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Mathematics 23 Online
OpenStudy (anonymous):

If A={1}, then find the no. of elements in P[P[P(A)]].

OpenStudy (anonymous):

P(A)?

OpenStudy (anonymous):

my book has the same question as i've posted.

OpenStudy (anonymous):

i cant remember what was P(A) :(

OpenStudy (anonymous):

P(A) = { {}, {1} }

OpenStudy (anonymous):

P[P(A)] = { {}, {{}}, {{1}}, { {}, {1} } }

OpenStudy (anonymous):

P[P(A)] has 4 elements. Next level of power set would have 2^4 elements

OpenStudy (anonymous):

is the no. of elements = 2 ?

OpenStudy (anonymous):

2^4 = 16

OpenStudy (anonymous):

in those 16 are - {} {1} and all other are in pair with {}.

OpenStudy (anonymous):

so it will have {} {1} {1}{1}{1}{1}{1}{1}{1}{1}{1}..16times

OpenStudy (anonymous):

We can list them but gets messy instead we use the formula Let S be a finite set with N elements. Then the powerset of S (that is the set of all subsets of S ) contains 2^N elements.

OpenStudy (anonymous):

it contains 16 elements {} {{}} {{{}}} {{{1}}} {{ {}, {1} }} . . . . . . . { {}, {{}}, {{1}}, { {}, {1} } }

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