Proof that no natural number can be subset of any of its elements. in order to proof that:
"no natural number can be subset of any of its elements" as showed in Halmos's book, is used the following passage:
if $n$ contains $n$ then $n$ can not be member of $n$ , and $n$ is a natural number, using the definition of number to be a collection of sets, with $0$ to be the empty set, and $n^{+}$ = $n$ (union) ${n}$, where $n^{+}$ is the sucessor of n.
 A: Specifically, Halmos’s argument is as follows. Let $S$ be the set of all natural numbers $n$ that are not subsets of any of their elements. $0$ has no elements, so it certainly isn’t a subset of one of its elements, and therefore $0\in S$. Now suppose that $n\in S$; we want to show that $n^+\in S$ (where $n^+=n\cup\{n\}$), so that we can appeal to the induction principle to conclude that $S=\omega$ and hence that no natural number is a subset of any of its elements.
By hypothesis $n\in S$, so $n$ is not a subset of any element of $n$. That is, for each $m\in n$ we know that $n\nsubseteq m$. But certainly $n\subseteq n$, so $n$ cannot be any of the elements $m\in n$, i.e., $n\notin n$. Now $n^+=n\cup\{n\}$, so $n\in n^+$, and therefore $n^+\nsubseteq n$: $n$ is an element of $n^+$ but not of $n$. Can $n^+$ be a subset of some element $m\in n$? No: $n\subseteq n^+$, so if $n^+$ were a subset of $m$, we’d have $n\subseteq m$, contradicting the hypothesis that $n\in S$. Since $n^+=n\cup\{n\}$, the members of $n^+$ are $n$ and the members of $n$. We’ve just seen that $n^+$ is not a subset of any of these sets, so by definition $n^+\in S$. This shows that the set $S$ is inductive and hence is equal to $\omega$: every natural number belongs to $S$, as desired.
