SET
A well defined collection of distinct
objects is called a set.
When we say, ‘well defined’, we mean that there must be given a rule or rules with the help of which we should
readily be able to say that whether a particular object is a member of the set or is not a member of the set. The sets are generally denoted by capital
letters A,B,C,….X,Y,Z . The members of a set are called its elements. The elements of a set are denoted by small letters
a, b, c,..…,x, y, z. If an element a belong to a set A than
we
write a ∈
A and if a does
not belong to set A then we write
a ∉ A.
LAW OF ALGEBRA OF SETS
1.
Idempotent Laws : For any set A,
(i)
A
∪ A
= A
(ii)
A
∩ A
= A
.
2.
Identity
Laws : For any set A,
(i)
A
∪ö = A
(ii)
A
∩U = A
i.e., ö and are identity elements
for union and intersection respectively.
3.
Commutative Laws : For any two sets A and B,
(i) A
∪ B
= B ∪ A
(ii)
A
∩ B
= B ∩ A
i.e. union and intersection are commutative.
4.
Associative Laws : If A,
B and C are any three sets, then
(i) ( A ∪ B) ∪ C = A ∪ ( B ∪ C )
i.e. union and intersection
are associative.
(ii) A ∩ ( B ∩ C ) = ( A
∩ B) ∩ C
5.
Distributive Laws : If A, B and
C are any three sets, the
(i) A ∪ ( B ∩ C ) = ( A ∪ B) ∩ ( A ∪ C
)
(ii) A ∩ ( B ∪ C ) = ( A ∩ B) ∪ ( A ∩ C )
i.e. union and intersection
are distributive over intersection and union respectively.
6.
De-Morgan’s Laws : If A
and B are any two
sets, then
(i)
( A ∪ B)′ = A′ ∩ B
(ii) ( A ∩ B)′ = A′ ∪ B′
7.
More results on operations on sets
If A
and B are any two
sets, then
(i)
A
− B
= A
∩ B′
(ii) B − A = B ∩ A′
(iii) A − B = A ⇔ A ∩ B = ö
(iv) ( A − B) ∪ B = A ∪ B
(v) ( A − B) ∩ B = ö
(vi) A ⊆ B ⇔ B′ ⊆ A′
(vii) ( A − B) ∪ ( B − A) = ( A ∪ B) − ( A ∩ B) .
If
A, B and C are any three sets, then
(i)A − ( B ∩ C ) = ( A − B) ∪ ( A − C )
(ii)A − ( B ∪ C ) = ( A − B) ∩ ( A − C )
(iii)A ∩ ( B − C ) = ( A ∩ B) − ( A ∩ C )
(iv)A ∩ ( B ∆ C ) = ( A ∩ B) ∆ ( A ∩ C )
8. Important
results on number of elements
in sets
If
A, B and C are finite sets, and U be the finite
universal set, then
(i)n ( A ∪ B) = n ( A) + n ( B) − n ( A ∩ B)
(ii)n
( A
∩ B) = n ( A) + n ( B) ⇔ A, B
are disjoint non