# Cardinality and Powersets

Hi there, I'm having alot of trouble understanding this particular problem, and I hope the fine people of this forum can help me out =)

1. Homework Statement

Let A and B be infinite sets with the same cardinality. Prove that P(A) and P(B) have the same cardinality. Do this by giving explicitly a bijective function from P(A) to P(B). You must also prove that your function is indeed a bijection.

2. Homework Equations

3. The Attempt at a Solution

To be honest, I have absolutely no idea on how to even approach this problem. After the 3+ hours of Office hours (bless my T.A's heart, being patient with me and all), all I could come up with is:

We know that g:A -> B is a bijection and we have to show that f:P(A) ->P(B) is a bijection. I also know that I have to somehow find this bijective function for P(A) and P(B), and when I do that, I have show that the function is one-to-one and onto.

So since g:A->B is bijective we know we could have something like

g:
a1 ---> b1
a2 ---> b2 and so on.

My T.A. also wrote down:

{a} ---> {b}
{a1, a2} ---> {b1, b2}
{b2} ---> {b2}
Which he said could possibly help in finding the bijective function (which he said should probably be in set builder form). Any help would be greatly appreciated.

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Dick
Homework Helper
That's like, pretty easy. P(A) is the set of all subsets of A, ditto for P(B). If you have a bijection from A to B, it should be pretty easy to write down a bijection from a subset of A to a subset of B. Map all of the elements of A in the subset to elements of B, creating a subset of B. Now show it's a bijection.

I really don't get what you're trying to say. So if I take the elements of the Subset of A and map them to elements of B, making a new subset for B, how does help with the Powerset of A mapping to the powerset of B, as well as creating a bijective function? My only guess as to why you could do this is if there exists a bijection within subsets, then the set itself must be a bijection(but that doesn't sound right at all to me, I will skim my notes and the chapter again just to make sure). Again, I apologize for my lack of understanding, for some reason this is just like a brick wall to me.

since f:A->B is a bijection f restricted to to any subset of A is a bijection to a subset of B. this is a mapping from P(A) -> P(B). proving that the sum of all these bijections is a bijection is probably easy so if you want i'll fill it in the rest of the way.

Last edited:
Dick