Here is the hint that my text gave: maybe you could help put it into easier terms for me. This proving that for all non-empty sets A, P(A) > A.

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Hint: Indirect proof! Suppose to the contrary that f:A->P(A) is a one-to-one correspondence between A and P(A) and consider the set

. Then

so it must correspond to some element

. Is

?

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I was going to try to say as much as I understood, but I actually don't know what the colon means when defining a set. Is set B have x in A and x not in f(x)? That seems impossible.