Complement of a set: The complement of the set A denoted by Ac or |A̅| or A’ is formed by removing each element of A from the universal set .Thus is μ is the universal set,
Ac = μ – A
Ac = {x | x ∉ A}
By definition A ∩ Ac = φ
A∩Ac = μ
|Ac|= |μ|-|A|
|Ac|+ |A|= |μ|
Prove that: A-B = A ∩ Bc
A-B ={x |x ϵ A and x ∉ B}
= {x |x ϵ A and x ∉ Bc}
= A ∩ Bc
B-A = B ∩ Ac (imp results)
A-B = A ∩ Bc
De Morgan’s Laws: For any two sets A, B
(i) (A U B)c = Ac ∩ Bc
(ii) (A ∩ B)c = Ac U Bc
(iii) A – (B U C) = (A – B) ∩ (A – C)
(iv) A – (B ∩ C) = (A – B) U (A – C)
(v) |Ac|c = A
Given two non-empty sets A, B which are not disjoint then represents the following sets in venn diagram.
(i) Ac
(ii)(A ∩ B)c = Ac U Bc
(iii) Ac ∩ Bc = (A U B)c
Note: |A U B U C|= |A ∩ Bc ∩ Cc |+ | Ac ∩ B ∩ Cc| + Ac ∩ Bc ∩ C) + |A U B U Cc | + |A ∩ Bc ∩ Cc |+ | Ac ∩ B ∩ C| + |A ∩ B ∩ C|
For any 3 set A, B, C
|A U B U C|= |A|+ |B|+|C|- |A∩B|- |B∩C|-|A∩C|+|A∩B∩C|
A, B, C, D are any 4 sets
|A U B U C U D|= | (A U B) U (C U D)|
=|A U B|+ |C U D|- | (A U B) ∩ (C U D)|
= |A|+ |B| – |A∩B|+ |C|+|D|- |C ∩ D|- |A ∩ D| – |B ∩ C|- |A ∩ B ∩ C|+|A ∩ B ∩ D|+ |A ∩ C ∩ D|+|B ∩ C ∩ D|- |B ∩ D|- |A ∩ C|- |A ∩ B ∩ C ∩ D|
Note: |A1 U A2 U A3 — U An|
\(\sum\limits_{n=1}^{i}{\left| Ai \right|-\sum\limits_{i\le j\le k}{\left| Ai\cap Aj \right|+}\sum\limits_{i\le j\le n}{\left| Ai\cap Aj\cap Ak \right|}+{{\left( -1 \right)}^{n-1}}\left| {{A}_{1}}{{A}_{2}}{{A}_{3}}—-{{A}_{n}} \right|}\).
Principle of inclusion and exclusion:-
|AC1 ∩ AC2 ∩ — ACn|= |μ|- |A1 U A2 U — U An|