- #1
opt!kal
- 19
- 0
Hi there, I'm having a lot of trouble understanding this particular problem, and I hope the fine people of this forum can help me out =)
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.
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.
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.
Homework Equations
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.