Show the set of all functions from $\mathbb{N}$ to $\{0, 1\}$ is uncountable using a contradiction. This is what I have written: 
By contradiction, assume it is countable. Write $S=\{\text{all functions } \mathbb{N} \rightarrow \{0,1\} \}$. Then, we can find a bijection $\mathcal{H}: S \rightarrow \mathbb{N}$. Now, I would like to check how to incorporate Cantor's method to find the contradiction. Would it be right to think of each function as a binary representation (because they map to either $0$ or $1$)? So, I will write 
$f(1) \mapsto a_{11}a_{12}a_{13}...$
$f(2) \mapsto a_{21}a_{22}a_{23}...$
$f(3) \mapsto a_{31}a_{32}a_{33}...$
where $a_{ij} \in \{0,1\}$.
and so on. So for example, $f(1)$ has input any natural number, so it will spit out either a $0$ or a $1$, and I have written all possibilities in a list. 
Then, I define a function in $S$ that is $0$ if a string value is $1$ and $1$ if the string value is $0$. 
I have one more question: what is the meant by the notation $\{0,1\}^{\mathbb{N}}$?
Thank you.
 A: You can get contradiction by defining another $f(x)$ that can never be in list as follow:
$f(x)=b_{11}b_{22}\cdots, \hspace{2 mm} b_{ii}\neq a_{ii}$
Note that $f(x)$ can not be in the list anyway because if there is a $j$ such that $f(j)=f(x)$, then $b_{jj}=a_{jj}$, a contradiction. So no $1-1$ mapping is possible. 
$\{0,1\}^{\mathbb{N}}$ is the set of all functions that map $\mathbb{N}$ to $\{0,1\}$.
A: Your idea is correct, Using the binary representation actually makes the explanation way easier. You can always get a binary number that is not in the list and obtain a contradiction using cantor's diagonal method
A: This is a classic application of Cantor's argument, first instead of thinking about functions lets just think about sequences of 0's and 1's. That is if $f: \mathbb{N} \to \{0,1\}$ and $f(1)=0, f(2)=1,f(3)=1, \ldots$ I will just write the sequence $011\ldots$. Suppose that the set is countable and list the elements
$$a_{1}=a_{11}a_{12}a_{13} \ldots\\
a_{2}=a_{21}a_{22}a_{23} \ldots\\
a_{3}=a_{31}a_{32}a_{33} \ldots\\
\vdots
$$
Now we construct an element $b$ that we have not listed by taking
$$b_{n}=1 \text{ if }a_{nn}=0 \text{ or }b_{n}=1 \text{ if }a_{nn}=0$$
$b_{n}$ is not in the list because it differs from $a_{n}$ in the $n$th position for all $n$.
