Since A and B have the same cardinality, there must exist a bijection f:A -> B.
Now
Now define f:P(A) -> P(B). This must be a bijection
You can fill the details
I've been racking my brain on this one and I have no idea on where to even start, so any help would be appreciated!
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 (show that it is 1-1 and onto).
\forall X \in P(A) , f(X) \in P(B)
Okay so I understand what that is trying to show, but is that my bijective function that I am trying to give? So let's say I wanted that in set builder notation, how do I go about that? Also, what steps do you recommend in showing that it is one-to-one and onto? (again I apologize, for some reason I just cant wrap my head around this problem).
Okay I think I'm starting to see what you say.
So for the one to one case since we know that the power set is just the set of all subsets, we can take certain subsets of A, and map them to elements in B, creating a subset itself. So for example:
{a1} ---> {b1}
{a2} ---> {b2}
{a2, a3} ---> {b2, b3}
.
.
.
and so on right? So we have shown that the subsets of A can be mapped to elements to B in a one-to-one fashion. Since we know that all Subsets of A are one-to-one , we know that it must have the same cardinality. And since the set of all subsets is the Powerset, we now know that P(A) is one to one with P(B)? Does that make any sense? But that onto case still slips me. Thanks in advance.