I. Tell under what conditions on sets \( A \) and \( B \) that we would have each of the following: 1. \( \mathrm{A} \cup \mathrm{B}=\varnothing \) 2. \( \mathrm{A} \cap \mathrm{B}=\varnothing \) 3. \( \mathrm{A} \cap \mathrm{B}=\mathrm{U} \) 4. \( \mathrm{A} \cap \mathrm{B}=\mathrm{A} \) 5. \( \mathrm{A} \cup \mathrm{B}=\mathrm{U} \) 6. \( \mathrm{A} \cup \mathrm{B}=\mathrm{A} \) 7. \( \mathrm{A}-\mathrm{B}=\varnothing \) 8. \( \mathrm{A}-\mathrm{B}=\mathrm{A} \) 9. \( \mathrm{B}=\mathrm{A} \) 10. \( \mathrm{A}=\mathrm{A}-\mathrm{B} \)

Let's analyze each condition: 1. \( \mathrm{A} \cup \mathrm{B}=\varnothing \) - This occurs when both \( \mathrm{A} \) and \( \mathrm{B} \) are empty sets, i.e., \( \mathrm{A} = \varnothing \) and \( \mathrm{B} = \varnothing \). 2. \( \mathrm{A} \cap \mathrm{B}=\varnothing \) - This happens when there are no elements that are common to both \( \mathrm{A} \) and \( \mathrm{B} \). In other words, \( \mathrm{A} \) and \( \mathrm{B} \) are disjoint sets. 3. \( \mathrm{A} \cap \mathrm{B}=\mathrm{U} \) - This is true when \( \mathrm{A} \) and \( \mathrm{B} \) are the same set, i.e., \( \mathrm{A} = \mathrm{B} \). 4. \( \mathrm{A} \cap \mathrm{B}=\mathrm{A} \) - This condition is met when every element of \( \mathrm{A} \) is also an element of \( \mathrm{B} \), which means \( \mathrm{A} \) is a subset of \( \mathrm{B} \), i.e., \( \mathrm{A} \subseteq \mathrm{B} \). 5. \( \mathrm{A} \cup \mathrm{B}=\mathrm{U} \) - This occurs when the union of \( \mathrm{A} \) and \( \mathrm{B} \) contains all elements of the universal set \( \mathrm{U} \). This means that \( \mathrm{A} \) and \( \mathrm{B} \) together cover the entire universal set, i.e., \( \mathrm{A} \cup \mathrm{B} = \mathrm{U} \). 6. \( \mathrm{A} \cup \mathrm{B}=\mathrm{A} \) - This is true when \( \mathrm{B} \) is a subset of \( \mathrm{A} \), i.e., \( \mathrm{B} \subseteq \mathrm{A} \). 7. \( \mathrm{A}-\mathrm{B}=\varnothing \) - This happens when there are no elements in \( \mathrm{A} \) that are not in \( \mathrm{B} \), which means \( \mathrm{A} \) is a subset of \( \mathrm{B} \), i.e., \( \mathrm{A} \subseteq \mathrm{B} \). 8. \( \mathrm{A}-\mathrm{B}=\mathrm{A} \) - This condition is met when there are no elements in \( \mathrm{A} \) that are not in \( \mathrm{B} \), which means \( \mathrm{A} \) and \( \mathrm{B} \) have no elements in common, i.e., \( \mathrm{A} \cap \mathrm{B} = \varnothing \). 9. \( \mathrm{B}=\mathrm{A} \) - This is straightforward: \( \mathrm{B} \) and \( \mathrm{A} \) are the same set. 10. \( \mathrm{A}=\mathrm{A}-\mathrm{B} \) - This condition is true when there are no elements in \( \mathrm{A} \) that are not in \( \mathrm{B} \), which means \( \mathrm{A} \) is a subset of \( \mathrm{B} \), i.e., \( \mathrm{A} \subseteq \mathrm{B} \). However, this is not the same as \( \mathrm{A} = \mathrm{A} - \mathrm{B} \) in general. The correct interpretation is that \( \mathrm{A} \) is equal to the set of elements in \( \mathrm{A} \) that are not in \( \mathrm{B} \), which is always true because \( \mathrm{A} \) is a subset of itself.