Show that if A B and C are sets then A union B union C A B

Show that if A, B, and C are sets, then |A union B union C| = |A| + |B| + |C|-|A intersection B|-|A intersection C|-|B intersection C| + |A intersection B intersection C|.

Solution

Let A and B be finite sets. If A B = , then
(i) n ( A B ) = n ( A ) + n ( B ) ... (1)
The elements in A B are either in A or in B but not in both as A B = . So, (1) follows immediately.
In general, if A and B are finite sets, then
(ii) n ( A B ) = n ( A ) + n ( B ) – n ( A B ) ... (2)
Note that the sets A – B, A B and B – A are disjoint and their union is A B (Fig 1.11). Therefore
n ( A B) = n ( A – B) + n ( A B ) + n ( B – A )
= n ( A – B) + n ( A B ) + n ( B – A ) + n ( A B ) – n ( A B)
= n ( A ) + n ( B ) – n ( A B), which verifies (2)
(iii) If A, B and C are finite sets, then
n ( A B C ) = n ( A ) + n ( B ) + n ( C ) – n ( A B ) – n ( B C)
– n ( A C ) + n ( A B C ) ... (3)
In fact, we have n ( A B C ) = n (A) + n ( B C ) – n [ A ( B C ) ] [ by (2) ]
= n (A) + n ( B ) + n ( C ) – n ( B C ) – n [ A ( B C ) ] [ by (2) ]
Since A ( B C ) = ( A B ) ( A C ), we get
n [ A ( B C ) ] = n ( A B ) + n ( A C ) – n [ ( A B ) (A C)]
= n ( A B ) + n ( A C ) – n (A B C)
Therefore
n ( A B C ) = n (A) + n ( B ) + n ( C ) – n ( A B ) – n ( B C)
– n ( A C ) + n ( A B C )
This proves (3).