Let $k\neq 0$ cardinal number.How can we prove that doesn't exist the set of all elements with cardinal number $k$?I try to prove it for the cases : (i) $k$ is finite (ii) $k=\aleph_0$ but i can't use that ever infinite set has infinite countable subset for this case : $k$ uncountable infinite .How can we solve this problem with an alternative way?
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How did you prove it in cases i) and ii)? – saulspatz May 16 '20 at 12:25
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In first case let that this set exists,let K.We have $k \neq 0$ so $k=Sn$ with $n\in \omega$ so for every set $A$ we have that ${A}\cup n\in K$.Now from the union axiom $A\in \bigcup K$ and we khow that doesn't exist the set of all sets.With the same method can you show the second case. – KBi7700 May 16 '20 at 12:37
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If $k$ is infinite, let $S$ be any set of cardinality $k$. Then for every set $A$, $S\cup{A}\in K$. Make the same argument as you did in case i). In future, it will be to your advantage to put clarifications in the body of the question, not in the comments. Many people browsing questions will vote to close for lack of detail without looking at the comments. – saulspatz May 16 '20 at 12:56
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Thank you very much i will fix it. – KBi7700 May 16 '20 at 12:59