# Homework Help: If A~B, then P(A)~P(B) proof

1. Sep 1, 2008

### Unassuming

1. The problem statement, all variables and given/known data
If A~B, then P(A)~P(B), which means the same as,

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

2. Relevant equations

3. 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

2. Sep 1, 2008

### morphism

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)?

3. Sep 1, 2008

### Unassuming

This is a try,

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

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

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

Would this be a bijection?

4. Sep 2, 2008

### 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.)

5. Sep 2, 2008

### Unassuming

This is another try,

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

Would this be a bijection?

6. Sep 5, 2008

### morphism

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