Prove that the cardinality of $\{0,1\}^R$ is not equal to that of $R$ From what I understood $\{0,1\}^R$ is the set of all functions from $\{0,1\}$ to $R$. I would be happy not only for the proof but a good and maybe simplified explanation of the concept of Aleph's and its relation to $\{0,1\}$ I've read many questions on mathSE on this topic and been to lectures. There's something not clicking with this whole set theory class! 
Thanks
 A: Note: As Hagen noted, $\{0,1\}^R$ is the set of functions $R\to\{0,1\}$ not the other way around. The set of functions $\{0,1\}\to R$ does have the same cardinality as $R$ whenever $R$ is infinite (and assuming the axiom of choice).

With this correction, what you need is Cantor's theorem: Let $R$ be any set, and let $f$ be any function from $R$ to $\{0,1\}^R$. Then $f$ is not surjective.
Proof. Define $g\in\{0,1\}^R$ by $$g(r)=1-\bigl[f(r)\bigr](r).$$ Then $g$ is not in the image of $f$. Namely, if there is an $r_0$ such that $g=f(r_0)$, then
$$g(r_0)=1-f(r_0)(r_0) = 1-g(r_0)$$
which cannot be true no matter whether $g(r_0)=0$ or $g(r_0)=1$.
A: There is an error that I suspect is widespread among mathematicians: that "aleph"s have much to do with things like $\{0,1\}^R$.
$\aleph_0$ is the cardinality of the set of all finite ordinals.
$\aleph_1$ is the cardinality of the set of all countable ordinals.
$\aleph_2$ is the cardinality of the set of all ordinals of cardinality $\aleph_1$.
Those definitions have been standard since Cantor established them.
The proof that the set of all countable ordinals is uncountable is different from the proof that $\{0,1\}^S$ must have a larger cardinality than that of $S$.
So "alephs" are a different topic from Cantor's theorem that $|\{0,1\}^S|>|S|$ for every set $S$.
That $2^{\aleph_0}$ is even comparable with any "alephs" besides $\aleph_0$ cannot be proved in Zermelo--Fraenkel set theory.
The set $\{0,1\}^S$ is the set of all fuunctions from $S$ into $\{0,1\}$.  Suppose for each $s\in S$ we have a corresponding $f\in\{0,1\}^S$.  Then for each $s\in S$ let
$$
g(s) = 1-f_s(s).
$$
Then $g\in\{0,1\}^S$ and for each $s\in S$, $g\ne f_s$.  Therefore, every way of assigning to each $s\in S$ a function $f_s\in\{0,1\}^S$ must fail to include all such funcntions.  Hencec $|S|<\{0,1\}^S$.
This gives rise to the "beths":
$$
\begin{align}
\beth_0 & = \aleph_0 \\
\beth_1 & = 2^{\beth_0} \\
\beth_2 & = 2^{\beth_1} \\
& {}\  \vdots
\end{align}
$$
