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Show that collection of all sets with cardinality $\kappa\neq0$, is not set.

I'll state my approach and I need to see whether this idea is precise/precisable or not :

First let $K$ be the set with $card(K)=\kappa.$

Then let $C$ be the collection of all sets with cardinality $\kappa$. By contrary, suppose it's a set.

Now it can be shown that there's a bijection $\varphi:C\rightarrow F$, which $F$ is the collection of all one-to-one function like $f$, with $dom(f)=K$.

I don't know whether $F$ is now a set or not !

But if $F$ is a set, then I think $\displaystyle\bigcup ran(f)$ would be a good set to be considered and to construct set of all sets.

Every guidance (even a completely new approach) or correction to my idea is very appreciated.

Fardad Pouran
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    Any set is an element of a set of cardinality $\kappa$. Now let $C$ be as in the OP, and use the axiom of Union. – André Nicolas Jun 19 '14 at 07:17
  • @AndresCaicedo, I've visited that before, but i need help people focus on my idea to make it precise ! I never just wanted the final solution – Fardad Pouran Jun 19 '14 at 07:17
  • It was really straightforward. Thank You @AndréNicolas. – Fardad Pouran Jun 19 '14 at 07:22
  • You are welcome. – André Nicolas Jun 19 '14 at 07:26
  • I'm sorry, i just have one question: Can we say a collection $F$ is a set, whenever it's equinumerous to a set $C$ ? – Fardad Pouran Jun 19 '14 at 07:28
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    If you have a function $f:C\to F$ (where we don't know whether $F$ is a set yet, but $C$ is) which is a bijection (this is the closest you can get to equinumerous in your setting), then by the axiom of replacement, $F$ is contained in some set, and then by comprehension, $F$ is indeed a set. – Arthur Jun 19 '14 at 07:41
  • @Arthur, Yes it was better to use bijection phrase. Anyway thank you. Axiom of replacement is perfect ! – Fardad Pouran Jun 19 '14 at 07:46
  • I made a mistake in my phrasing, I meant to say that $F$ was a subclass of some set, not that it was contained in some set. Replacement yields a set $B$ which contains all elements of $F$, not a set which contains $F$. I was too late to edit. – Arthur Jun 19 '14 at 07:50
  • it's Ok. I got it, thanks – Fardad Pouran Jun 19 '14 at 07:55

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