91Ó°ÊÓ

In set theory, the union of sets is a fundamental operation that unites elements from multiple sets. The union combines all elements from each set, ensuring each element appears only once.
To perform a union, list all distinct elements from the involved sets. For instance, consider sets:

• \(A = \{1, 3, 5, 7, 9\}\)
• \(B = \{1, 2, 4, 7, 8\}\)
• \(C = \{2, 4, 6, 8\}\)

The union of \(B \cup C\) means we gather all elements from both \(B\) and \(C\), so \(B \cup C = \{1, 2, 4, 6, 7, 8\}\).
When performing a full union with \(A\), we find that \(A \cup (B \cup C) = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}\). Notice there's no repetition, each element is listed once. This approach confirms that the order in which unions are performed does not alter the result, reflecting the associative property.

Intersection of sets involves identifying elements common to multiple sets, listing only shared elements in the intersection result. For example, consider the same sets:

• \(A = \{1, 3, 5, 7, 9\}\)
• \(B = \{1, 2, 4, 7, 8\}\)
• \(C = \{2, 4, 6, 8\}\)

Applying the intersection operation, \(B \cap C\) identifies elements present in both \(B\) and \(C\), which are \(\{2, 4, 8\}\).
Performing \(A \cap (B \cap C)\), no common elements exist between \(A\) and the result of \(B \cap C\). Thus, we have \(A \cap (B \cap C) = \emptyset\), indicating an empty set. Similarly, \((A \cap B) \cap C\) results in \(\emptyset\). Hence, the intersection is associative; the order of operations doesn't affect the final outcome when intersecting three sets.

The associative property is a key concept in mathematics, applicable to various operations, including set theory. It demonstrates that the grouping of elements doesn't affect the result for union and intersection operations.
For the union of sets, consider the expression \((A \cup B) \cup C\). First, combining \(A\) and \(B\) yields \(\{1, 2, 3, 4, 5, 7, 8, 9\}\), and uniting with \(C\) gives \(\{1, 2, 3, 4, 5, 6, 7, 8, 9\}\). Similarly, \(A \cup (B \cup C)\) directly achieves the same set, proving the associative law via a common result.
For intersections, \((A \cap B) \cap C\) and \(A \cap (B \cap C)\) both lead to the empty set \(\emptyset\).
Therefore, associative property confirms that regardless of how elements are grouped in these operations, the end result remains consistent, simplifying complex set expressions.