Question

Using the definitions of union and intersection show that, when viewed as binary operations on sets, union and intersection demonstrate, for all sets \(\mathrm{X}, \mathrm{Y}\) and \(\mathrm{Z}\) : (A) Commutativity (B) Associativity (C) Distributivity of intersection over union.

Short answer

Union and intersection of sets demonstrate commutativity, associativity, and distributivity by the following proofs: (A) Commutativity: (i) Union: \(X ∪ Y = Y ∪ X\) as \( x \in X ∪ Y \) if and only if \( x \in Y ∪ X \). (ii) Intersection: \(X ∩ Y = Y ∩ X\) as \( x \in X ∩ Y \) if and only if \( x \in Y ∩ X \). (B) Associativity: (i) Union: \((X ∪ Y) ∪ Z = X ∪ (Y ∪ Z)\) as \(x \in (X ∪ Y) ∪ Z\) if and only if \(x \in X ∪ (Y ∪ Z)\). (ii) Intersection: \((X ∩ Y) ∩ Z = X ∩ (Y ∩ Z)\) as \(x \in (X ∩ Y) ∩ Z\) if and only if \(x \in X ∩ (Y ∩ Z)\). (C) Distributivity of intersection over union: \(X ∩ (Y ∪ Z) = (X ∩ Y) ∪ (X ∩ Z)\) as \(x \in X ∩ (Y ∪ Z)\) if and only if \(x \in (X ∩ Y) ∪ (X ∩ Z)\).

Step-by-step solution

Proof of Commutativity: Union

To prove that union is commutative, let's show that X ∪ Y = Y ∪ X for all sets X and Y. We will show that a general element x is in X ∪ Y if and only if it is in Y ∪ X. \(x \in X ∪ Y\) if and only if \(x \in X\) or \(x \in Y\). Similarly, \(x \in Y ∪ X\) if and only if \(x \in Y\) or \(x \in X\). As both conditions are the same, we can conclude that X ∪ Y = Y ∪ X, proving the commutativity of union.

Proof of Commutativity: Intersection

To prove that intersection is commutative, let's show that X ∩ Y = Y ∩ X for all sets X and Y. We will show that a general element x is in X ∩ Y if and only if it is in Y ∩ X. \(x \in X ∩ Y\) if and only if \(x \in X\) and \(x \in Y\). Similarly, \(x \in Y ∩ X\) if and only if \(x \in Y\) and \(x \in X\). As both conditions are the same, we can conclude that X ∩ Y = Y ∩ X, proving the commutativity of intersection.

Proof of Associativity: Union

To prove that union is associative, let's show that (X ∪ Y) ∪ Z = X ∪ (Y ∪ Z) for all sets X, Y, and Z. We will show that a general element x is in (X ∪ Y) ∪ Z if and only if it is in X ∪ (Y ∪ Z). \(x \in (X ∪ Y) ∪ Z\) if and only if \(x \in (X ∪ Y)\) or \(x \in Z\), which means that \(x \in X \) or \( x \in Y \) or \( x \in Z \). Similarly, \(x \in X ∪ (Y ∪ Z)\) if and only if \(x \in X\) or \(x \in (Y ∪ Z)\), which also means that \(x \in X \) or \( x \in Y \) or \( x \in Z \). As both conditions are the same, we can conclude that (X ∪ Y) ∪ Z = X ∪ (Y ∪ Z), proving the associativity of union.

Proof of Associativity: Intersection

To prove that intersection is associative, let's show that (X ∩ Y) ∩ Z = X ∩ (Y ∩ Z) for all sets X, Y, and Z. We will show that a general element x is in (X ∩ Y) ∩ Z if and only if it is in X ∩ (Y ∩ Z). \(x \in (X ∩ Y) ∩ Z\) if and only if \(x \in (X ∩ Y)\) and \(x \in Z\), which means that \(x \in X \) and \( x \in Y \) and \( x \in Z \). Similarly, \(x \in X ∩ (Y ∩ Z)\) if and only if \(x \in X\) and \(x \in (Y ∩ Z)\), which also means that \(x \in X \) and \( x \in Y \) and \( x \in Z \). As both conditions are the same, we can conclude that (X ∩ Y) ∩ Z = X ∩ (Y ∩ Z), proving the associativity of intersection.

Proof of Distributivity of Intersection over Union

To prove that intersection distributes over union, let's show that X ∩ (Y ∪ Z) = (X ∩ Y) ∪ (X ∩ Z) for all sets X, Y, and Z. We will show that a general element x is in X ∩ (Y ∪ Z) if and only if it is in (X ∩ Y) ∪ (X ∩ Z). \(x \in X ∩ (Y ∪ Z)\) if and only if \(x \in X\) and \(x \in (Y ∪ Z)\), which means that \( x \in X\) and (\(x \in Y\) or \( x \in Z\)). On the other hand, \(x \in (X ∩ Y) ∪ (X ∩ Z)\) if and only if \(x \in (X ∩ Y) \) or \( x \in (X ∩ Z) \), which also means that \( x \in X\) and (\(x \in Y\) or \( x \in Z\)). As both conditions are the same, we can conclude that X ∩ (Y ∪ Z) = (X ∩ Y) ∪ (X ∩ Z), proving the distributivity of intersection over union.

Key concepts

Commutativity

Commutativity is a fundamental property seen in operations like addition and multiplication in arithmetic. In set theory, commutativity applies to both union and intersection. But what does it really mean for these operations to be commutative?

For any two sets, say \(X\) and \(Y\), the union operation is commutative if \(X \cup Y\) equals \(Y \cup X\). This means that the order in which you join the sets doesn’t matter. If you think about the definition of union, where an element belongs to the union if it belongs to either of the sets, the order clearly makes no difference. Thus:

• Union: \(X \cup Y = Y \cup X\)

Similarly, for intersection, if \(X \cap Y\) equals \(Y \cap X\), the order of the sets doesn’t change the elements you find in both sets. Intersection is all about what the sets share, and switching the order doesn’t change the answer, giving us:

• Intersection: \(X \cap Y = Y \cap X\)

Commutativity ensures that no matter which set you place first in your notation, the result remains constant. This property can simplify calculations by reassuring us that the arrangement of sets won't affect the results.

Associativity

Associativity is another important concept that plays a crucial role in both mathematics and set theory. An operation is said to be associative if the outcome is the same regardless of how the operands are grouped.

For the union of sets, we say that union is associative if \((X \cup Y) \cup Z = X \cup (Y \cup Z)\). This means that when you're combining three sets, it doesn't matter which two you combine first, the result will be the same. Essentially:

• Union: \((X \cup Y) \cup Z = X \cup (Y \cup Z)\)

In the case of intersection, associativity means that combining the intersection of two sets with a third yields the same result however you group them. So, for three sets \(X, Y,\) and \(Z\), the property can be expressed as:

• Intersection: \((X \cap Y) \cap Z = X \cap (Y \cap Z)\)

Associativity helps in systematically planning computations without rearranging parentheses, making complex operations easier to manage and understand.

Distributivity

Distributivity is a unique and intriguing property that relates different operations. In set theory, it involves both union and intersection. To say that intersection is distributive over union means we have the following relationship:

\(X \cap (Y \cup Z) = (X \cap Y) \cup (X \cap Z)\). This formula shows us how intersection can "spread over" union in a useful way. For any element to be in \(X \cap (Y \cup Z)\), it must be in \(X\) and in the union of \(Y\) and \(Z\). This condition can be further examined and split into two possibilities: it either belongs to both \(X\) and \(Y\), or to both \(X\) and \(Z\), leading to:

• \(X \cap (Y \cup Z)\) = \((X \cap Y) \cup (X \cap Z)\)

Distributivity is powerful as it helps transform expressions and simplify logical statements or proofs by breaking down complex relationships into simpler ones.

Union and Intersection of Sets

Union and intersection are core operations in set theory, each with its own unique role. Understanding these operations can greatly enhance your ability to solve problems in set theory and beyond.

The union of two sets, denoted as \(X \cup Y\), is a set containing all distinct elements present in either set. This is akin to combining all elements and eliminating duplicates. It’s like creating a list of all unique participants in two different groups, leading to the rule:

• Element belongs to \(X \cup Y\) if it’s in \(X\) or \(Y\) (or both).

Intersection, written as \(X \cap Y\), involves finding elements common to both sets. It’s similar to pinpointing shared members between the groups. If you’re attending events represented by two sets, the intersection identifies which events you must attend both to be considered a participant:

• Element belongs to \(X \cap Y\) if it’s in both \(X\) and \(Y\).

These concepts provide the foundation for more complex set operations, aiding in logical proofs and understanding relationships between different sets.