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Let's say $\Im = \{$ $A$ $|$ $A$ is an inductive set $\}$. With that, I mean $\Im $ is the "set" of all inductive sets. When I was reading induction is a right way of proving, the author proved it in a different way and said that $\Im$ is not a set. I searched and tried to prove that $\Im$ is not set but I couldn't get anything.

Thanks for any help.

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    $\begingroup$ Which definition of inductive set do you refer to? $\endgroup$ – Hagen von Eitzen Nov 15 '13 at 16:25
  • $\begingroup$ Ask if $J$ is inductive, and try to get a contradiction, both when it's inductive, and when it isn't inductive. $\endgroup$ – Noy Soffer Nov 15 '13 at 16:31
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    $\begingroup$ @NoySoffer $J$ is not an inductive class because $\omega \in J$ but $\omega \cup \{\omega\} = \omega+1 \notin J$. Even if we assume toward a contradiction that $J$ is a set, I don't see how the extra assumption "$J$ is not inductive" would help obtain a contradiction. $\endgroup$ – Trevor Wilson Nov 15 '13 at 18:55
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HINT: Show that every set is a member of an inductive set.

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If we call a set $A$ inductive if $a\in A$ implies $a\cup\{a\}\in A$, then among others all limit ordinals are inductive. The class $L$ of all limit ordinals is proper. For if it were a set, then $\bigcup L$ would be the class of all ordinals which is proper (because otherwise it would itslef be an ordinal - contradiction).

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