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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

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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 $|A|=|A_x|$.

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  • $\begingroup$ Beat me to it... $\endgroup$ – goblin Mar 29 '14 at 6:19
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    $\begingroup$ And I'm sleep deprived too. Tsk tsk tsk! :-) $\endgroup$ – Asaf Karagila Mar 29 '14 at 6:20
  • $\begingroup$ 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? $\endgroup$ – goblin Mar 29 '14 at 6:27
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    $\begingroup$ @user18921 Is my answer here what you meant by a proof using axiom of replacement? $\endgroup$ – Martin Sleziak Mar 29 '14 at 8:17
  • $\begingroup$ @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. $\endgroup$ – goblin Mar 29 '14 at 8:52
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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 $\bigcup R$?

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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.

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