What is wrong with my "disproof" of Cantor's Theorem?

I cannot figure out what is wrong:

We will attempt to show that $\mathcal{P} (\mathbb{N})$ is countable. We use the following corollary from Rudin's Principles of Mathematical Analysis, p. 29:

Suppose $A$ is at most countable, and, for every $\alpha\in A$, $B_{\alpha}$ is at most countable. Put

$$T=\bigcup_{\alpha \in A}B_{\alpha}$$

Then $T$ is at most countable.

"Proof" 1:

Let $A = \mathbb{N}$ and for every $\alpha \in A$ let $B_{\alpha}=\{S \in \mathcal{P} (\mathbb{N})| \text{the sum of the elements of } S \text{ is } \alpha \}$. $A$ is countable and for every $\alpha \in A$, $B_{\alpha}$ is finite. Therefore

$$\bigcup_{\alpha \in A}B_{\alpha}$$

is countable. But $\displaystyle \bigcup_{\alpha \in A}B_{\alpha}=\mathcal{P} (\mathbb{N})$, so $\mathcal{P} (\mathbb{N})$ is countable.

"Proof" 2:

Let $A= \mathbb{N}$ and for every $\alpha \in A$ let $B_{\alpha}=\{ S \in \mathcal{P} (\mathbb{N}): |S| = \alpha \}$. I think that I can show by induction (if requested) that for each $\alpha \in A$, $B_{\alpha}$ is countable. Thus

$$\bigcup_{\alpha \in A}B_{\alpha}$$

is countable. But again, $\bigcup_{\alpha \in A}B_{\alpha} = \mathcal{P} (\mathbb{N})$

Solution 1:

In both your "proofs", it is not true that $\displaystyle \bigcup_{\alpha \in A}B_{\alpha}=\mathcal{P} (\mathbb{N})$. Indeed, if $S\in\mathcal{P}(\mathbb{N})$ is any infinite set, then $S$ is not in any $B_\alpha$ (by either definition).

What both your arguments show correctly is that the set of all finite subsets of $\mathbb{N}$ is countable.