# Set theory question

1. Oct 23, 2006

### JonF

If P(X) denotes the power set of X. Is |P(A)| = |P(B)| iff A=B true? If so, I have no idea how to prove the |P(A)| => |P(B)| iff A=B direction, so any hints would be great.

2. Oct 24, 2006

### marcmtlca

does |A| mean the cardinality of A? If so, it is clearly false. For example: the sets {0,1,2} and {3,4,5} have power sets with the same cardinality but they are not equal.

3. Oct 24, 2006

### ircdan

No.

If A = B, then P(A) = P(B) so of course we have |P(A)| = |P(B)|.

But if |P(A)| = |P(B)|, A need not be equal to B.
Just take say A = {1}, B = {2}, then P(A) = {{},{1}} and P(B) = {{}, {2}}, so |P(A)| = |P(B)| but A != B.

4. Oct 25, 2006

### CRGreathouse

Does |P(A)| = |P(B)| imply |A| = |B| for infinite sets without the CH?

5. Oct 30, 2006

### JonF

Ok never mind on help trying to prove this, is it even true? I suspect it might be equivalent to the SCH.

6. Oct 31, 2006

### CrankFan

Yes.

Any set $$A$$ is equinumerous to the set $$A_s = \{ \{x\} \in P(A) | x \in A \}$$

For any set $$A$$ Let $$f_A$$ denote the bijective map from $$A$$ to $$A_s$$.

If $$|P(A)| = |P(B)|$$ then there is a bijection $$g$$ from $$P(A)$$ to $$P(B)$$ and the restriction of $$g$$ to the set $$A_s$$, is a bijection from $$A_s$$ to $$B_s$$. Thus a bijection from $$A$$ to $$B$$ is given by the composition of functions

$$f_B^{-1} \circ g|A_s \circ f_A$$

Where $$g|A_s$$ is the restriction of g to the set $$A_s$$ and $$f_B^{-1}$$ is the inverse of the function mapping $$B$$ onto $$B_s$$.

Last edited: Oct 31, 2006
7. Oct 31, 2006

### Hurkyl

Staff Emeritus
Why do you know that? That's false for most bijections $\mathcal{P}(A) \rightarrow \mathcal{P}(B)$.

For example, suppose that A and B actually are both equal to the set {0, 1}. Then, one example of a bijection $\mathcal{P}(A) \rightarrow \mathcal{P}(B)$ is:

g({}) = {0}
g({0}) = {}
g({1}) = {0, 1}
g({0, 1}) = {1}

Here, the image of $g \circ f_A$ never takes values in $B_s$.

8. Oct 31, 2006

### CrankFan

I guess I said that because I wasn't thinking clearly. Thanks for the correction.