# Thread: Computing number of elements of finite union of finite sets.

1. ## Computing number of elements of finite union of finite sets.

Hi:
Let M:= {A_1, A_2, ..., A_m} be a set of a subsets of a given finite set G, and let E be a subset of G of only one element, that is |E|=1. If for any two elements of M their intersection is E, then the number of elements of the union over M, that is C:= A_1 U A_2 U ... U A_m, is |A_1| - 1 + |A_2| - 1 + ... + |A_m| -1 + 1 (that is, |C| = |A_1| + |A_2| + ... + |A_m| - m + 1).

This is so obvious to me that I can't find a way to prove it. For every i, A_i \ E has |A_i| - 1 elements. And the A_i \ E together with E form a partition of C. That is, U (A_i \ E) U E = C and for any i, j, A_i \ E and A_j \ E are disjoint sets. Hence |C| = |A_1 \ E| + ... + |A_m \ E| + |E| which agrees with the statement. OK. But does the fact that the A_i \ E together with E form a partition of C not need to be proved? This is my question then.

In fact what I need to prove is that given a finite group G and a collection of m subgroups of G, each having b elements, such that any two of them intersect trivially, then the number of elements of the union is m (b-1) + 1. But I think this is a particular case of the statement above.

EDIT: I think I've got it. That A_i \ E and A_j \ E are disjoint is trivial (too obvious). And U (A_i \ E) U E = U (A_i \ E) U E U E U ... U E (m times) = U ((A_i \ E) U E) (by the associativity of the union) = U (A_i U E) = U A_i (because E = A_i intersection A_j subset of A_i and then A_i U E = A_i) = C. Hence the A_i \ E together with E are indeed a partition of C.