Thread: P(A)..Power set of A.

Hey all,

Given
Is
?

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

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.