. How many competitors competed in all three events considering that 130 students competed in at least one event?

Let 𝐴 = #𝑆ℎ𝑜𝑡 − 𝑃𝑢𝑡, 𝐵 = #100𝑚 𝑆𝑝𝑟𝑖𝑛𝑡 and 𝐶 = #𝐻𝑖𝑔ℎ 𝐽𝑢𝑚𝑝. Then:


#(𝐴 ∪ 𝐵 ∪ 𝐶) = 130 = 60 + 80 + 50 − 30 − 25 − 15 + #(𝐴 ∩ 𝐵 ∩ 𝐶) =

130 = 120 + #(𝐴 ∩ 𝐵 ∩ 𝐶)

10 = #(𝐴 ∩ 𝐵 ∩ 𝐶)

Therefore 10 students competed in all three events.


In order to receive a scholarship students must all three tests in Mathematics, Economics and English. Suppose that 1300 people participate in the contest for the scholarship and 200

𝑨 𝑩

The union of 𝐴 𝐵 and 𝐶 includes all the elements and represented in the diagram as the space of 𝐴,𝐵,𝐶 and 𝐴 ∩ 𝐵 ∩ 𝐶.

The intersection of the two sets is simply 𝐴 ∩ 𝐵 ∩ 𝐶.

𝑆 represent the space in which these sets occur. This is

𝑨 ∩ 𝑩 ∩ 𝑪

𝑺

𝑪


Inside ECONOMICS

pass in Mathematics, 300 pass Economics and 600 Pass English. In addition 100 pass Mathematics and Economics, 150 pass Mathematics and English and 200 pass both Economics and English. If 800 Candidates passed at least one subject then how many candidates passed all three tests?

Let 𝐴 = #𝑀𝑎𝑡ℎ𝑒𝑚𝑎𝑡𝑖𝑐𝑠, 𝐵 = #𝐸𝑐𝑜𝑛𝑜𝑚𝑖𝑐𝑠 and 𝐶 = #𝐸𝑛𝑔𝑙𝑖𝑠ℎ, then


#(𝐴 ∪ 𝐵 ∪ 𝐶) = 800 = 200 + 300 + 600 − 100 − 150 − 200 + #(𝐴 ∩ 𝐵 ∩ 𝐶) =

800 = 650 + #(𝐴 ∩ 𝐵 ∩ 𝐶)

150 = #(𝐴 ∩ 𝐵 ∩ 𝐶)

Therefore 150 students passed all three tests and are eligible for a scholarship.


How many numbers between 1 and 100 are multiples of 5,7 or 9?

𝑎5 = 20

𝑎7 = 14

𝑎9 = 11

𝑎35 = 2

𝑎45 = 2

𝑎63 = 1

𝑎315 = 0


20 + 14 + 11 − 2 − 2 − 1 + 0 = 40

Therefore there are 40 number between 1 and 100 that are multiples of 5,7 or 9

Basic Set Theory - inside-economics.cominside-economics.com/s/Mathematics For Economists/1....

Documents

Transcript of Basic Set Theory - inside-economics.cominside-economics.com/s/Mathematics For Economists/1....