Proving that for any cardinal number, there doesn't exist a set containing containing all sets of that cardinality. 
Let $\mathcal{K}$ be a nonzero cardinal number. Show that there does not exist a set to which every set of $\mathcal{K}$ belongs. 

Let the set containing all sets of cardinality $\mathcal{K}$ be $A$. Let $S\subset A$ such that $S$ contains all sets of $A$ that do not contain themselves. Now select $R\subset S$ such that $\text{card } R=\mathcal{K}$. It can now easily be proven that $R\notin A$.


*

*Is the argument above correct?

*How can we ensure that $\text{card }S\geq \mathcal{K}$, in order to create a subset $R$ of $S$ or cardinality $\mathcal{K}$?


Thanks 
 A: The argument you give is not correct. Even if you can prove that such $S$ exists, the fact that $R\subseteq S$ does not mean that $R\notin A$. It might be that $R\in A$ and we just have $R\in S\setminus R$.
The crux of your error is in the words "easily be proven".

Instead, show that there is no set of singletons (HINT: the axiom of union); then use this fact and the fact that given a non-empty set $A$ and an object $x$, there is a set $A_x$ such that $x\in A_x$ and $|A|=|A_x|$.
A: Hint: Assume for a contradiction that there is a collection $R$ of all sets having cardinality $\kappa$ (a non-zero cardinal). What can we say about $\bigcup R$?
A: Another proof would run thus: let K be any nonzero cardinal number, and suppose A is the set containing all sets of cardinality K. Let a be any set whatsoever. Then there is some set X such that a belongs to X and cardX = K. Thus UA would be a set containing all sets. But such a set does not exist. Consequently, A is not a set.
A: The proof uses a consequence of the axiom of regularity:
(1) $\neg (A\in B\wedge B\in A).$
Proof.
Suppose the opposite is true.Let $\mathcal{K}$ be a nonzero cardinal number, $S$ be the set of all sets of cardinality $\mathcal{K}$, $C$ be a set of cardinality $\mathcal{K}$.Then $C$ is nonempty and $C\in S$, and by (1) $S\not\in C$. Let $y\in C$, then there exists a bijection $f$ from $C$ to $(C\setminus\{y\})\cup\{S\}$:
$$
f(x)=
\begin{cases}
x,& \text{if $x\in C\setminus\{y\}$}\\
S,& \text{if $x=y$}
\end{cases} 
$$
So $|(C\setminus\{y\})\cup\{S\}|=|C|=\mathcal{K}$, then $(C\setminus\{y\})\cup\{S\}\in S$, which contradicts (1) because $S\in (C\setminus\{y\})\cup\{S\}$.
Consequently,the set $S$ does not exist.
