Class of all sets with cardinality of $\kappa$

Question

Let $\kappa$ be a nonzero cardinal number. Show that there does not exist a set to which every set of cardinality $\kappa$ belongs. I've seen some other posts like this and also this. But I didn't understand. Can someone give me a complete solution?

Answer 1

It seems you are confused by the other answers, so I will do my best to give as elementary of an answer as I can. Fix $\kappa=1$ for the most simplicity.

Let $X$ be a set. Then $\{X\}$ is a set of cardinality $\kappa=1$.

Let $\lambda$ be any cardinal. Then for any ordinal $\alpha<\lambda$, we have a singleton set $S_\alpha=\{\alpha\}$. In other words, the family $\{S_\alpha\}_{\alpha<\lambda}$ is a family of singleton sets of cardinality $\lambda$. (Note that if there were a set $X$ of all sets of cardinality $1$, then the family $\{S_\alpha\}_{\alpha<\lambda}$ would be a subset of $X$, since every element of it is a singleton.)

Now let $\lambda$ get arbitrarily large. Then the $\{S_\alpha\}_{\alpha<\lambda}$ constructed above is a set of singleton sets of arbitrarily large cardinality. This shows that if the set of all singletons were to exist, its cardinality would be bigger than every cardinal $\lambda$. This is impossible.

In general, the idea is to cook up a collection $\{S_\alpha\}$ of sets of the required size which can get arbitrarily large. (This is equivalent to providing an injective class function $\mathrm{Card}\to V_\kappa$, where $\mathrm{Card}$ is the (proper) class of all cardinals and $V_{\kappa}$ is the (a posteriori proper) class of all sets of cardinality $\kappa$.) This is not any harder in the case that $\kappa$ is bigger than $1$.

Answer 2

Let $A$ be the proper class of all sets, and let $B$ be the class of all sets of cardinality $\kappa$. Further, for every set $X$, let $K(X)$ be a set containing $\kappa$ many disjoint copies of $X$ (explicitly, you can fix a set $C$ with $|C|=\kappa$ and let $K(X)=\{X\times \{c\}: c\in C\}$). Then the map $X \mapsto K(X)$ is an injection $A \rightarrow B$. Since $A$ is not a set, neither is $B$.

See https://proofwiki.org/wiki/Set_is_Not_Element_of_Itself for a quick proof that $A$ is not a set.