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 $\Bbb{N}$.

Answer 3

First we prove Cantor's theorem $f:A\rightarrow \text{pow}(A)$ is not surjective.\ The approach is show that there exists a set $T\in \text{pow}(A)$ that is not in the range of $f(A)$. In other words, set T in the codomain is not mapped from the domain. For each $a \in A$ its mapping to subset $f(a)$ could either contain $a$ as a set member or doesn't. For example $f:1\rightarrow \{2,9\}$ doesn't but $f:2\rightarrow \{1,2,9\}$. Thus to construct such a T we simply let it be equal $A_f$ which we define to be $\{a\in A_f | a\notin f(a)\}$. Once again, this is well defined subset of A and hence a member of pow(A)

Why? Consider pow({1,2,3}). Say if 1 is not a member of f(1) and 2 is not a member of f(2), surely (1,2) is neither a member of f(1) or f(2).Thus it is obviously not equal to f(1) or f(2). If 3 is a member of f(3), we simply don't include it in T because there is a chance that it equals f(3) given that it contains 3. For example f(3) could be {1,2,3},{1,3},{3} etc. This implies $\forall a \in A, f(a)\neq A_f $ which is exactly what we want.

More precisely, consider the case if $\exists a \in A_f$ where $f(a)=A_f$ implies $a \in f(a)$ which is a contradiction to the definition of $A_f$. Now consider if $\exists a \notin A_f$ where $f(a)=A_f$. However that is impossible as $a \in f(a)$. Therefore $f(a)$ contains $a$ which is not a member of $A_f$. Therefore proposing $f(a)=A_f$ leads to a contradiction. Now let $A$ be $\mathbb{N}$ and we are done