Download Some extra 7.1 exercises for class 1. (T/F) 3 ∉ {2, 3, 5}

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.