Important theorem of algebra of a set theory for BSc 1st semester.

An important theorem of the algebra of a set theory for BSc 1st semester and 1st paper.

In this article, I have provided notes on an important theorem of the algebra of a set theory for BSc 1st semester and 1st paper according to the DDU syllabus. It is also profitable for class 11th students, Jee mains, and Jee advance aspirants. I provide free handwritten notes for set theory and related important theorems based on the algebra of set theory. I say it more confidently because these notes are trusted by thousands of aspirants every year. If you are a BSc mathematics, Class 11th student, or Jee mains aspirant, you can follow us to boost your preparation levels.

Dear student, do you search for important questions of set theory based on theorem? Do you search for De-Morgan's Laws and related questions? then you are at the right place. Here, I provided all of the above topics in the easiest way. 

I have provided complete handwritten notes on the theorem of the algebra of set theory that will help in fast-track revision. I hope all concepts, derivations and solutions to all questions clear your doubt regarding the set theory. If you have any relevant doubts regarding the given topics and questions. You should immediately consult with our expert teams.

First, let's derive the theorem of the set theory which provides a more convenient way to understand the related questions and the previous year's questions. 

All these topics are based on the New syllabus of DDU BSC 1st semester and first paper.

Let's understand all concepts one by one.

Some theorem on Subsets

Theorem 1: Every set is a subset of itself.

Proof: 

Let A be any set. 

then, each element or component of A is contained in A itself.

⇒A ⊆ A

For Example A = { 1, 2, 3, 4 }

⇒A ⊆ A.

Theorem 2: The empty set is a subset of every set.

Proof:

Let A be any set and ф be the empty set. 

To show that ф ⊆ A, we must show that every element of ф is an element of A or contained in A.

But, ф does not contain any element.

So, we can say that every element or component of ф is contained in A.

Hence, ф ⊆ A.

Theorem 3: The total number of subsets of a finite set containing n element is 2ⁿ.

Proof:

Let A be a finite set containing n elements. Let 0 ≤ r ≤ n.

Consider those subsets of A that have r elements each.

We know that the number of ways, r elements can be chosen out of n elements is ⁿC_r.

Therefore, the number of subsets having r elements each is ⁿC_r. 

Hence, the total number of subsets of A is

ⁿC₀ + ⁿC₁ + ⁿC₂ + ... + ⁿC_r = (1 + 1)ⁿ = 2ⁿ 

[using binomial theorem]

The Law of Algebra of Sets Theory.

Theorem 1: (Idempotent Laws) For any set A, (1) A⋃A = A   (2) A ⋂A = A.

Proof:

(1) A⋃A = A 

⇒A⋃A = { x : x ∊ A or x ∊ A} = { x : x ∊  A} = A.

For example:

if A = { 1, 2, 3 }

⇒A⋃A = { 1, 2, 3 } = A.

(2) A⋂A = A

⇒A⋂A = { x : x ∊ A and x ∊ A } = { x : x ∊ A} = A.

For example:

if A = { 1, 2, 3 }

⇒A⋂A = { 1, 2, 3 } = A.

Theorem 2: (Identity Laws) For any A, (1) A⋃ ф = A   (2) A ⋂ U = A, i.e. ф and U are identity elements for union and intersection respectively.

Proof:

(1) A⋃ф = A

⇒A⋃ф = {x: x ∊ or x ∊ ф} = {x: x ∊ A} = A.

we know that ф ⊆ A. ф is the empty set. A⋃ф = A

(2) A⋂U = A

⇒A ⋂U = {x: x ∊ A and x ∊ U} = {x: x ∊ A} = A.

For example:

If U = {1, 2, 3, 4, 5, 6, 7} and A = {1, 2, 3, 4, 5}

⇒A⋂U = {1, 2, 3, 4, 5} = A.

Theorem 3: (Commutative Laws) for any two sets A and B.  (1) A⋃B = B⋃A    (2)A⋂B = B⋂A. i.e. union and intersection are commutative.

Proof:

Let's two sets A and B are equal iff A ⊆ B and B ⊆ A. Also, A ⊆ B if every element of A belongs to B.

 

(1) A⋃B = B⋃A.

Let x be an arbitrary element of A⋃B. Then,

⇒x ∊ A⋃B

⇒x ∊ A or x ∊ B

⇒x ∊ B or x ∊ A

⇒x ∊ B⋃A

∴ A⋃B ⊆ B⋃A.

Similarly,

let y be an arbitrary element of B⋃A. Then,

⇒y ∊ B⋃A

⇒y ∊ B or y ∊ A

⇒y ∊ A or y ∊ B

⇒y ∊ A⋃B

∴ B⋃A ⊆ A⋃B.

Hence, A⋃B = B⋃A.

(2) A⋂B = B⋂A.

Let x be an arbitrary element of A⋂B. Then,

⇒x ∊ A⋂B

⇒x ∊ A or x ∊ B

⇒x ∊ B or x ∊ A

⇒x ∊ B⋂A

∴ A⋂B ⊆ B⋂A.

Similarly,

Let y be an arbitrary element of B⋂A. Then,

⇒y ∊ B⋂A

⇒y ∊ B or y ∊ A

⇒y ∊ A or y ∊ B

⇒y ∊ A⋂B

∴ B⋂A ⊆ A⋂B.

Hence,

A⋂B = B⋂A.

Theorem 4: (Associate Laws) If A, B and C are any three sets, then (1) (A⋃B)⋃C = A⋃(B⋃C)    (2) (A⋂B)⋂C = A⋂(B⋂C). i.e. union and intersection are associative.

Proof:

(1) (A⋃B)⋃C = A⋃(B⋃C)

Let x be an arbitrary element of (A⋃B)⋃C. Then,

⇒x∊ (A⋃B)⋃C

⇒x∊ (A⋃B) or x ∊ C

⇒(x ∊ A or x ∊ B) or x ∊ C

⇒x∊ A or  (x ∊ B or x ∊ C)

⇒x∊ A or x ∊ (B⋃C)

⇒x ∊ A⋃(B⋃C)

∴(A⋃B)⋃C ⊆ A⋃(B⋃C)

Similarly,

let y be an arbitrary element of A⋃(B⋃C). Then,

⇒y∊ A⋃(B⋃C)

⇒y∊ A or y ∊ (B⋃C)

⇒y ∊ A or (y ∊ B or y ∊ C)

⇒(y ∊ A or  y ∊ B) or y ∊ C

⇒(y∊ A or y ∊ B)⋃C

⇒y ∊ (A⋃B)⋃C

∴A⋃(B⋃C) ⊆ (A⋃B)⋃C

Hence, 

(A⋃B)⋃C = A⋃(B⋃C).

(2) (A⋂B)⋂C = A⋂(B⋂C)

Let x be an arbitrary element of (A⋂B)⋂C. Then,

⇒x∊ (A⋂B)⋂C

⇒x∊ (A⋂B) or x ∊ C

⇒(x ∊ A or x ∊ B) or x ∊ C

⇒x∊ A or  (x ∊ B or x ∊ C)

⇒x∊ A or x ∊ (B⋂C)

⇒x ∊ A⋂(B⋂C)

∴(A⋂B)⋂C ⊆ A⋂(B⋂C)

Similarly,

let y be an arbitrary element of A⋂(B⋂C). Then,

⇒y∊ A⋂(B⋂C)

⇒y∊ A or y ∊ (B⋂C)

⇒y ∊ A or (y ∊ B or y ∊ C)

⇒(y ∊ A or  y ∊ B) or y ∊ C

⇒(y ∊ A or y ∊ B)⋂C

⇒y ∊ (A⋂B)⋂C

∴A⋂(B⋂C) ⊆ (A⋂B)⋂C 

Hence, 

(A⋂B)⋂C = A⋂(B⋂C).

Theorem 5: (Distributive Laws) If A, B, and C are any three sets, then (1) A⋃(B⋂C) = (A⋃B)⋂(A⋃C)    (2) A⋂(B⋃C) = (A⋂B)⋃(A⋂C). i.e. union and intersection are distributive over intersection and union respectively.

Proof:

(1) A⋃(B⋂C) = (A⋃B)⋂(A⋃C)

Let x be an arbitrary element of A⋃(B⋂C). Then,

⇒x ∊ A⋃(B⋂C)

⇒x ∊ A or x ∊ (B⋂C)

⇒x ∊ A or (x ∊ B and x ∊ C)

⇒(x ∊ A or x ∊ B) and (x ∊ A or x ∊ C)

⇒x ∊ (A⋃B) and x ∊ (A⋃C)

⇒x ∊ (A⋃B)⋂(A⋃C)

∴ A⋃(B⋂C)⊆ (A⋃B)⋂(A⋃C)

Similarly,

let y be an arbitrary element of (A⋃B)⋂(A⋃C). Then,

⇒y ∊ (A⋃B)⋂(A⋃C)

⇒y ∊ (A⋃B) and y ∊ (A⋃C)

⇒(y ∊ A or y ∊ B) and (y ∊ A or y ∊ C)

⇒y ∊ A or (y ∊ B and y ∊ C)

⇒y ∊ A or y ∊ (B⋂C)

⇒y ∊ A⋃(B⋂C)

∴(A⋃B)⋂(A⋃C)⊆ A⋃(B⋂C)

Hence, A⋃(B⋂C) = (A⋃B)⋂(A⋃C).

(2) A⋂(B⋃C) = (A⋂B)⋃(A⋂C)

Let x be an arbitrary element of A⋂(B⋃C). Then,

⇒x ∊ A⋂(B⋃C)

⇒x ∊ A and x ∊ (B⋃C)

⇒x ∊ A and (x ∊ B or x ∊ C)

⇒(x ∊ A and x ∊ B) or (x ∊ A and x ∊ C)

⇒x ∊ (A⋂B) or x ∊ (A⋂C)

⇒x ∊ (A⋂B)⋃(A⋂C)

∴ A⋂(B⋃C)⊆ (A⋂B)⋃(A⋂C)

Similarly,

let y be an arbitrary element of (A⋂B)⋃(A⋂C). Then,

⇒y ∊ (A⋂B)⋃(A⋂C)

⇒y ∊ (A⋂B) or y ∊ (A⋂C) 

⇒(y ∊ A and y ∊ B) or (y ∊ A and y ∊ C)

⇒y ∊ A and (y ∊ B or y ∊ C)

⇒y ∊ A and y ∊ (B⋃C)

⇒y ∊ A⋂(B⋃C)

∴(A⋂B)⋃(A⋂C) ⊆ A⋂(B⋃C)

Hence, A⋂(B⋃C) = (A⋂B)⋃(A⋂C).

Theorem 6: (De Morgan's Laws) If A and B are any two sets, then (1) (A ⋃ B)' = A'⋂ B'      (2) (A ⋂ B)' = A'⋃ B'.

[VERY IMPORTANT]

Proof:

(1) (A⋃B)' = A'⋂ B'

Let x be an arbitrary element of (A ⋃ B)'. Then,

⇒x ∊ (A ⋃ B)'

⇒x ∉ (A ⋃ B)

⇒x ∉ A and x ∉ B

⇒x ∊ A' and x ∊  B'

⇒x ∊ A' ⋂ B'

∴ (A⋃B)' ⊆ A'⋂ B'

Similarly,

let y be an arbitrary element of A'⋂ B'. Then,

⇒y ∊ A'⋂ B'

⇒y ∊ A' and y ∊ B'

⇒y ∉ A and y ∉ B

⇒y ∉ (A ⋃ B)

⇒y ∊ (A⋃ B)'

∴ A'⋂ B'  ⊆ (A⋃B)'

Hence, (A⋃B)' = A'⋂ B'

(2) (A ⋂ B)' = A'⋃ B'.

Let x be an arbitrary element of (A⋂B)'. Then,

⇒x ∊ (A⋂B)'

⇒x ∉ (A⋂B)

⇒x ∉ A or x ∉ B

⇒x ∊ A' or x ∊ B'

⇒x ∊ A' ⋃ B'

∴ (A ⋂ B)'⊆ A'⋃ B'.

Similarly,

let y be an arbitrary element of A'⋃ B'. Then,

⇒y∊ A'⋃ B'

⇒y ∊ A' or y ∊ B'

⇒y ∉A or y∉ B

⇒y ∉ (A⋂B)

⇒y ∊ (A⋂B)'

∴ A'⋃ B' ⊆ (A ⋂ B)'

Hence, (A ⋂ B)' = A'⋃ B'.

More Important Results on Operations on Sets

Theorem 1: If A and B are any two sets, then

(1) A - B = A⋂B' 
(2) B - A = B⋂A' 
(3) A - B = A ⇔ A⋂B = ф 
(4) (A - B)⋃B = A⋃B 
(5) (A - B)⋂B = ф 
(6) A ⊆ B ⇔ B' ⊆ A' 
(7) (A - B)⋃(B - A) = (A⋃B) - (A⋂B)

Proof:

(1) A - B = A⋂B'

Let x be an arbitrary element of A - B. Then,

⇒x ∊ (A - B)

⇒x ∊ A and x ∉ B

⇒x ∊ A and x ∊ B'

⇒x ∊ A ⋂ B'

ஃ A - B  ⊆ A⋂B'

Similarly,

let y be an arbitrary element of A⋂B'. Then,

⇒y ∊ A⋂B'

⇒y ∊ A and y ∊ B'

⇒y ∊ A and y ∉ B

⇒y ∊ (A - B)

ஃ A⋂B' ⊆ A - B

Hence, A - B = A⋂B'

(2) B - A = B⋂A'

Let x be an arbitrary element of B - A. Then,

⇒x ∊ (B - A)

⇒x ∊ B and x ∉ A

⇒x ∊ B and x ∊ A'

⇒x ∊ B ⋂ A'

ஃ B - A  ⊆ B ⋂ A'

Similarly,

let y be an arbitrary element of B⋂A'. Then,

⇒y ∊ B⋂A'

⇒y ∊ B and y ∊ A'

⇒y ∊ B and y ∉ A

⇒y ∊ (B - A)

ஃ B⋂A' ⊆ B - A

Hence, B - A = B⋂A'

(3) A - B = A ⇔ A⋂B = ф

There are two prospective ways to prove the above operations.

1. A - B = A ⇒A⋂B = ф
2. A⋂B = ф = A - B = A.

(1) A - B = A ⇒A ⋂ B = ф

 Let A - B = A. Then we have to prove that A ⋂ B = ф. 

If it is possible, let A ⋂ B ≠ ф. then,
⇒A ⋂ B ≠ ф

⇒There exists x ∊ A⋂B

⇒x ∊ A and x ∊ B

⇒x ∉ A - B and x ∊ B

⇒(x ∊ and x ∉  B) and x ∊ B                                                                            [∵ A - B =A]

⇒x ∊ A and (x ∉  B and x ∊ B)                                                                         [By definition of A - B]

But x ∉ B and x ∊ B both can never be possible simultaneously. 

Thus, we reach a contradiction. So, our supposition is wrong. 

Therefore, A ⋂ B = ф.

Hence,

⇒A - B = A ⇒A⋂B = ф

2. A ⋂ B = ф = A - B = A.

Conversely,  let A⋂B = ф. Then we have to prove that A - B = A.

For this, we shall show that

⇒A - B ⊆ A and A ⊆ A - B.

Let x be an arbitrary element of  A - B. Then,

⇒x ∊ A - B 

⇒x ∊ A and x ∉ B

⇒x ∊ A

∴ A - B ⊆ A.

Similarly,

let y be an arbitrary element of A. Then,

⇒y ∊ A 

⇒y ∊ A and y ∉ B                                                                                           [∵ A ⋂ B = ф]

⇒y ∊ A - B

⇒A ⊆ A - B.                                                                                                    [By definition of A - B]

∴ A ⊆ A - B

So, we have A - B ⊆ A and  A ⊆ A - B. 

∴ A - B = A.

Thus,

A⋂B = ф = A - B = A.

(4) (A - B)⋃B = A⋃B

Let x be an arbitrary element of (A - B)⋃B. Then,

⇒x ∊ (A - B)⋃B

⇒x ∊ (A - B) or x ∊ B

⇒x ∊ A and x ∉ B) or x ∊ B

⇒(x ∊ A or x ∊ B) and (x ∉ B or x ∊ B)

⇒x ∊ A⋃B

ஃ(A - B)⋃B ⊆ A⋃B

Similarly,

Let y be an arbitrary element of A⋃B. Then,

⇒y ∊ A⋃B

⇒y ∊ A or y ∊ B

⇒(y ∊ A or y ∊ B) and (y ∉ A or y ∊ B)

⇒(y ∊ A and y ∉ B) or y ∊ B

⇒y ∊ (A - B)⋃B

⇒A⋃B (A - B)⋃B

ஃ A⋃B ⊆ (A - B)⋃B

Hence, (A - B)⋃B = A⋃B

(5) (A - B)⋂B = ф

If possible let (A - B)⋂B ≠ ф. Then,

at least one element x exists in (A - B)⋂B.

⇒x ∊ (A - B)⋂B 

⇒x ∊ (A - B) and x ∊ B

⇒(x ∊ A and x ∉ B) and x ∊ B

⇒x ∊ A and (x ∉ B and x ∊ B)

But, x ∉ B and x ∊ B can never be possible simultaneously. 

Thus, we conclude with a contradiction.

So, our supposition is wrong.

Hence,

(A - B)⋂B = ф. 

(6) A ⊆ B ⇔ B' ⊆ A'

Let A ⊆ B. Then we have to prove that B' ⊆ A'

Let x be an arbitrary element of B'.

Then,

⇒x ∊ B'

⇒x ∉ B

⇒x ∉ A                                                                                                                            [∵ A ⊆ B]

⇒x ∊ A'

ஃ B' ⊆ A'

Thus,

⇒A⊆B 

⇒B'⊆A'

Conversely, let B'⊆A'. Then, we have to prove that A⊆B. 

Let y be an arbitrary element of A.

Then,

⇒y ∊ A

⇒y ∉ A'

⇒y ∉ B'                                                                                                                            [∵B'⊆A']

⇒y ∊ B

ஃ A'⊆B

Thus,

⇒B'⊆A'

⇒A⊆B

Hence,

A ⊆ B ⇔ B' ⊆ A'

(7) (A - B)⋃(B - A) = (A⋃B) - (A⋂B)

Let x be an arbitrary element of (A - B)⋃(B - A). Then,

⇒x ∊ (A - B)⋃(B - A)

⇒x ∊ A - B or x ∊ B - A

⇒(x ∊ A and x ∉ B) or ( x B and x ∉ A)

⇒(x ∊ A or x ∉ B) and (x ∉ B or x ∉ A)

⇒x ∊ (AB) and x ∉ (AB)

⇒x ∊ (A B) - (AB)

ஃ (A - B)⋃(B - A)⊆(A⋃B) - (A⋂B).

Similarly,

let y be an arbitrary element of (A⋃B) - (A⋂B). Then,

⇒y ∊ (A⋃B) - (A⋂B)

⇒y ∊ (A⋃B)  and y (A⋂B)

⇒(y ∊ A or y ∊ B) and (y ∉ A or y ∉ B)

⇒(y ∊ A and y ∉ B) or (y ∊ B and y ∉ A)

⇒y ∊ (A - B) or y ∊ (B - A)

⇒y ∊ (A - B)⋃(B - A).

ஃ (A⋃B) - (A⋂B)⊆(A - B)⋃(B - A).

Hence,

(A - B)⋃(B - A) = (A⋃B) - (A⋂B)

Theorem 2: If A, B, and C are any three sets, then prove that:

1. A - (B⋂C) = (A - B)⋃(A - C)
2. A - (B⋃C) = (A - B)⋂(A - C)
3. A⋂(B-C) = (A⋂B) - (A⋂C)
4. A⋂(B∆C) = (A⋂B)∆(A⋂C)

Proof:

1. A - (B⋂C) = (A - B)⋃(A - C)

Let x be any element of A - (B⋂C). Then,

⇒x ∊ A - (B⋂C)

⇒x ∊ A and x ∉ (B⋂C)

⇒x ∊ A and (x ∉ B or x ∉ C)

⇒(x ∊ A and x ∉ B) or (x ∊ A and x ∉ C)

⇒x ∊ (A - B) or x ∊ (A - C)

⇒x ∊ (A - B)⋃(A - C)

ஃ A - (B⋂C) ⊆ (A - B)⋃(A - C)

Similarly,

let y be any element of (A - B)⋃(A - C). Then,

⇒y ∊ (A - B)⋃(A - C)

⇒y ∊ (A - B) or y ∊ (A - C)

⇒(y ∊ A and y ∉ B) or (y ∊ A and y ∉ C)

⇒y ∊ A and (y ∉ B or y ∉ C)

⇒y ∊ A and y ∉ (B⋂C)

⇒y ∊ A - (B⋂C)

ஃ (A - B)⋃(A - C) ⊆ A - (B⋂C)

Hence,

A - (B⋂C) = (A - B)⋃(A - C)

2. A - (B⋃C) = (A - B)⋂(A - C)

Let x be any arbitrary element of A - (B⋃C). Then,

⇒x ∊ A - (B⋃C)

⇒x ∊ A and x ∉ (B⋃C)

⇒x ∊ A and (x ∉ B and x ∉ C)

⇒x ∊ A and x ∉ B) and (x ∊ A and x ∉ C)

⇒x ∊ (A - B) and x ∊ (A - C)

⇒x ∊ (A - B)⋂(A - C)

ஃ A - (B⋃C) ⊆ (A - B)⋂(A - C)

Similarly,

Let y be any arbitrary element of (A - B)⋂(A - C). Then,

⇒y ∊ (A - B)⋂(A - C)

⇒y ∊ (A - B) and y ∊ (A - C)

⇒y ∊ A and y ∉ B) and (y ∊ A and y ∉ C)

⇒y ∊ A and (y ∉ B and y ∉ C)

⇒y ∊ A and y ∉ (B⋃C)

⇒y ∊ A - (B⋃C)

ஃ (A - B)⋂(A - C) ⊆ A - (B⋃C)

Hence,

(A - B)⋂(A - C) = A - (B⋃C)

3. A⋂(B-C) = (A⋂B) - (A⋂C)

Let x be any arbitrary element of A⋂(B-C). Then,

⇒x ∊ A⋂(B-C)

⇒x ∊ A and x ∊ (B-C)

⇒x ∊ A and (x ∊ B and x ∉ C)

⇒(x ∊ A and x ∊ B) and (x ∊ A and x ∉ C)

⇒x ∊ (A⋂B)  and x ∉ (A⋂C)

⇒x ∊ (A⋂B) - (A⋂C)

ஃ  A⋂(B-C) ⊆ (A⋂B) - (A⋂C)

Similarly,

let y be any arbitrary element of (A⋂B) - (A⋂C). Then,

⇒y∊ (A⋂B) - (A⋂C)

⇒y ∊ (A⋂B)  and y ∉ (A⋂C)

⇒(y ∊ A and y ∊ B) and (y ∊ A and y ∉ C)

⇒y ∊ A and (y ∊ B and y ∉ C)

⇒y ∊ A and y ∊ (B-C)

⇒y ∊ A⋂(B-C)

ஃ (A⋂B) - (A⋂C) ⊆ A⋂(B-C)

Hence,

A⋂(B-C) = (A⋂B) - (A⋂C)

4. A⋂(B∆C) = (A⋂B)∆(A⋂C)

⇒A⋂(B∆C) = [A⋂(B-C)]⋃[A⋂(C-B)]                                                       [by distributive law]

          =[(A⋂B)-(A⋂C)]⋃[A⋂C)-(A⋂B)]

         = (A⋂B)Δ(A⋂C)