Does a function that maps from (almost) any natural number to its set of prime factors is surjective?

Question

Let $A$ be the set of all prime numbers, let $B = \mathbb{N} - \{0,1\}$, and let $f : B \to P(A)$ be the function that maps any $b \in B$ its set of prime factors. e.g, $f(70)=\{2,5,7\}$.

Is $f$ surjective over $P(A)$?

What I think: Yes, it's surjective, because we can take any prime-factors subset in $P(A)$, and the product of its members is a natural number.

However, I'm not quite sure, since $A$ itself is in $P(A)$, and $A$ is an infinite set, and I don't really know if we can represent some $x \in B$ with some "infinite quality" as well...Honestly, using Infinity is currently out of my knowledge (or this homework question).

I'd be glad to understand why is it\not surjective, and should I take into account the "infinite quality" of $A$ or\and $B$.

Answer 1

Your argument is fine: the map $f$ is not surjective since, for each $n\in B$, $f(n)$ is a finite subset of $A$, and therefore no infinite subset of $A$ belongs to the range of $f$. On the other hand, you proved correctly that every finite subset of $A$ belongs to the range of $f$.

Answer 2

Your concern is exactly right. If you consider the subset $A$ of $A$, i.e., $$ A = \{2, 3, 5, \ldots \} $$ then there's no natural number whose prime factorization corresponds to $A$, for every natural number has a unique factorization into a product of finitely many (possible zero!) prime factors, while the set $A$ is infinite. In particular, the prime factors of $n$ are all no greater than $n$, but there's always a prime greater than any natural number $n$.