Set Theory
10.0 Venn Diagrams and operation on sets
10.0 Venn Diagrams and operation on sets
The diagrams which are used to represent the relationships between the sets are called Venn diagrams.
Named after John Venn, these diagrams consist of a rectangle that represent the universal set and circles that represent different subsets.
- Subset:
A subset is represented by a circle within another circle that acts as a superset.
- Union of sets
Let $A$ and $B$ be two sets. The union of $A$ and $B$ is the set of all those elements which belong either to $A$ or to $B$ or to both.
The notation used is $$A \cup B$$
This is read as $A$ union $B$
$$\begin{equation} \begin{aligned} A \cup B = \{ x:x \in A\;\;or\;\;x \in B\} \\ x \in A \cup B \Leftrightarrow x \in A\;or\;x \in B \\ x \notin A \cup B \Leftrightarrow x \notin A\;and\;x \notin B \\\end{aligned} \end{equation} $$
It is evident from the definition that, $$A \subseteq A \cup B\;\;and\;\;B \subseteq A \cup B$$
i. If $A$ and $B$ are two sets, such that, $A \subset B$ then $A \cup B = B$ .
ii. If $A$ and $B$ are two sets, such that, $B \subset A$ then $A \cup B = A$.
Illustration 20. Consider the following sets, $A = \{ c,m,n,r,s,v,w,x,z\} $ , $B = \{ b,d,l,f,h,k,t\} $ , $C = \{ g,j,p,q,y\} $ and $D = \{ a,e,i,o,u\} $. Find
i. $A \cup B$
ii. $C \cup B$
iii. $B \cup D$
iv. $A \cup D$
v. $A \cup B \cup C$
vi. $C \cup B \cup D$
Solution:
$\begin{equation} \begin{aligned} A = \{ c,m,n,r,s,v,w,x,z\} \\ B = \{ b,d,l,f,h,k,t\} \\ C = \{ g,j,p,q,y\} \\ D = \{ a,e,i,o,u\} \\\end{aligned} \end{equation} $
$$\begin{equation} \begin{aligned} i.A \cup B = \{ c,m,n,r,s,v,w,x,z,b,d,l,f,h,k,t\} \\ ii.C \cup B = \{ g,j,p,q,y,b,d,l,f,h,k,t\} \\ iii.B \cup D = \{ b,d,l,f,h,k,t,a,e,i,o,u\} \\ iv.A \cup D = \{ c,m,n,r,s,v,w,x,z,a,e,i,o,u\} \\ v.A \cup B \cup C = \{ c,m,n,r,s,v,w,x,z,b,d,l,f,h,k,t,g,j,p,q,y\} \\ = \{ x:x\;is\;a\;{\mkern 1mu} consonant\;in\;English\;alphabet\} \\ vi.C \cup B \cup D = \{ b,d,l,f,h,k,t,g,j,p,q,y,a,e,i,o,u\} \\\end{aligned} \end{equation} $$
- Intersection of sets
Let $A$ and $B$ be two sets. The intersection of $A$ and $B$ is the set of all those elements that belong to both $A$ and $B$.
The intersection of $A$ and $B$ is denoted by $$A \cap B$$
This is read as $A$ intersection $B$
$$\begin{equation} \begin{aligned} A \cap B = \{ x:x \in A\;\;and\;\;x \in B\} \\ x \in A \cap B \Leftrightarrow x \in A\;and\;x \in B \\ A \cap B \subseteq A\;\;and\;\;A \cap B \subseteq B \\\end{aligned} \end{equation} $$
i. If $A$ and $B$ are two sets, such that, $A \subset B$ then $A \cap B = A$ .
ii. If $A$ and $B$ are two sets, such that, $B \subset A$ then $A \cap B = B$ .
Illustration 21. Consider the following sets, $A = \{ x:x = 2n\,,\;n \in N\} $, $B = \{ x:x \in Z\} $, $C = \{ x:x = 2n - 1,\;n \in N\} $, $D = \{ x:x = 2n + 1,\,n \in N\} $ . Find
i. $A \cap B$
ii. $C \cap B$
iii. $B \cap D$
iv. $A \cap D$
v. $A \cap B \cap C$
vi. $C \cap B \cap D$
Solution: $\begin{equation} \begin{aligned} A = \{ x:x = 2n,\;n \in N\} \\ B = \{ x:x \in Z\} \\ C = \{ x:x = 2n - 1,\;n \in N\} \\ D = \{ x:x = 2n + 1,n \in N\} \\\end{aligned} \end{equation} $
$i.A \cap B$
Elements of $A$ are $2,4,6,8,10,12,.......$
Elements of $B$ are $........-4,-3,-2,-1,0,1,2,3,4.......$
The elements common to both are, $2,4,6,8,10,12,.......$
Hence, $$A \cap B = \{ 2,4,6,8,....\} = \{ x:x = 2n{\mkern 1mu} ,\;n \in N\} = A$$
$ii.C \cap B$
Elements of $C$ are $1,3,5,7,9,11,.......$
Elements of $B$ are $........-4,-3,-2,-1,0,1,2,3,4.......$
The elements common to both are, $1,3,5,7,9,11,.......$
Hence, $$C \cap B = \{ 1,3,5,7,9,.....\} = \{ x:x = 2n - 1,\;n \in N\} = C$$
$iii.B \cap D$
Elements of $D$ are $3,5,7,9,11,.......$
Elements of $B$ are $........-4,-3,-2,-1,0,1,2,3,4.......$
The elements common to both are, $3,5,7,9,11,.......$
Hence, $$B \cap D = \{ 3,5,7,9,.....\} = \{ x:x = 2n + 1,\;n \in N\} = D$$
$iv.A \cap D$
Elements of $A$ are $2,4,6,8,10,12,.......$ , i.e. the set of even natural numbers.
Elements of $D$ are $3,5,7,9,11,.......$ , i.e. the set of all odd natural numbers except $1$.
There are no elements common to both.
Hence, $$A \cap D = \emptyset $$.
$v.A \cap B \cap C$
From the above result $(i)$, we get, $$A \cap B = \{ 2,4,6,8,....\} = \{ x:x = 2n{\mkern 1mu} ,\;n \in N\} = A$$
Thus, this can be rewritten as, $$A \cap B \cap C = A \cap C$$
Elements of $A$ are $2,4,6,8,10,12,.......$ , i.e. the set of even natural numbers.
Elements of $C$ are $1,3,5,7,9,11,.......$ , i.e. the set of all odd natural numbers.
There are no elements common to both.
Hence, $$A \cap B \cap C = A \cap C = \emptyset $$
$vi.C \cap B \cap D$
From the above result $(ii)$, we get, $$C \cap B = \{ 1,3,5,7,9,.....\} = \{ x:x = 2n - 1,\;n \in N\} = C$$
Thus, this can be rewritten as, $$C \cap B \cap D = C \cap D$$
Elements of $C$ are $1,3,5,7,9,11,.......$ , i.e. the set of all odd natural numbers.
Elements of $D$ are $3,5,7,9,11,.......$ , i.e. the set of all odd natural numbers except $1$.
The elements common to both are, $3,5,7,9,11,.......$
Hence, $$C \cap B \cap D = C \cap D = \{ 3,5,7,9,.....\} = \{ x:x = 2n + 1,\;n \in N\} = D$$
- Disjoint sets
Two sets $A$ and $B$ are said to be disjoint if, $$A \cap B = \emptyset $$
If $A \cap B \ne \emptyset $, then $A$ and $B$ are said to be intersecting or overlapping sets.
Illustration 22. Classify the following as disjoint and intersecting sets
i. $A = \{ x:x = 2n\,,\;n \in N\} $ and $B = \{ x:x \in Z\} $
ii. $A = \{ x:x < 0\} $ and $B = \{ x:x = {n^3},n \in N\} $
iii. $A = \{ x:x = 2n\,,\;n \in N\} $ and $B = \{ x:x = 2n - 1,\;n \in N\} $
iv. $B = \{ x:x \in Z\} $ and $A = \{ x:x \in C\} $
v. $A = \{ x:x < 0\} $ and $B = \{ x:x = {n^3},n \in Z\} $
vi. $A = \{ x:x = 2n + 1,\,n \in N\} $ and $B = \{ x:x = {n^2},n \in N\} $
Solution:
| SETS | CLASSIFICATION |
| i. $A = \{ x:x = 2n\,,\;n \in N\} $ and $B = \{ x:x \in Z\} $ | Elements of $A$ are $2,4,6,8,10,12,.......$ , i.e. the set of even natural numbers. Elements of $B$ are $.....,-3,-2,-1,0,1,2,3,....$, The elements common to both are, $2,4,6,8,10,12,.......$ Therefore, $$A \cap B \ne \emptyset $$ Hence they are intersecting sets. |
| ii. $A = \{ x:x < 0\} $ and $B = \{ x:x = {n^3},n \in N\} $ | Set $A$ is the set of negative numbers. $B$ has the elements as cube of natural numbers. Hence, all elements are positive. There are no elements in common to both. Therefore, $$A \cap B = \emptyset $$ Hence they are disjoint sets. |
| iii. $A = \{ x:x = 2n\,,\;n \in N\} $ and $B = \{ x:x = 2n - 1,\;n \in N\} $ | Elements of $A$ are $2,4,6,8,10,12,.......$ , i.e. the set of even natural numbers. Elements of $B$ are $1,3,5,7,9,11,.......$ , i.e. the set of odd natural numbers. There are no elements in common to both. Therefore, $$A \cap B = \emptyset $$ Hence they are disjoint sets. |
| iv. $B = \{ x:x \in Z\} $ and $A = \{ x:x \in R\} $ | We know that, $Z \subset R$ Hence, $$A \cap B = Z$$ Therefore they are intersecting sets. |
| v. $A = \{ x:x < 0\} $ and $B = \{ x:x = {n^3},n \in Z\} $ | Set $A$ is the set of negative numbers. $B$ has the elements as cube of integer numbers. Hence, the elements are positive or negative. The negative elements belonging to $B$ are common to both. Therefore, $$A \cap B \ne \emptyset $$ Hence they are intersecting sets. |
| vi. $A = \{ x:x = 2n + 1,\,n \in N\} $ and $B = \{ x:x = {n^2},n \in N\} $ | Elements of $A$ are $3,5,7,9,11,.......$ , i.e. the set of even natural numbers. Elements of $B$ are $1,4,9,16,.......$, The odd square numbers are common to both, Therefore, $$A \cap B \ne \emptyset $$ Hence they are intersecting sets. |
- Difference of sets
Let $A$ and $B$ be two sets. The difference of $A$ and $B$ is the set of all those elements of $A$ which do not belong to $B$, written as $$A - B$$
$$\begin{equation} \begin{aligned} A - B = \{ x:x \in A\;\;and\;\;x \notin B\} \\ x \in A - B \Leftrightarrow x \in A\;and\;x \notin B \\\end{aligned} \end{equation} $$
Similarly, $$B - A$$
is set of all elements belonging to $B$ that do not belong to $A$
$$\begin{equation} \begin{aligned} B - A = \{ x:x \notin A\;\;and\;\;x \in B\} \\ x \in B - A \Leftrightarrow x \notin A\;and\;x \in B \\\end{aligned} \end{equation} $$
- Symmetric Difference of two sets
Let $A$ and $B$ be two sets. The symmetric difference of sets $A$ and $B$ is the set $$(A - B) \cup (B - A)$$
denoted as
$$A\;\Delta \;B$$
Then,
$$A\;\Delta \;B = (A - B) \cup (B - A) = \{ x:x \notin A \cap B\} $$
- Complement of a set
Let $U$ be the universal set and let $A$ be a set such that $A \subset B$.
Then, the complement of $A$ with respect to $U$ is defined as the set of all elements of $U$ which are not in $A$.
It denoted as $$A'\,\;or\,\;{A^c}\;\;or\;\,U - A$$
$$\begin{equation} \begin{aligned} {A^c} = \{ x \in U:x \notin A\} \\ x \in {A^c} \Leftrightarrow x \notin A \\\end{aligned} \end{equation} $$
Illustration 23. Find the results of the following:
| i. ${U'}$ | ii. $(A')'$ | iii. $A \cap A'$ | iv. $A \cup A'$ | v. ${\emptyset '}$ |
Solution: According to the definition
$$A' = \{ x \in U:x \notin A\} $$
i. $$U' = \{ x \in U:x \notin U\} = \emptyset $$
as, an element cannot both, belong and not belong to a set.
ii. $$(A')' = \{ x \in U:x \notin A'\} = \{ x \in U:x \in A\} = U \cap A = A$$
an element belongs to universal set such that it belongs the complement of the complement of $A$ means it belongs to set $A$.
iii. $$A \cap A' = \{ x:x \in A\,and\,x \notin A\} = \emptyset $$
If $x$ does not belong to $A$, it cannot belong to $A$. This is contradictory.
iv. $$A \cup A' = \{ x:x \in A\,or\,x \notin A\} = U$$
In a universal set, an element can either belong to $A$ or not belong to $A$.
v. $$\emptyset ' = \{ x \in U:x \notin \emptyset \} = U$$
If an element belongs to universal set, it cannot be in an empty set.
Note: The above results are standard results for every set.
