Respuesta :
Answer:
a) P({a, b, {a, b}})=[tex] 2*2*2=8[/tex]
b) P({∅, a, {a}, {{a}}})=[tex] 2*2*2*2=16[/tex]
c) P(P(∅))=P(P{∅})=2
Step-by-step explanation:
Previous concepts
For this case we need to remember the concpts of subset and the power set.
A is a subset of B if every element of A is an eelement if B and we write [tex]A \subseteq B[/tex]
The power set of a set R is th set of all the possible subsets of R and we writ this as P(R)
Solution to the problem
Part a
P({a, b, {a, b}}) as we can see this set contains 3 elements a,b and {a,b}
And since the power set contains all the possible subsets of the elements. For this case for each element we have 3-1 = 2 options in order to combine, then the number of possible subsets would be equal to the product of possible options and we got:
P({a, b, {a, b}})=[tex] 2*2*2=8[/tex]
Part b
P({∅, a, {a}, {{a}}}) as we can see this set contains 4 elements ∅,a,{a} and {{a}}
And since the power set contains all the possible subsets of the elements. For this case for each element we have 3-1 = 2 options in order to combine, then the number of possible subsets would be equal to the product of possible options and we got:
P({∅, a, {a}, {{a}}})=[tex] 2*2*2*2=16[/tex]
Part c
P(P(∅)) for this case we need to remember that P(∅) have all the possible subsets empty, but the power set for the empty set is also empty.
P(∅)={∅}
And we see that this last one set have just one element.
For this special case we have again 2 options since we can have the element in the subset or the element that is not on the subset, so then we have this:
P(P(∅))=P(P{∅})=2
The number of elements of the set are 8, 16, and 2.
How to solve the elements?
It should be noted that the objects in a set are known as the elements or members of the set.
The number of elements in the set will be calculated thus:
P({a, b, {a, b} = 2³ = 2 × 2 × 2 = 8
P({∅, a, {a}, {{a}}) = 2⁴ = 16
P(P(∅)) = 2¹ = 2
In conclusion, the correct options are 8, 16, and 2.
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