Hey all,

Given

Is

?

And also how can I prove that for all sets and we have

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- March 16th 2011, 08:51 PMOilerP(A)..Power set of A.
Hey all,

Given

Is

?

And also how can I prove that for all sets and we have - March 16th 2011, 09:00 PMTinyboss
What is the cardinality of A (denoted |A|)? The cardinality of P(A) is 2^|A|, so what you wrote is missing a few. Also, the elements of P(A) are

*subsets*of A, so there's no way that 1 is an element of P(anything), since it's not a set.

I think once you get the first part straightened out, the second part won't give you much trouble.