Define A relation S for subset of R by S=[{A,B:ACR,B is improper

Subset to R,[A/B]u[B/A]is finite prove that s is an equivalent relation and the family of finite subset of R one equivalent class

To prove that the relation S is an equivalence relation, we need to show that it satisfies three properties: reflexivity, symmetry, and transitivity.

1. Reflexivity: We need to show that every set A in R is related to itself, meaning ASA. In this case, since we are given that B is an improper subset of A, we know that A ⊂ B is not true. Therefore, A cannot be equal to B, which implies that ASA is false. Hence, the relation S is not reflexive.

2. Symmetry: We need to show that if A is related to B, then B is related to A, which can be represented as ASB implies BSA. Here, we are given that [A/B] ∪ [B/A] is finite. To prove symmetry, let's consider two sets, A and B, where A is related to B. This means that either A ⊂ B or B ⊂ A. Without loss of generality, let's assume A ⊂ B. Now, we need to show that B is related to A, i.e., B ⊂ A. To show this, we can use proof by contradiction. If B is not a subset of A, then we can always find an element x ∈ (B \ A) such that x ∉ A. But this contradicts the given fact that [A/B] ∪ [B/A] is finite because we can construct infinitely many x's. Therefore, B ⊂ A, and we have shown that ASB implies BSA, proving symmetry for the relation S.

3. Transitivity: We need to show that if A is related to B and B is related to C, then A is related to C, which can be represented as ASB and BSC implies ASC. Let's assume that A is related to B, which means either A ⊂ B or B ⊂ A. Similarly, let's assume B is related to C, which means either B ⊂ C or C ⊂ B. We need to prove that A is related to C, i.e., either A ⊂ C or C ⊂ A. We consider four possible cases:
a) If A ⊂ B and B ⊂ C, then A ⊂ C.
b) If A ⊂ B and C ⊂ B, then C ⊂ B ⊂ A.
c) If B ⊂ A and C ⊂ B, then C ⊂ B ⊂ A.
d) If B ⊂ A and B ⊂ C, then C ⊂ B ⊂ A.

In all cases, we have either A ⊂ C or C ⊂ A, which means A is related to C, proving transitivity.

Now, let's discuss the second part of the question: "the family of finite subsets of R forms one equivalence class."

For a set A in R, the equivalence class [A] consists of all sets B such that B is an improper subset of A and [A/B] ∪ [B/A] is finite. To show that the family of finite subsets of R forms one equivalence class, we need to establish that every finite subset of R is related to every other finite subset of R.

Consider two finite subsets of R, A and B. Since both subsets are finite, the union A ∪ B is also finite. Therefore, [A/B] ∪ [B/A] is finite, satisfying the condition for the equivalence class. Hence, any two finite subsets of R belong to the same equivalence class.

In conclusion, the relation S is not an equivalence relation since it doesn't satisfy reflexivity. However, the family of finite subsets of R forms one equivalence class.