# 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 $R\notin A$.

1. Is the argument above correct?
2. 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

• How do you know that such an $R$ exists? – user642796 Mar 29 '14 at 5:56
• @ArthurFischer- I suppose if we can ensure that $\text{card }S\geq\mathcal{K}$, then we can use the Axiom of Choice (?) to select a $\mathcal{K}$-subset. – algebraically_speaking Mar 29 '14 at 6:02
• As a matter of fact, if you want to check the soundness of an argument, it's a good idea to pay special attention to words like "easily proven", "clear" or "trivial". – Najib Idrissi Mar 29 '14 at 6:57
• I think @nik gave an advice which is clearly a good advice (the proof of this claim is trivial). – Asaf Karagila Mar 29 '14 at 7:03
• – Martin Sleziak Mar 29 '14 at 8:11

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 $|A|=|A_x|$.

• Beat me to it... – goblin Mar 29 '14 at 6:19
• And I'm sleep deprived too. Tsk tsk tsk! :-) – Asaf Karagila Mar 29 '14 at 6:20
• A question for when you're less sleep-deprived: can we do it without the axiom of union, using in particular either separation or replacement? – goblin Mar 29 '14 at 6:27
• @user18921 Is my answer here what you meant by a proof using axiom of replacement? – Martin Sleziak Mar 29 '14 at 8:17
• @MartinSleziak, yep all is clear now. I just wanted to make sure that we weren't using the axiom of union in a fundamental way. – goblin Mar 29 '14 at 8:52

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 $\bigcup R$?

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.