Disjoint Subset Cardinality and Ordered Pairs: A Comprehensive Analysis

Introduction: In combinatorics, understanding the cardinality of sets is fundamental. This article explores a specific problem where the cardinality of a set [S] is determined to be 3^n. Here, [S] is the collection of all ordered pairs of disjoint subsets of the set {1, 2, ..., n}, and n is a non-negative integer.

The Core Question

The problem at hand is to establish why the cardinality of a set [S] is 3^n, where [S] is the collection of all ordered pairs of disjoint subsets of {1, 2, ..., n}. To solve this, we rely on a general result from combinatorics which posits that the number of functions from a finite set X to a finite set Y is Y^X.

Applying the General Result

We begin by assuming S {1, 2, ..., n}. To choose an ordered pair of disjoint subsets from S is equivalent to defining a function from S to the set A {1, 2, 3}. This equivalence is based on the following:

If we have a function f: S to A, then the subsets S_1 {s in S : f(s) 1} and S_2 {s in S : f(s) 2} are disjoint and constitute the ordered pair (S_1, S_2). By considering the partition of each element of S into three possibilities: it can either be in S_1 (corresponding to f(s) 1), in S_2 (corresponding to f(s) 2), or in neither set (corresponding to f(s) 3).

To apply the result, let X S n and Y A 3. This implies that the number of such functions, and hence the number of ordered pairs of disjoint subsets, is 3^n.

Alternative Counting Method

An equivalent method of counting the number of ordered pairs of disjoint subsets involves considering each element of S as having three choices: it can be in subset A, in subset B, or in neither. This gives a total of 3^n possibilities, aligning with our previous reasoning.

Removing the “Ordered” Condition

What happens if we remove the “ordered” condition, and instead consider unordered pairs of disjoint subsets? In this case, the partitions are now indistinguishable based on the label of the subsets. We need to count the number of unique partitions.

Unordered Pairs of Disjoint Subsets

Consider defining a function from S to a set B {0, 1, 2} where:

0 indicates the element is in neither subset. 1 indicates the element is in subset A. 2 indicates the element is in subset B.

The total number of such functions is 3^n, but because the order of the subsets does not matter, we need to consider the symmetry. Essentially, each unordered pair is counted 2 times in the ordered pairs (since (A, B) and (B, A) are the same but counted separately).

To correct for this, we divide the number of ordered pairs by 2, giving us:

[frac{3^n}{2}]

However, this simplistic division does not always work, as the exact count can depend on whether the set is empty or if the sets are indistinguishable.

Conclusion

This comprehensive analysis has shown that the cardinality of the set of all ordered pairs of disjoint subsets of a set {1, 2, ..., n} is 3^n. When the “ordered” condition is removed, the counting method must account for the indistinguishability of the subsets, leading to a more complex partitioning process.

References

For further exploration and detailed proofs, refer to the following resources:

Combinatorics and Graph Theory by John M. Harris, Jeffry L. Hirst, and Michael J. Mossinghoff. Discrete Mathematics and Its Applications by Kenneth H. Rosen.