How to show that the class of all sets of a particular cardinality ,say $h$ is not a set.
My argument:
I assume that I've shown the following lemma.
Lemma: If $X$ is an infinite set of cardinality $h$ , $a\ne x$ for every $x\in X$ then $\{a\}\cup X$ is of cardinality $h$ too i.e. adding new element to an infinite set doesn't change cardinality.
Let $h$ be an infinite cardinal number. Let $T_h=\{x:\text{x is a set of cardinality h}\}$. Suppose $T_h$ is a set.
Let $a$ be any set and consider any $t\in T_h$. Let $R=t\cup \{a\}$. Using the above lemma, $R\in T_h$.
So, $a\in R\subset \bigcup T_h$. As $T_h$ is a set then $\bigcup T_h$ is also a set. As $a$ was arbitrary set then we have, $a\in\bigcup T_h$ for every set $a$. Hence, the set $\bigcup T_h$ contains all sets which meas that the collection of all sets is a set, contradiction.
If $h$ is non-zero finite cardinal number. Define $T_h$ the same way as above. Let $a$ be any set, Let $A_1=a, A_{n+1}=A_n\times a$. Consider $A=\{A_1,A_2,...,A_h\}$. So, $A\in T_k$. Hence, $a=A_1\in \bigcup T_h=K$. So K is a set containing all sets $a$ since $a$ was arbitary,contradiction.
My main question is, Is there any flaws in my argument? If yes, point it out please.
If the argument is valid, Is there any standard proof to show the same fact?
Your argument is good for infinite $h$, but the claim is true for finite non-zero $h$ too. To see this you need to adjust your lemma to say that for any non-zero cardinal $h$ and for any set $a$, there is a set of cardinality $h$ that has $a$ as a member. To prove this note that if $x$ has $|x| = h > 0$ and $a \not\in x$, then $x = \{y\} \cup z$ for some $z$ with $y \not\in z$, but then $|\{a\} \cup z| = h$ and $a \in \{a\} \cup z$.