Counting subsets of different sizes of a set Let $A$ be a non-empty set, and $\mathcal{P}^*(A)$ denote the power set of $A$ excluding empty set. There is a natural equivalence relation on $\mathcal{P}^*(A)$: 

for $S_1,S_2\in \mathcal{P}^*(A)$ (i.e. $S_1,S_2\subseteq A$), we say $S_1\sim S_2$ if there is a bijection between $S_1,S_2$.

The question I am concerning here is about the number of equivalence classes of $\mathcal{P}^*(A)$ under $\sim$. When $A$ is a finite set, then the number of equivalence classes is easily seen to be $|A|$; this can also be proved if $A$ is countable, by providing a bijection between $A$ and the equivalence classes of $\mathcal{P}^*(A)$. 
Question: What is the number of equivalence classes of $\mathcal{P}^*(A)$ under $\sim$ if $A$ is uncountably infinite?
The number of equivalence classes in $\mathcal{P}^*(A)$ is intuitively seems to be $|A|$, but I am not getting any direction in case $A$ is uncountably infinite. I didn't find such thing (question/fact) in some standard texts of set theory. 
 A: It depends on what $A$ is.  In fact, (assuming the axiom of choice) the number of equivalence classes of $\mathcal{P}^*(A)$ for $A$ uncountable can be any infinite cardinality at all, by choosing $A$ appropriately.  By definition, if $|A|=\aleph_{\alpha}$, the infinite cardinals that are less than or equal to $|A|$ exactly the cardinals $\aleph_\beta$ for $\beta\leq\alpha$.  So the number of different cardinalities of subsets of $A$ is $\aleph_0+|\alpha|$.  Here $\alpha$ can be any ordinal at all, so $|\alpha|$ can be any cardinal.
Note that in particular, for instance, ZFC cannot answer this question for $A=\mathbb{R}$, since ZFC cannot determine which aleph number $|\mathbb{R}|$ is (it cannot even determine the cardinality of the ordinal $\alpha$ such that $\aleph_\alpha=|\mathbb{R}|$; all that can really be said is that $|\alpha|\leq|\mathbb{R}|$).
A: If $|A|=\omega_\alpha$, there are $\omega+|\alpha|$ cardinal numbers less than or equal $|A|$ and hence $\omega+|\alpha|$ equivalence classes under $\sim$. Thus, the answer is $\max\{\omega,|\alpha|\}$. (I am assuming the axiom of choice here, as otherwise matters get really messy.)
