Cardinality of the set of all subsets of $E$ equipotent to $E$ I'm trying to prove the following statement (an exercise in Bourbaki's Set Theory): 
If $E$ is an infinite set, the set of subsets of $E$ which are equipotent to $E$ is equipotent to $\mathfrak{P}(E)$. 
As a hint, there is a reference to a proposition of the book, which reads: 
Every infinite set $X$ has a partition $(X_\iota)_{\iota\in I}$ formed of countably infinite sets, the index set $I$ being equipotent to $X$. 
I don't have any idea how that proposition might help. 
If $E$ is countable, then a subset of $E$ is equipotent to $E$ iff it is infinite. But the set of all finite subsets of $E$ is equipotent to $E$. So its complement in $\mathfrak{P}(E)$ has to be equipotent to $\mathfrak{P}(E)$ by Cantor's theorem. Hence the statement is true if $E$ is countable. Unfortunately, I don't see a way to generalize this argument to uncountable $E$.
I'd be glad for a small hint to get me going. 
 A: Using the axiom of choice, every infinite set $X$ can be divided into two disjoint sets $X_0\sqcup X_1$, both of which are equinumerous with $X$. (Just well-order $X$, and take every other point in the enumeration.) 
Now, consider all sets of the form $X_0\cup A$ for any $A\subset X_1$. There are $2^X$ many such $A$ and hence $2^X$ many such sets, and each is equinumerous with the original set $X$. So we've got $2^X$ many sets as desired, and there cannot be more than this, so this is the precise number. 
Incidently, the stated answer to this question does in fact depend on the axiom of choice, since it is known to be consistent with $ZF+\neg AC$ that there are infinite Dedekind finite sets, and these are not equinumerous with any proper subsets of themselves. So for such an infinite set $X$, there would be only one subset to which it is equinumerous. 
A: Another approach would make use of the fact that $\kappa_1 + \kappa_2 = \max(\kappa_1,\kappa_2)$ when $\kappa_1,\kappa_2$ are cardinals at least one of which is infinite. From this it follows that, for any subset of $S \subset X$, either $S$ or its complement has cardinality $|X|$. Since more straightforward cardinal arithmetic shows there are as many complementary pairs of subsets as there are subsets of $X$, we get the result.
By the way, JDH's approach could allow you to show the (slightly) stronger statement that there are $2^X$ sets $S \subset X$ with both of $S$ and $X - S$ equipotent to $X$. Simply write $X$ as a disjoint union $X = X_0 \cup X_1 \cup X_2$ with all parts equipotent to $X$ and consider subsets of the form $X_0 \cup A$ where $A \subset X_1$. 
An interesting follow-up question might be to see if you can produce $2^X$ sets $S \subset X$ so that $|S|=|X|$, but $|X-S| < |X|$. 
