On cardinality of binary representation of natural numbers

Question

Let $\mathbb N = \{0, 1, 2, \cdots\}$, $A = \{2^n: n \in \mathbb N\} = \{1, 2, 4, 8, ...\}$ and $B = \mathscr P(A) = \{ \emptyset, \{1\}, \{2\}, \{1, 2\}, \{4\}, \{1, 4\}, \cdots \}$. Then $|A| = |\mathbb N| = \aleph_0$ and $|B| = 2^{|A|} = 2^{\aleph_0}$. But isn't $$f: B \to \mathbb N, S \mapsto \sum_{n \in S} n$$ an injection, implying $|B| \leq |\mathbb N|$?

Answer 1

$B$ also contains subsets which are infinite, while a natural number only has finitely many $1$'s in its binary representation. Specifically, your $\sum n$ might fail to exist.