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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

algebraically_speaking
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  • How do you know that such an $R$ exists? – user642796 Mar 29 '14 at 05:56
  • @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 06:02
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    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 06:57
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    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 07:03
  • Similar questions: http://mathoverflow.net/questions/34098/is-the-isomorphism-class-of-a-fixed-cardinality-a-set, http://math.stackexchange.com/questions/3815/why-does-the-set-of-all-singleton-sets-not-exist and http://math.stackexchange.com/questions/21973/class-of-sets-of-a-given-infinite-cardinality – Martin Sleziak Mar 29 '14 at 08:11

4 Answers4

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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 $|A|=|A_x|$.

Asaf Karagila
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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 $\bigcup R$?

goblin GONE
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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.

Babalu
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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.

Imperton
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