Cantor's theorem shows us that the power set of the natural numbers is uncountably infinite. But today (and before remembering Cantor's proof) I was trying to prove the incorrect version: that the power set of the natural numbers is actually countably infinite. My work obviously has to be wrong, so I'm hoping that someone can spot a false implication.
So, to prove that the set of natural numbers is of the same cardinality as the power set of the natural numbers, I intend to show that a bijection $f: \mathcal{P}(\mathbb{N}) \to \mathbb{N}$ exists.
Now, it suffices for $f$ to be injective; Afterwards, we can just sort the image of $f$ and number the results starting from 1 to make it surjective as well.
So here is my $f$: For any set $A$ consisting of natural numbers, $f(A) = p_1^{\delta(1)}p_2^{\delta(2)}p_3^{\delta(3)}\cdots$ where $p_i$ is the $i$-th prime number and $ \delta(x) = \begin{cases} 1, & x\in A\\ 0, & x\not \in A \end{cases} $
So, as an example, if $A = \{1,3,4\}$, then $f(A) = 2^1\cdot5^1\cdot7^1 = 70$. This function should be one-to-one, since each integer has a unique prime factorization. Therefore, any set in $\mathcal{P}(\mathbb{N})$ will be associated with a unique natural number. We sort these natural numbers are re-number them starting with 1, producing a bijection between the power set of naturals and the naturals, meaning that the power set of the naturals is actually countably infinite.
But this is wrong. Where is the weakness in my proof?