Prove that if $n \in \omega$ then $n \notin n$ Prove that if $n \in \omega$ then $n \notin n$. 
I'm trying to do it by induction. Consider $S=\{n \in \omega : n \notin n\}$


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*$0 \in S$: $\emptyset \notin\emptyset $

*$i\in S \Rightarrow s(i) \in S$: $i \notin i \Rightarrow \{i\} \notin \{i\}$ (if $\{i\} \in \{i\} \text{ then } i=\{i\}, \text{ therefore } i \in i$). I want show now that $\{i\} \notin\{i\} \Rightarrow s(i)=\{i\} \cup i \notin s(i)=\{i\} \cup i$. Suppose by absurdity that $\{i\} \cup i \in\{i\} \cup i$ then: 


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*or $\{i\} \cup i \in \{i\} \Rightarrow \{i\}\cup i = i \text{ then } i \in i$, contradition.

*or $\{i\} \cup i \in i$. I do not know to get in a contradition. I do not have the axim of regularity.

For any ordinal number $\alpha$, we cannot have $\alpha \in \alpha$. If so, $\in_\alpha=\{\langle a,b\rangle|\, a,b\in\alpha \,\wedge \, a\in b\}$ would not be a (strict) linear order of $\alpha$, just because we would have $\langle\alpha,\alpha\rangle\in\,\in_\alpha$, contrary to the fact that $\in_\alpha$ is irreflexive.
P.S. Note that by definition, for any ordinal $\alpha$, the natual membership order relation, $\in_\alpha$, must be a well-ordering of the set $\alpha$.
