A if 1 pa then 1 a but 1 a b if a b and c are sets such that...

(a) If {1} ∈ P(A), then 1 ∈ A but {1} ∉ A. (b) If A, B and C are sets such that A ⊂ P(B) ⊂ C and |A| = 2, then |C| can be 5 but |C| cannot be 4. (c) If a set B has one more element than a set A, then P(B) has at least two more elements than P(A). (d) If four sets A, BC and D are subsets of {1, 2, 3} such that |A| = |B| = |C| = |D| = 2, then at least two of these sets are equal

Transcript

00:01 Hi, let's start the solution in part a we have given let curly bracket 1 or species set 1 is belong to power set a then 1 belongs to a but set 1 doesn't belongs to a yes our answer is true set 1 belongs to power set, but it doesn't belongs to a next b part let a comma b comma c are sets such that power set this is a is a subset of power set b is a subset of c and cardinality of a is 2 then cardinality c can be 5 but cardinality c can't be 9 so we verify that a is a subset of probability sorry power set of b is a subset of c.

01:25 So that is cardinality a is less than equal to cardinality of b which is less than equal to cardinality of c so that is 2 is less than equal to cardinality of p b which is less than equal to cardinality of c so that implies cardinality of c is greater than equal to 2 so cardinality of c is equal to 4 that is possible so answer of b part is false next part c let cardinality of a is n then cardinality of b is n plus 1 so cardinality of power set b is 2 power cardinality of b that is equal to 2 power n plus 1 and cardinality of power set a that is equal to 2 power cardinality of a which is equal to 2 power n so we get p b power set of b is 2 into 2 power n that is equal to 2 into power set of a so answer of part c is true and next part d let a comma b comma c and d are subsets of 1 comma 2 comma 3 such that cardinality of a is cardinality of b is cardinality of c is cardinality of d all cardinalities of a b c d are equal which is equal to 2 now take a is 1 comma 2 b is 2 comma 3 c is set 1 comma 3 and d is 1 comma 3 so here we get a is therefore we get a is equal to d since there is some correction set c is equal to d not a c is equal to d since they are three elements in one two three set so we can create three distinct set by choosing two elements from set 1 comma 2 comma 3 so the fourth set or we say the last set should be equal to anyone of the previous three set so answer of part d is true...

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