91Ó°ÊÓ

The union of sets is a fundamental concept in set theory. It combines all elements from the given sets, avoiding duplicates. If we take three sets: \(A_{1}\), \(A_{2}\), and \(A_{3}\), the union, written as \(A_{1} \cup A_{2} \cup A_{3}\), contains every element that is in \(A_{1}\), \(A_{2}\), or \(A_{3}\). For example, if:

• \(|A_{1}| = 100\)
• \(|A_{2}| = 1000\)
• \(|A_{3}| = 10,000\)

The process of finding the union depends on the relationships between these sets.
In part (a) of the exercise, we see that \(A_{1} \subseteq A_{2}\) and \(A_{2} \subseteq A_{3}\). This means every element in \(A_{1}\) is in \(A_{2}\), and every element in \(A_{2}\) is in \(A_{3}\). So, the union of these sets, \(A_{1} \cup A_{2} \cup A_{3}\), is simply the largest set in the hierarchy, which is \(A_{3}\), containing 10,000 elements.
In part (b), the sets are disjoint, meaning they do not share any elements. So, to find the union, we add up the elements in all sets: \(100 + 1000 + 10,000 = 11,100\).
In part (c), we apply more complex set operations using the inclusion-exclusion principle.

The inclusion-exclusion principle is a technique to calculate the size of the union of overlapping sets. This principle helps us avoid counting elements multiple times when they appear in more than one set.
For three sets, the formula is:

\(|A_{1} \cup A_{2} \cup A_{3}| = |A_{1}| + |A_{2}| + |A_{3}| - |A_{1} \cap A_{2}| - |A_{1} \cap A_{3}| - |A_{2} \cap A_{3}| + |A_{1} \cap A_{2} \cap A_{3}|\)

In part (c) of the exercise, we know:

• 
  \(|A_{1} \cap A_{2}| = 2\)
  
• 
  \(|A_{1} \cap A_{3}| = 2\)
  
• 
  \(|A_{2} \cap A_{3}| = 2\)

and

• The intersection of all three sets \(|A_{1} \cap A_{2} \cap A_{3}| = 1\).
  

So, we use the inclusion-exclusion principle to calculate the union:

\(|A_{1} \cup A_{2} \cup A_{3}| = 100 + 1000 + 10,000 - 2 - 2 - 2 + 1 = 11,095\).This result accounts for all the overlapping elements correctly.

Disjoint sets are sets that have no elements in common. If sets \(A_{1}\), \(A_{2}\), and \(A_{3}\) are disjoint, their intersections would be empty, implying:

• \(|A_{1} \cap A_{2}| = 0\)
  
• \(|A_{1} \cap A_{3}| = 0\)
  
• \(|A_{2} \cap A_{3}| = 0\)
  
• \(|A_{1} \cap A_{2} \cap A_{3}| = 0\)

n
This scenario is depicted in part (b) of the exercise. Because the sets are disjoint, we simply add the number of elements in each set to find the union:

\(|A_{1} \cup A_{2} \cup A_{3}| = |A_{1}| + |A_{2}| + |A_{3}|\).
Given:

• \(|A_{1}| = 100\)
  
• \(|A_{2}| = 1000\)
  
• \(|A_{3}| = 10,000\)

Thus, the union is:

• \(100 + 1000 + 10,000 = 11,100\).

By understanding disjoint sets, we see there are no overlaps, making calculations straightforward and simple.

A subset is a set where every element is also in another set. For example, \(A_{1} \subseteq A_{2}\) means every element in \(A_{1}\) is also in \(A_{2}\). This relationship influences how we calculate set unions.
In part (a) of the exercise, we observe:

• \(A_{1} \subseteq A_{2}\)
• \(A_{2} \subseteq A_{3}\)

This tells us every element in \(A_{1}\) is also in \(A_{2}\), and every element in \(A_{2}\) is also in \(A_{3}\). Because of these subset relationships, the union of these sets keeps only the largest set, \(A_{3}\). Hence,
\(|A_{1} \cup A_{2} \cup A_{3}| = |A_{3}|\), which is 10,000.
Understanding subsets help in simplifying set operations, as any subset relationships reduce redundant counting in unions.