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Some extra 7.1 exercises for class 1. (T/F) 3 ∉ {2, 3, 5} In words, “3 is not an element of the set containing the elements 2, 3 and 5.” But 3 is an element of that set so the original statement is FALSE 2. (T/F) {3} ∉ {2, 3, 5} In words, “The set containing the element 3 is not an element of the set containing the elements 2, 3 and 5.” Notice the IMPORTANT change from #1. Now we are no longer deciding if 3 is in {2,3,5} but whether the set {3} is in {2,3,5}. If {3} were an element of a set, it would actually appear in the set, for example, {2, {3} ,5}. Because {3} does not appear as an element of {2,3,5}, {3} is not an element of {2,3,5} which is what the original statement said. So, the original statement is TRUE Note: Also, don’t confuse element of “∈” with subset of “⊆”. 3. (T/F) { } ∈ {2,3,4} Using the same logic as in #2, { } does not appear as an element of {2,3,4}, so this statement is FALSE 4. (T/F) {4} ⊆ {4, 11, 17} In words, “The set containing the element 4 is a subset of the set containing the elements 4, 11 and 17.” In order for a set to be a subset of another set, every element of that set must also be an element of the other set. 4 is the only element of {4} and 4 is an element of {4,11,17} so the statement is TRUE 5. (T/F) {4} ⊂ {4, 11, 17} In words, “The set containing the element 4 is a proper subset of the set containing the elements 4, 11 and 17.” In order for a set to be a proper subset of another set, every element of that set must also be an element of the other set (as is true for just being an ordinary subset) plus a restriction that a set is not allowed to be a proper subset of itself. In other words subset and proper subset only differ in that a set cannot be a proper subset itself but a set is always a subset of itself. So, the original statement is TRUE 6. (T/F) {4, 11, 17} ⊆ {4, 11, 17} Every set is a subset of itself (see #5 for more info) so the original statement is TRUE 7. (T/F) {4, 11, 17} ⊂ {4, 11, 17} A set CANNOT be a proper subset of itself (see #5 for more info) so the original statement is FALSE 8. (T/F) 4 ⊆ {4} 4 is not a set so it cannot be a subset of any set. (4 is an element of {4} but not a subset of anything.) So, this statement is FALSE 9. (T/F) {4} ⊆ {4} Every set is a subset of itself (see #5 for more info) so the original statement is TRUE 10. (T/F) 0 ⊆ ∅ 0 is not a set so it cannot be a subset of any set. So, this statement is FALSE 11. A = {1, 3, 5, 7}, B = {2, 4, 6, 8}, C = {6, 7, 8} A∪B= The union of 2 sets is the set containing elements of either (or both) of the sets so A ∪ B = { 1, 2, 3, 4, 5, 6, 7, 8 } A ∪ B = { 1, 2, 3, 4, 5, 6, 7, 8 } 12. A = {1, 3, 5, 7}, B = {2, 4, 6, 8}, C = {6, 7, 8} A ∪ C = { 1, 3, 5, 6, 7, 8 } 13. A = {1, 3, 5, 7}, B = {2, 4, 6, 8}, C = {6, 7, 8} A∩C= The intersection of 2 sets is the set containing just the elements in both of the sets so A∩C={7} 14. A = {1, 3, 5, 7}, B = {2, 4, 6, 8}, C = {6, 7, 8} A∪∅= A∪∅=A∪{} A ∪ ∅ = {1, 3, 5, 7} ∪ { } A ∪ ∅ = {1, 3, 5, 7} 15. A = {1, 3, 5, 7}, B = {2, 4, 6, 8}, C = {6, 7, 8} A∩∅= A∩∅=A∩{} A ∩ ∅ = {1, 3, 5, 7} ∩ { } There are no elements in both so A∩∅=∅ 16. A = {1, 3, 5, 7}, B = {2, 4, 6, 8}, C = {6, 7, 8} (A ∪ C) ∩ B (A ∪ C) ∩ B = ({1, 3, 5, 7} ∪ {6, 7, 8} ) ∩ {2, 4, 6, 8} Working inside the parentheses first… (A ∪ C) ∩ B = {1, 3, 5, 6, 7, 8} ∩ {2, 4, 6, 8} (A ∪ C) ∩ B = {1, 3, 5, 6, 7, 8} ∩ {2, 4, 6, 8} (A ∪ C) ∩ B = {6, 8 } 17. U = {1, 2, 3, 4, 5, 6, 7, 8}, A = {2, 4, 6, 8} A’ = The complement of a set is the set of elements in the Universal Set (U) but not in the set itself. A’ = { 1, 3, 5, 7 } 18. If U is the set of U.S. colleges and A is the set of U.S. colleges with “City” in their name, then describe A’ in words. A’ is the set of U.S. colleges that do not have “City” in their name.