set cardinality

Question

Comments

P(∅) = {∅} = {{}} =  $2^0=1 $ element, we don't count $ \varnothing $ as element here

But ,

P({∅, {1}, {2}, {1,2}}) = 16 ,we count $\varnothing$ as element here,why so? 

Answer 1

P(∅) = {∅} = {{}} = $2^0 $= 1 element
P({∅}) = {∅, {∅}}   = $2^1$ = 2 elements
P({∅, {∅}}) = {∅, {∅}, {{∅}}, {∅, {∅}}} = $2^2$ = 4 elements

|R| = 4

P({1,2}) = {∅, {1}, {2}, {1,2}}
P({∅, {1}, {2}, {1,2}}) = 16 elements

$S = RXT = 4*16 = 64$ elements.

Comments

then it should be empty

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should we not consider that thing as well here?

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Here we are not doing cross product with empty set bhai . We are doing cross product with power set of element ie P(∅)={∅}.