Answer:
(a) No
(b) Yes
(c) No
(d) 4
(e) 3
(f) 3
Step-by-step explanation:
A = {red, blue, green, purple}
B = {red, red, green}
C = {red,{green}, red,{red, green},{green, green, red}}
D = {blue, blue, blue, green, green, purple}
U = {red, blue, purple, green}∪ P({red, blue, purple, green})
By removing duplicate elements, the sets become
A = {red, blue, green, purple}
B = {red, green}
C = {red, {green}, {red, green}}
D = {blue, green, purple}
U = {red, blue, purple, green} ∪ P({red, blue, purple, green})
(a) B ∉ A because B is a set while A contains elements that are not sets, even thought the elements of B are also in A. In fact, B is a subset of A.
(b) B ∈ C because {red, green} is a member of C.
(c) P(C) is the power set of C, the set of all subsets of C. The power set of C is
P(C) = {{}, {red}, {{green}}, {{red, green}}, {red, {green}}, {red, {red, green}}, {{green}, {red, green}}, {red, {green}, {red, green}}}
It is seen that B ∉ P(C).
(d) |A| is the cardinality of A or the number of distinct members of A.
Therefore, |A| = 4
(e) |C| = 3 (Note that the elements in the member set are not counted)
(f) |D| = 3