Proof by Indirect Method for the Union of n Sets Having at Most n Elements

In this article, we will delve into a proof by indirect method for a fascinating combinatorial problem in set theory. We will explore how among n sets each containing n-1 elements, any n-1 of these sets have at least one common element but the intersection of all n sets is empty. We aim to show that if the sets are distinct, the union of these sets contains at most n elements.

Understanding the Problem

Let's denote the given sets as G-sets. We consider their union U as a size-m set of values [1, m]. For any value v in [1, m], there must be a G-set that does not contain v. Additionally, for any G-set, there is some value that is an element of all the other n-1 G-sets.

Initial Assumptions and Setup

We will begin by assuming that m n. This means that the union U has more elements than the number of G-sets. For each G-set, we can define its complement in U, which we will call its O-set. Each O-set has m - (n-1) m - n 1 elements.

Intersections and Complements

The intersection of a G-set with another subset of U will eliminate the values that are in the O-set of the first G-set. This means that for each G-set, there must be a unique value that is not in any of the other n-1 O-sets. These values must be distinct and there must be exactly n of them. The remaining m - n 1 - n m - n - 1 values must be in all the O-sets and thus not in any of the G-sets.

Contradiction and Conclusion

However, if m n, the union of the G-sets would have to contain more values than there are distinct G-sets, which is a contradiction. Therefore, we must have m n.

Alternative Proof Using Complements

Another way to approach this problem is to consider the sets S_1, S_2, dots, S_n. For each j from 2 to n, there exists an element x_j in cap_{i eq j} S_i but x_j otin S_j. This element is unique to the set of all S_i except j. Therefore, S_1 must contain the remaining n-1 elements: x_2, x_3, dots, x_n. By similar logic, we can show that for each j geq 2, the set S_j contains exactly the elements x_1, x_2, dots, x_n with x_j removed.

Thus, the union of these sets is exactly the set {x_1, x_2, dots, x_n}, which has only n elements. This proves that the union of the sets contains at most n elements.

Conclusion

By both the indirect proof method and the alternative proof using complements, we have demonstrated that among n sets each containing n-1 elements, any n-1 of these sets have at least one common element but the intersection of all n sets is empty, and the union of these sets contains at most n elements.

This result highlights the intricate relationship between set intersections and unions in combinatorial set theory and provides a valuable tool for understanding the structure of such set systems.