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If  R=P(phi) and T=P({1,2}) where P is power set

Then cardinality for S=R * T is ?

What i know is phi *{Any set} = phi  so above cardinality must be zero too.

edited | 82 views
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P(phi) is not equal to phi

P(phi) = {{}}

Hence cardinality of above set should be 1*4 = 4
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cardinality of R is 1 because it is {{∅}}
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R = {$\phi$}   T = {$\phi$,{1},{2},{1,2}}

S = R*T = { ($\phi$,$\phi$), ($\phi$,{1}), ($\phi$,{2}), ($\phi$,{1,2}) }

|S| = 4
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Shivam Chauhan R is not {ϕ}

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Power set of $\phi$ = {{}} = {$\phi$}
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Mk Utkarsh you have done P(P(phi))
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so P(P(phi)) contains 2 elements? right?
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yes its 2

+1 vote
power set of empty set = $p(\phi )=\left \{ \Phi \right \} , powerset of singleton set=p(\left \{ \phi \right \})= \left \{ \phi ,\left \{ \phi \right \} \right \}$

as we know if a set contain n element then power set will contain $2^{n}$

so here empty set contain 0 element so p(empty set) contain $2^{0}=1$

and singleton set has 1 element so power set will contain $2^{1}=2$

so for given question |R|=1 and |T|=4 , S= 1*4=4 ans .
by Boss (17.1k points)
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