Is A~B equivalent to P(A)~P(B)?

[Unassuming]

Homework Statement

If A~B, then P(A)~P(B), which means the same as,

If |A|=|B|, then |P(A)|=|P(B)|

Homework Equations

The Attempt at a Solution

This problem is difficult for me because I am trying to learn the methods of proof while at the same time taking Intro Analysis. Anyway, here are my thoughts...

Assume f: A->B is a bijection. From Cantor's Thm. we know that |A|<|P(A)| and |B|<|P(B)|.

I also know that since |A|=|B|, |B|<|P(A)| and |A|<|P(B)|.

After this I feel like saying |P(A)|=|P(B)|, but I feel kind of guilty with that.

Any proof, insight, hint, or a one-liner would be appreciated. Thank you

 

[morphism]

Assume f: A->B is a bijection. From Cantor's Thm. we know that |A|<|P(A)| and |B|<|P(B)|.

I also know that since |A|=|B|, |B|<|P(A)| and |A|<|P(B)|.

After this I feel like saying |P(A)|=|P(B)|, but I feel kind of guilty with that.

You should! It's like saying 3=4 because 1<3 and 1<4!

Let's back-track. We know there is a bijection f:A->B. Can you use this to write down a bijection F:P(A)->P(B)?

 

[Unassuming]

This is a try,

Let a₁,a₂,...,a_i, be a sequence of elements of P(A).

Let b₁,b₂,...,b_i, be a sequence of elements of P(B).

Define F:P(A) --> P(B) as the function that takes a_n to b_n such that n $$\in$$ I, the index function of i.

Would this be a bijection?

 

[HallsofIvy]

?? You have not used the fact that A~ B! Also I don't know why you are dealing with sequences. If X is a member of P(A), what can you map it to in P(B)? (X is a set of things in A, and there is a mapping from A to B.)

 

[Unassuming]

This is another try,

Define F: A --> B, we know that F is bijective.
Define G:P(A) --> P(B). Let X_i be any set in P(A).
Then G(X_i) maps to an element of P(B), say Y_i.
Since A~B, then P(A)~P(B)

Would this be a bijection?

 

[morphism]

But it doesn't appear that you've actually defined what G is!