Do sets whose power sets have the same cardinality, have the same cardinality? [duplicate]

Possible Duplicate:
power set cardinal equality

Let $X$ and $Y$ be sets, and suppose that $|\mathscr{P}(X)| = |\mathscr{P}(Y)|$ (where $\mathscr{P}$ denotes the power set).

Does it follow that $|X|=|Y|$?

Remark: It's obviously true for finite sets, as $2^m=2^n$ implies $m=n$.

-

marked as duplicate by Asaf Karagila, Rudy the Reindeer, Martin Sleziak, Chris Eagle, J. M.May 10 '12 at 12:55

For all $A$, $|P(A)|=2^{|A|}$. – Salech Alhasov May 10 '12 at 12:34
@SalechAlhasov: How is that useful here? – Chris Eagle May 10 '12 at 12:49
@ChrisEagle: I really don't know. In future, I'll try to avoid with such comments. – Salech Alhasov May 10 '12 at 13:03
Thanks for pointing out the previous question. – Mark Grant May 10 '12 at 13:06

This assertion is not provable in ZFC (if ZFC is consistent, that is). If ZFC is consistent then it has a model in which this assertion is true, for example in Godel's constructible universe. However by forcing we can make the following true:

$$2^{\aleph_0}=2^{\aleph_1}=\aleph_2$$

Which is a clear contradiction to the statement that $\kappa\mapsto 2^\kappa$ is injective (this function is called the continuum function).

It is worth mentioning Easton's theorem which tells us that we can pretty much modify the behavior of the continuum function as long as we preserve the basic properties of this function (for example, $\kappa<\lambda$ then $2^\kappa\leq 2^\lambda$).

This means that we can have the assertion that the continuum function is injective fail everywhere; we can have it failing unboundedly high; we can have it true unboundedly high AND fail unboundedly high as well...

-
Y U answer duplicate? – Rudy the Reindeer May 10 '12 at 12:49
@MattN.: I had a few minutes to kill before going to teach. Doing that was better than doing nothing. – Asaf Karagila May 10 '12 at 12:50