Determine if the following relations are equivalence relations, anti-symmetric relations, asymmetric relations, or neither. Provide an explanation for each claim. 1. \( R=\subseteq \) 2. \( R=\subsetn

Question

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Answer 1

1. \( R=\subseteq \) is an anti-symmetric relation.

Explanation: For a relation to be anti-symmetric, if any two elements x and y are related and x ≠ y, then it cannot be the case that y is also related to x. In the case of \( R=\subseteq \), if x is a subset of y and x ≠ y, then it is not possible for y to be a subset of x. This is because if y were a subset of x, then x and y would be equal sets, contradicting the assumption that x ≠ y. Therefore, \( R=\subseteq \) satisfies the anti-symmetric property.

2. \( R=\subsetneq \) is neither an equivalence relation, anti-symmetric relation, nor an asymmetric relation.

Explanation:
- \( R=\subsetneq \) is not an equivalence relation because it does not satisfy all three properties of reflexivity, symmetry, and transitivity. For example, the relation includes pairs like (A, A) where A is a set, which violates symmetry as the relation implies A is a proper subset of itself. Additionally, the relation does not satisfy transitivity since if A is a proper subset of B and B is a proper subset of C, it does not necessarily mean that A is a proper subset of C.
- \( R=\subsetneq \) is also not an anti-symmetric relation because it allows for cases where two distinct sets are both proper subsets of each other. For example, if A = {1, 2} and B = {2, 3}, both A is a proper subset of B and B is a proper subset of A. Therefore, the relation fails the anti-symmetric property.
- \( R=\subsetneq \) is not an asymmetric relation because it allows for cases where if A is a proper subset of B, then B is not a proper subset of A. For example, if A = {1} and B = {1, 2}, A is a proper subset of B but B is not a proper subset of A. Therefore, the relation fails the asymmetric property.