The set $2^{\Bbb{N}}$ of all functions $f: \Bbb{N} \rightarrow \{0,1\}$ is not countable This question has been asked several times:
The set of all functions from $\mathbb{N} \to \{0, 1\}$ is uncountable?
The set of all functions from $\mathbb{N} \to \{0, 1\}$ is uncountable?
I just had some follow-up questions that I didn't understand. 
For the first one it says to supposed that $g: \Bbb{N} \rightarrow \{0,1\}$ is a bijection, define $b(k) = 1 - g(k)(k)$, then $b(k)$ is not in the image of $g(k)$ by the ways its constructed but I don't understand why $b(k)$ is an element of $2^{\Bbb{N}}$, I understand that $g(k)$ will evaluate to $0$ or $1$ but $g(k)(k)$ is not necessarily one or zero, it could be $g(20)(20) = 1*20 = 20$ for example.  
For the second question, it says to show that $F(f) = \{ n \in \Bbb{N} : f(n) = 1\}$ is a bijection $F(f): 2^{\Bbb{N}} \rightarrow P(\Bbb{N})$. I can see how the function is surjective, since by construction for every $n \in \Bbb{N} \ \exists f(n) \in 2^{\Bbb{N}}$ s.t. $f(n) = 1$, but I can not see how $F(f(n))$ is injective, if $f(n) = 0$, then that $n$ doesn't get mapped into $\Bbb{N}$, in this case wouldn't $F(f(n))$ just be the empty set, suppose for a certain $n \in \Bbb{N}$ there exists not function $f$, s.t. $f(n) = 1$, then wouldn't that n never be in $P(\Bbb{N})$. 
Thanks in advance
 A: I think you might be misunderstanding what "$2^\mathbb{N}$" means. It means the set of all functions from $\mathbb{N}$ to $\{0, 1\}$. Throughout your question, you seem to be conflating functions and numbers, and I think this is the source of the confusion.
Re: your first paragraph, we suppose $g$ is a bijection from $\mathbb{N}$ to $2^\mathbb{N}$; this is not what you have written (it should be obvious that there are no bijections from $\mathbb{N}$ to $\{0, 1\}$!). This means that - for each $k\in\mathbb{N}$ - $g(k)$ is not a number, but a function from $\mathbb{N}$ to $\{0, 1\}$. In particular, this means that it is not the case that "$g(k)$ will evaluate to $0$ or $1$" - $g(k)$ will be a function such that, when we plug a natural number into it, the result is either $0$ or $1$.
Re: your second paragraph, you are misunderstanding what sort of thing $F$ is. $F$ takes in a function from $\mathbb{N}$ to $\{0, 1\}$, and outputs a set of natural numbers; namely, $F(f)=\{n: f(n)=1\}$. For example, if $f$ is the function which maps every even number to $0$ and every odd number to $1$ (that is, $f$ is the parity function), then $$F(f)=\{n: f(n)=1\}=\{\mbox{odd numbers}\}.$$ So:


*

*To show that $F$ is injective, you need to show that if $F(f)=F(g)$, then $f=g$, for every $f, g\in 2^\mathbb{N}$. What does this mean? Well, you need to show that if two functions from $\mathbb{N}$ to $\{0, 1\}$ are equal to $1$ on exactly the same inputs, then they are the same function. Do you see why this is the case? (HINT: suppose $f$ is a function from $\mathbb{N}$ to $\{0, 1\}$; do you see why, if you know for which $n$ we have $f(n)=1$, you know all of $f$?)

*To show that $F$ is surjective, you need to show that for each set $A\subseteq\mathbb{N}$, there is some function $f$ such that $F(f)=A$. That is, $f$ should have the property that $f(n)=1$ when and only when $n\in A$. Do you see how to construct such a function? (HINT: what should $f(n)$ be when $n$ is not in $A$?)
