- #1

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Heres my logic

for the P({1}) on the inside

{EmptySet, {{1}}}

reason being, you always include the empty set, {1} is a part of the set. The cardinality is two

You have the set: {EmptySet, {{1}}}, and now you have to consider the outer "P"

the possibilities are:{EmptySet, {{1}}},EmptySet, {{1}}

I feel like there should be one more as its cardinality is three, when it should be four, and I don't think order matters