Set Theory Q: Is |P(A)| = |P(B)| iff A=B? Hints Needed

[JonF]

If P(X) denotes the power set of X. Is |P(A)| = |P(B)| iff A=B true? If so, I have no idea how to prove the |P(A)| => |P(B)| iff A=B direction, so any hints would be great.

 

[marcmtlca]

does |A| mean the cardinality of A? If so, it is clearly false. For example: the sets {0,1,2} and {3,4,5} have power sets with the same cardinality but they are not equal.

 

[ircdan]

No.

If A = B, then P(A) = P(B) so of course we have |P(A)| = |P(B)|.

But if |P(A)| = |P(B)|, A need not be equal to B.
Just take say A = {1}, B = {2}, then P(A) = {{},{1}} and P(B) = {{}, {2}}, so |P(A)| = |P(B)| but A != B.

 

[CRGreathouse]

Does |P(A)| = |P(B)| imply |A| = |B| for infinite sets without the CH?

 

[JonF]

Ok never mind on help trying to prove this, is it even true? I suspect it might be equivalent to the SCH.

 

[CrankFan]

Does |P(A)| = |P(B)| imply |A| = |B| for infinite sets without the CH?

Yes.

Any set A is equinumerous to the set A_s = \{ \{x\} \in P(A) | x \in A \}

For any set A Let f_A denote the bijective map from A to A_s.

If |P(A)| = |P(B)| then there is a bijection g from P(A) to P(B) and the restriction of g to the set A_s, is a bijection from A_s to B_s. Thus a bijection from A to B is given by the composition of functions

f_B^{-1} \circ g|A_s \circ f_A

Where g|A_s is the restriction of g to the set A_s and f_B^{-1} is the inverse of the function mapping B onto B_s.

 

[Hurkyl]

the restriction of g to the set A_s, is a bijection from A_s to B_s

Why do you know that? That's false for most bijections \mathcal{P}(A) \rightarrow \mathcal{P}(B).

For example, suppose that A and B actually are both equal to the set {0, 1}. Then, one example of a bijection \mathcal{P}(A) \rightarrow \mathcal{P}(B) is:

g({}) = {0}
g({0}) = {}
g({1}) = {0, 1}
g({0, 1}) = {1}

Here, the image of g \circ f_A never takes values in B_s.

 

[CrankFan]

I guess I said that because I wasn't thinking clearly. Thanks for the correction.