I'm sorry I still don't get it.
I don't understand what (c) and (d) are and how do their elements look like.
For c). The characteristic function is $\displaystyle {1_A}(x) = \left\{ {\begin{array}{rl} {1,}&{x \in A} \\ {0,}&{x \notin A} \end{array}} \right.$.
Look at the collection $\displaystyle \left\{1_{\{k\}}:k\in\mathbb{N}\right\}$
For d) define $\displaystyle f_k=\{(0,k),(1,k+1)\}$. Now for each $\displaystyle n\in\mathbb{N}$ the function $\displaystyle f_n$ mapps $\displaystyle \{0,1\}\to\mathbb{N}$.
If you need to visualize them to help you see the problem better, try visualizing them in terms of the suggestion made by DrSteve earlier in this thread.
(c) Functions from $\displaystyle \mathbb N$ to $\displaystyle \{0,1\}$ are essentially infinite sequences of $\displaystyle 0\text{'s}$ and $\displaystyle 1\text{'s}$ – e.g. $\displaystyle (1,0,1,0,1,\ldots),$ $\displaystyle (0,0,1,0,0,1,\ldots),$ $\displaystyle (1,1,1,1,\ldots),$ etc. Can you visualize an infinite list of such infinite sequences?
(d) Functions from $\displaystyle \{0,1\}$ to $\displaystyle \mathbb N$ can be regarded as sets of two ordered pairs of the form $\displaystyle \{(0,m),(1,n)\}$ where $\displaystyle m,n$ are natural numbers. Can you visualize an infinite list of such sets?
To answer your question about irrational, algebraic and transcendental numbers:
An algebraic number is the solution of a polynomial equation with integer coefficients. For example, the fraction m/n is a solution of nx-m=0. This shows that all rational numbers are algebraic numbers. But there are also lots of algebraic numbers that are irrational. For example, the polynomial equation x^2-c=0 gives two algebraic numbers that are irrational whenever c is not a perfect square. It is not hard to show that the set of algebraic numbers is countable. Therefore, there are uncountably many transcendental numbers.
Thanks DrSteve, and thank you all for your time and your patience,
I believe that the source of confusion came along with something I read somewhere, where it said that "all the Transcendental numbers are not Algebraic" therefore I concluded that all the Irrationals should be Transcendental however, now I understand my mistake since $\displaystyle {x^2}-2=0$ gives two irrational numbers that are not transcendental since they solve the last polynomial equation.
by the way I figured out (c) and (d):
(c) $\displaystyle \{f\in\{0,1\}^\mathbb{N}|f(n)=\begin{cases}{1, n\in\{n\}} \\ {0, n\notin\{n\}}\end{cases}\}$.
This should generate a set with the following sequences: $\displaystyle \{(1,0,0,0,...),(0,1,0,0,...),(0,0,1,0,...),(0,0,0 ,1,...),\}$
(d)$\displaystyle \{f\in\mathbb{N}^{\{0,1\}}|f(\{n\})=\{(0,n),(1,n+1 )\}\}$.
This should generate a set which contains sets of the form: $\displaystyle \{\{(0,1),(1,2)\},\{(0,2),(1,3)\},\{(0,3),(1,4)\}, ...\}$
Is that correct?
The idea is correct, but the notation is a bit off.
If $\displaystyle f(n)=\begin{cases}{1, n\in\{n\}} \\ {0, n\notin\{n\}}\end{cases}\}$, then f(n) = 1 for all n because $\displaystyle n\in\{n\}$ for all n
There is a type error here because if $\displaystyle f\in\mathbb{N}^{\{0,1\}}$, then f cannot be applied to {n}.
Specifying a countable subset of some set X is more easily done by giving a sequence of elements of X (i.e., a function from $\displaystyle \mathbb{N}$ to X) than by using a set-builder notation {x ∈ X | ...}. So a sequence of function $\displaystyle f_0, f_1,\dots$ in $\displaystyle \{0,1\}^{\mathbb{N}}$ can be defined by
$\displaystyle f_k(n)=\begin{cases}1& n=k \\0& n\ne k\end{cases}$
A sequence of function $\displaystyle f_0, f_1,\dots$ in $\displaystyle \mathbb{N}^{\{0,1\}}$ can be defined by
$\displaystyle f_k=\{(0,k),(1,k+1)\}$
The notation
$\displaystyle f_k(n)=\begin{cases}1 & n=k \\0 & n\ne k\end{cases}$
means (in this context) that f is a function from $\displaystyle \mathbb{N}\times\mathbb{N}$ to {0, 1}: f(k, n) = 1 or 0 depending on whether n = k. We write f(k, n) as $\displaystyle f_k(n)$. The arguments (k, n) of f can be any pair from $\displaystyle \mathbb{N}\times\mathbb{N}$.