Proving Power Set of $\mathbb N$ is Uncountable

Question

I'm getting hung up on a proof that I remember being fairly easy... Showing that the power set of $\mathbb N$ is uncountable. Supposing it's countable, say $A=\{A_1,...\}$, we choose a set $B$ composed of elements $b_i$, where $b_i\notin A$. How to we guarantee that $B$ isn't the naturals itself, an element of its own power set? Thanks!

Answer 1

It does not matter if $B$ is empty, $\Bbb N$, or anything. The point is that $B$ is not one of the $A_i$'s.

Just to clarify, $B$ should be defined as $\{i\in\Bbb N\mid i\notin A_i\}$, rather than involving $b_i$'s.

Answer 2

It is better proved in this way.

Enumerate all natural number as $a_1,a_2,\cdots,a_n\cdots$. Suppose $A=\{a_{n_i}|\:i\in\Bbb{N}\}$ be a subset of $a_n$.

Now let $$ r=c_1c_2\cdots c_n\cdots\quad\text{where }c_n=\begin{cases}1, \quad\text{if }a_n\in A\\0,\quad\text{if }a_n\notin A\end{cases} $$ Then it is clear that each $A$ corresponds to a $r$. We prove that $r$ can not be enumerated.

If not, let $r_k=c_{1k}c_{2k}\cdots c_{nk}\cdots$ be an enumeration, where $k\in\Bbb{N},\:c_{nk}\in\{0,1\}$. Let $r'=d_{1}\cdots d_{n}\cdots, \:d_{n}\ne c_{nn}$. Then $\forall k\:r'\ne r_k\:$ for $d_k\ne r_{kk}$, which means $r'$ is not in the enumeration.

The contradiction means $r$ can not be enumerated and must be uncountable, and so is all the number of $A$, or power set of $\Bbb{N}$.