I'm having trouble developing my intuition about the need for a particular form of proof that Halmos presents in Naive Set Theory.
The theorem is on page 53 in the Arithmetic section is about the natural numbers:
In other words, if $n \in \omega$ then $n$ is not equivalent to a proper subset of $n$.
I'll present my attempt at a proof, and then Halmos's more detailed proof. My question is:
what error have I made in my proof? What assumptions have I made that are not allowed, and that require the Halmos version?
The theorem's truth seems like an obvious consequence of the definition of equivalence and an earlier theorem about the natural numbers:
Every proper subset of a natural number $n$ is equivalent to some smaller natural number (i.e., to some element of n).
It seems trivial to state that if $n$ were equivalent to a proper subset of $n$, then because of this latter theorem it would also be equivalent to a smaller $m$, which is not possible because there are strictly fewer elements in that smaller $m$ ($n \cap m\neq \emptyset$)
Side note - equivalence is defined on page 53:
Two sets $E$ and $F$ are called equivalent, in symbols $E \sim F$, if there exists a one-to-one-correspondence between them.