Consider the following statements:A: The set of all reflexive relations are closed under the operation of set union.B: The set of all irreflexive relations are not closed under the operation of set union.C: The set of all reflexive relations are not closed under set difference.D: The set difference of all reflexive relations is not irreflexive.Which of the given statements are false?a)Both B and Db)Both B and Cc)Both A and Dd)Both C and DCorrect answer is option 'A'. Can you explain this answer?

Question

Answer

To determine which of the given statements are false, let's analyze each statement one by one.

Statement A: The set of all reflexive relations are closed under the operation of set union.

- A relation R is reflexive if every element in a set A is related to itself. In other words, (a, a) ∈ R for every a ∈ A.
- The set of all reflexive relations on a set A is denoted by R(A).
- The operation of set union (∪) combines two sets and forms a new set that contains all distinct elements from both sets.
- To check if the set of all reflexive relations is closed under set union, we need to verify if the union of two reflexive relations is also reflexive.
- Let's consider two reflexive relations R1 and R2 on a set A. Since R1 and R2 are reflexive, (a, a) ∈ R1 and (a, a) ∈ R2 for every a ∈ A.
- Now, let's take the union of R1 and R2, denoted by R1 ∪ R2. Since R1 and R2 have the same elements, the union will still contain (a, a) for every a ∈ A.
- Therefore, the set of all reflexive relations is closed under set union, and statement A is true.

Statement B: The set of all irreflexive relations are not closed under the operation of set union.

- A relation R is irreflexive if no element in a set A is related to itself. In other words, (a, a) ∉ R for every a ∈ A.
- The set of all irreflexive relations on a set A is denoted by I(A).
- Similar to statement A, we need to check if the union of two irreflexive relations is also irreflexive.
- Let's consider two irreflexive relations R1 and R2 on a set A. Since R1 and R2 are irreflexive, (a, a) ∉ R1 and (a, a) ∉ R2 for every a ∈ A.
- Now, let's take the union of R1 and R2, denoted by R1 ∪ R2. Since R1 and R2 have the same elements, the union can potentially contain (a, a) for some a ∈ A.
- Therefore, the set of all irreflexive relations is not closed under set union, and statement B is true.

Statement C: The set of all reflexive relations are not closed under set difference.

- The operation of set difference (A - B) between two sets A and B results in a new set that contains elements from A that are not in B.
- To check if the set of all reflexive relations is closed under set difference, we need to verify if the difference between two reflexive relations is still reflexive.
- Let's consider two reflexive relations R1 and R2 on a set A. Since R1 and R2 are reflexive, (a, a) ∈ R1 and (a, a) ∈ R2 for every a ∈ A.
- Now, let's take the difference between R1 and R2, denoted by R1 - R2. The resulting set will contain elements from R1 that are not in R2. However, there is

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