For two sets A and B. Suppose|2^A| = |2^B| (cardinality of power sets of A and B), does |A|=|B| ?


(It is easy to see that|A|=|B| if we assume generalized continuity hypothesis. Do we have the same result without it?)



Equality of Cardinality of Power Set 
The answer is that if the axioms of set theory are consistent, then you cannot prove that conclusion. Although it seems very reasonable to expect that a smaller
set must have strictly fewer subsets, which is another way
of stating your property, in fact this property is
independent of ZFC.


(The fact that many people find this surprising is the reason I posted [this
answer](https://mathoverflow.net/questions/1924/what-are-some-reasonable-sounding-statements-that-are-independent-of-zfc/#6594)
to the MO question requesting examples of
reasonable-sounding statements that are independent of
ZFC.)


You are asking whether the continuum function
$\kappa\mapsto 2^\kappa$ is injective, and it turns out that if ZFC is consistent, then this assertion is neither provable nor refutable in ZFC.


On the one hand, the property is relatively consistent with
ZFC, as you observe, since it follows easily from the GCH.


On the other hand, it is known by the method of [forcing](http://en.wikipedia.org/wiki/Cohen_forcing) to be relatively consistent that the property fails. Specifically, in Cohen's original model of
$\text{ZFC}+\neg\text{CH}$, he
forced over a model of GCH to add $\omega\_2$ many Cohen
reals, and the result is $2^\omega=2^{\omega\_1}=\omega\_2$ in his model,
which violates your property. Cohen's model has an uncountable set $B\subset\mathbb{R}$ that has the same number of subsets as the countable set $A=\mathbb{N}$. This is usually one of the
first nontrivial forcing arguments that set-theorists
learn, when first exposed to the technique, and when teaching this, I invariably find the situation somewhat magical.


It is also a consequence of [Martin's
Axiom](http://en.wikipedia.org/wiki/Martin%27s_axiom) that
$2^\kappa=2^\omega$ for all $\kappa\lt\frak{c}$, and if one
has MA plus $\neg\text{CH}$, which is known to be
relatively consistent by the forcing method, then there are
again counterexamples to the requested property.


Meanwhile, one can show by forcing that the injectivity of
the continuum function is not equivalent to GCH, since by
[Easton's
theorem](http://en.wikipedia.org/wiki/Easton's_theorem),
one can find a forcing extension (of any model of GCH) in
which $2^{\aleph\_n}=\aleph\_{n+2}$ for every natural number
$n$, and otherwise the GCH holds. Such a model exhibits the
desired injectivity property, but does not satisfy GCH. One
can use Easton's theorem more generally to make even more
extravagant violations of GCH, while still ensuring an
injective continuum function.


