Differentiation (Part-38) | S N De

A great German mathematician  George Cantor  developed the theory of set. This is one of the most fundamental concepts of mathematics and is now used in almost all branches of science, technology, commerce and economics. There is hardly any doubt that modern  mathematics is systematized and becomes more logical because of the advent of set theory. The theory has established co-ordination among different branches of mathematics namely geometry, algebra, calculus etc. The solutions of complicated problems of science and commerce become easier by the application of set theory. Not only that, the theory helps to base the subject probability on a solid logical foundation. This chapter deals with the basic concepts of sets.

Today we are going to discuss Short answer type questions of S.N. Dey. 

1. Write short notes: [Each question carries 4 marks]

(i) Concepts of sets, subset, equality of two sets; universal set and null set, finite and infinite sets.

(ii) Union, intersection and difference of two sets.

(iii) Universal set and subset.

(iv) The three set operations (union, intersection and complementation).

Sol. (i) Concepts of sets :  A collection of objects is defined to be a set when

$\,\,a.\,\,$ the collection is well-defined

$\,\,b.\,\,$ objects belonging to the collection are different

$\,\,c.\,\,$ objects of the collection are independent of the order of their arrangements.

A well-defined collection of objects is said to be a set if any two objects of the collection are different. Thus, the collection $\{1, 2, 3, 4, 5\}$ of five different integers, defines a set but the collection $\{1, 2, 2, 3, 4, 2, 5, 3\}$ of eight integers does not define a set as any two integers of this collection are not different.

A collection of well-defined distinct objects is said to form a set if the objects are independent of the order of their arrangements. Thus, the set of vowels of English alphabet can be represented as $\{a, e, i, o, u\}$ or $\{e, o, i, a, u\}$ or in any other order.

Subset : 

If $\,A\,$ and $\,B\,$ are two sets such that every element of $\,B\,$ is also an element of $\,A\,$, then $\,B\,$ is said to be a subset of $\,A\,$ and is written as $\,B \subseteq A\,$ [read as, $\,B\,$ is contained in A] or A 2 B [read as, A contains B].

Symbolically, B is a subset of $\,A\,$ [i.e., $\,B \subseteq A\,$], if $\,\,x\in B ⇒ x\in A\,$ [i.e., $\,x\,$ belongs to $\,B\,$ implies that $\,x\,$ belongs to $\,A\,$].

Equality of two sets :

Two sets $\,A\,$ and $\,B\,$ are said to be equal if every element of $\,B\,$ is also an element of $\,A\,$ and every element of $\,A\,$ is also an element of $\,B\,$  and written as $\,\,A = B\,$. Symbolically, $\,\,A = B\,\,$ if $\,B \subseteq A\,\,$ and $\,\,A \subseteq B\,\,$ i.e., if $\,x \in B⇒ x \in A\,\,$ and $\,y \in A⇒y\in B\,\,$. 

Universal set : 

In the discussion of set theory all the sets defined in a particular context are considered as subsets of a fixed set. This fixed set is called the universal set or universe of discourse and is denoted by $\,U\,$ or $\,S\,$. This fixed set may be a finite set as  well as an infinite set. 

Null set :

Suppose a set is defined stating some rules or properties. The set is called a null set if there is no element in the set satisfying the given rules. A null set is also known as empty set or void set and is denoted by the symbol or $\{\}$. The concept of null set is very important in set theory.

Example of a null set : $\,\,\phi=\{x: x\,\,\text{is an integer and}\,\,2<x<3\}$

Finite and Infinite sets : 

A set is said to be finite if the number of elements contained in the set can be counted and the counting process has an end. In other words, a set is said to be finite if it contains a specific number of elements. On the contrary, a set is said to be infinite if the number of elements contained in the set cannot be determined by counting; clearly, an infinite set does not contain a specific number of elements. 

(ii) Union or join of two sets :

The union or join of two given sets $\,A\,$ and $\,B\,$ is the set of all elements which belong either to $\,A\,$ or to $\,B\,$ or to both $\,A\,$  and $\,B\,$ and is denoted by $\,A \cup B\,$. Symbolically, $\,\,A \cup B=\{x: x \in A \vee x \in B\}$

Intersection or meet of two sets :

The intersection or meet of two given sets $\,A\,$ and $\,B\,$ is the set of all elements which are common to both $\,A\,$ and $\,B\,$. In other words, the intersection of two given sets $\,A\,$ and $\,B\,$ is the set of all elements which belong to both A and B and is denoted by $\,A \cap B\,$. Symbolically, $\,A \cap B=\{x: x \in A \wedge x \in B\}$

The difference of two sets :

The difference of two given sets $\,A\,$  and $\,B\,$  is the set of all elements of $\,A\,$  which do not belong to $\,B\,$  and is denoted by $\,A-B\,$ or, $\,A \sim B\,$. Symbolically, $\, A \sim B=\{x: x \in A \wedge x \not\in B\}$

(iii) In the discussion of set theory all the sets defined in a particular context are considered as subsets of a fixed set. This fixed set is called the universal set or universe of discourse and is denoted by $\,U\,$ or $\,S\,$. This fixed set may be a finite set as well as an infinite set. The concept of universal set is very important in set theory. See the following example:

EXAMPLES

Let $\,A = \{0, 1, 2, 3, 4\}; \,\,B = \{1, 3, 5, 7\};\\ C = \{1, 2, 3, 4, \cdots \}~~\text{and}~~ D \\= \{0,-1,-2, -3, -4, \cdots\}~$  be four given sets. Clearly, all the sets $\,A, B, C\,$ and $\,D\,$ are subsets of the set

$U = \{\cdots, -4,-3, -2,-1, 0, 1, 2, 3, 4, \cdots \}$.

Therefore, $\,U\,$ is the universal set of the sets $\,A, B, C\,$ and $\,D\,$ and it is an infinite set.

Again note that both $\,A\,$ and $\,B\,$ are subsets of the set $\,U = \{-1, 0, 1, 2, 3, 4, 5, 6, 7, 8\}$ [or, of the set $S = \{-2,-1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$]. 

Here $\,U\,$ (or $\,S\,$) is the universal set of the sets $\,A\,$ and $\,B\,$ and $\,U\,$ (or $\,S\,$) is a finite set.

(iv)  The three set operations (union, intersection and complementation).

Union and intersection property have been already discussed earlier. 

Complement of a set (or negation of a set) :

Complement of a set $\,A\,$ is defined with respect to its universal set $\,U\,$. It is the set of all elements of the universal set $\,U\,$ which do not belong to $\,A\,$, that is, the difference of the universal set $\,U\,$ and the set $\,A\,$

Since, the complement of a set $\,A\,$ is the set of elements not belonging to $\,A\,$, it is also called negation of $\,A\,$. We shall denote the complement of $\,A\,$ by $\,A'\,$ or $\,\bar{A}\,$ or $\,A^c\,$ or $\, \sim A.\,$ 

Symbolically, if $\,A^c\,$ is the complement of $\,A\,$, then $\,\,A^c=U-A=\{x: x\in U \wedge x \not\in A\}$

2. State De Morgan's laws of sets. Verify the laws in terms of Venn diagram. 

De Morgan's laws of sets : For any two sets $\,A\,$ and $\,B\,$ , we have $\,(i)\, (A \cup B)^c=A^c \cap B^c \\ (ii)\,(A \cap B)^c=A^c \cup B^c.$

For 2nd part, follow S.N. De Math book (refer : Pg-14)

3. State laws of algebra of sets :  For detailed explanation follow this . 

4. If $\,A\,$ is a finite set and contains $\,n\,$ elements , prove that the power set of $\,A\,$ has $\,2^n\,$ elements.

Sol. We know that power set of a given set $\,A\,$ is the set of all its subsets and is denoted by $\,P(A)\,.$ Symbolically, $\,\,P(A)=\{X | X \subseteq A\}.$

Now, coming back to the question, we see the null set $\,\phi\,$ is a subset of $\,A.$ 

No. of subsets of $\,A\,$ containing $\,1\,$ element $=^{n}C_1$,

No. of subsets of $\,A\,$ containing $\,2\,$ element $=^{n}C_2$,

No. of subsets of $\,A\,$ containing $\,3\,$ element $=^{n}C_3$ and so on. 

So, no. of subsets of $\,A\,$ containing $\,n\,$ element $=^{n}C_n$

Hence, the number of elements in the power set of $\,A\,$ i.e. , in $\,P(A)\,$ is : $\,\,=1+^{n}C_1+^{n}C_2+^{n}C_3+ \cdots +^{n}C_n\\=(1+1)^n\\=2^n$

5.  Let $\,\, A=\{a,b,c\}, \,\, B=\{a,b\},\,\, C=\{a,b,d\},\\ D=\{c,d\}\,\,\text{and}\,\, E=\{d\}.$ 

State which of the following statements are correct and give reasons:

(i) $\,\, B \subset A$ : This  statement is True (Correct) as by observing the given set , we note that all elements of $B$ belong to $A$  and there is an element $c \in A \,\,$ which does not belong to $B$ and so we can conclude that $B$ is a proper subset of $A.$

(ii) $\,\, D \not \supset E$ : This  statement is False (Incorrect)as by observing the given set , we note that all elements of $E$ belong to $D$ and there is an element $c \in D \,\,$ which does not belong to $E$ and so we can conclude that $E$ is a proper subset of $D.$

(iii) $\,\,D \subset B$ : This  statement is False (Incorrect)as by observing the given set , we note that none of elements of $D$ belong to $B$ and  $\,\, D \cap B=\phi,  \,\,$null set.

(iv) $\,\, \{a\} \in A : $ This  statement is True  (Correct)as by observing the given set , we note that $\,\, a \in A$.

6. Let $\,\, A=\{a,b,c,d,e,f,g,h,i\}, \,\, B=\{b,d,f,h\},\\ C=\{a,c,e,g,i\},D=\{c,d,e\} \\ \text{and}\,\, E=\{c,e\}.$ 

Which set can equal $\,X\,$ if we are given the following information?

(i) $\, X \,$ and $\,B\,$ are disjoint : For this statement to be true , it means none of the elements of $\,B\,$ belong to $\,X\,$ and that means  , $X$ must be a set which will be disjoint to the set $\,B\,$and from the given questions $\,\,C,E\,\,$ are two sets which are disjoint to $\,B\,$ and so $X=C \,\text{or}\, X=E.$

(ii) $\, X \subset A \,\,\text{but}\,\, X \not \subset C: $ From this statement , it is evident that there is an element in $\,\,X\,\,$ which does not belong to $\,C\,$ and from the given question $\,\,B,D \subset A\,\,$ are two sets which are not subset of $\,C\,$ and so $X=B \,\text{or}\, X=D.$

(iii)$\,\, X \subset D \,\,\text{but}\,\, X \not \subset B : $ For this statement to be true , we have to find the sets which are not subset of $\,\,B$  but that set will be subset of $\,\,D.\,\,$ From the given sets we see that $\,\,E\,\,$ satisfies both the conditions and so $\,\,X=E.$ 

(iv)$\,\, X \subset C \,\,\text{but}\,\, X \not \subset A : $ For this statement to be true , we have to find the sets which are not subset of $\,\,A$  but that set will be subset of $\,\,C.\,\,$ From the given sets we see that $\,\,C \subset A\,\,$ and so the given the condition can not be fulfilled. So, $\,X\,$ is not equal to any of the given sets. 

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