Cardinality of the power set of the set of all primes Please show me what I am doing wrong...
Given the set $P$ of all primes I can construct the set $Q$ being the power set of P.
Now let $q$ be an element in $Q$. ($q = \{p_1,p_2,p_3,\ldots\}$ where every $p_n$ is an element in $P$.)
Now I can map every $q$ to a number $k$, where $k$ is equal to the product of all elements of $q$.  ($k = p_1p_2p_3\ldots$) (for an empty set $q$, $k$ may be equal to one)  
Let the set $K$ consist of all possible values of $k$.
Now because of the uniqueness of the prime factorization I can also map every number $k$ in $K$ to a $q$ in $Q$. (letting $k=1$ map to $q=\{\}$)
Thus there exists a bijection between $Q$ and $K$. But $K$ is a subset of the natural numbers which are countable, and $Q$, being the power set of $P$, needs to be uncountably infinite (by Cantor's theorem), since $P$ is countably infinite.  
This is a contradiction since there cannot exist a bijection between two sets of different cardinality. What am I overlooking?

Some of the elements of $Q$ are infinite sets, and for these your proposed $k$ is not well-defined.
The subset $Q'$ of $Q$ where every element of $Q'$ is a finite set of primes is indeed countable, and your argument is valid for this subset.
