Showing that the natural numbers are totally ordered with respect to set membership Working with the usual set theoretic construction of the natural numbers, denoted $\omega$ for now.
I am trying to show that $\omega$ is totally ordered with respect to set membership, that is, $n<m$ if and only if $n\in m$ is a total ordering on $\omega$. I can show irreflexivity and transitivity, but I am struggling to show totality.
Here is an attempt using induction:
Fix $m\in\omega$. Define $W=\{n\in\omega:[n \neq m\Longrightarrow (n\in m)\vee (m\in n)]\vee n=m\}$.
If $m=\emptyset$, then $\emptyset\in W$. If $m\neq\emptyset$, then $\emptyset\in m$ so $\emptyset\in W$.
Suppose that $n\in W$ for some $n\in\omega$. Now either $m=n\cup\{n\}$ or $m\neq n\cup\{n\}$. If $m=n\cup\{n\}$, then $n\cup\{n\}\in W$.
Otherwise, $n\cup\{n\}\neq m$.
If $n=m$, then $m\in n\cup\{n\}$ so $n\cup\{n\}\in W$.
If $n\neq m$, then $n\in m$ or $m\in n$. If $m\in n$, then $m\in n\cup\{n\}$. If $n\in m$, then $n\subset m$ and $\{n\}\subseteq m$ so $n\cup\{n\}\subset m$.
This is as far as I can get. I have been unable to show that $n\subset m$ implies $n\in m$, so my proof comes to a halt at the last case. If I had that fact, then I would be able to say $n\cup\{n\}\in m$ which would complete the proof. Trying to show that fact inductively has felt more difficult than it should be. What else would I need to show in order to complete the proof, or what other route could I take to show that set membership totally orders $\omega$?
 A: I'm not completely sure if the following form of induction is compatible with your construction, but here we go.

Curiously, I needed the following lemma which is the exact converse of the goal. If you already know its proof, feel free to skip it.
Lemma: For natural numbers $m,n, n\in m\Rightarrow n\subset m$. (In fact $n$ can be any set)
Proof: Induct on $m$. Obviously true when $m$ is empty. If it holds for $m$, then consider any $n\in m\cup\{m\}$: either $n\in m$, which implies $n\subset m$ by induction hypothesis; or $n\in\{m\}$, which implies $n=m\subset m\cup\{m\}$. Hence done.

Claim: For natural numbers $m,n, n\subset m\text{ and }n\neq m\Rightarrow n\in m$.
Proof: Base case: vacuously true when $m=\emptyset$.
Induction step: assume the claim holds for $m$, we now prove it holds for $m\cup\{m\}$. Fix any $n\subset m\cup\{m\}$. We divide into two cases: $m\in n$ and $m\notin n$.
If $m\in n$, this is where the lemma is used: we immediately get $m\subset n$. But $m\in n$ means $\{m\}\subset n$, so $m\cup\{m\}\subset n$, hence $n=m\cup\{m\}$, a contradiction.
If $m\notin n$, then $n\subset m$, so $n\in m$ by induction hypothesis, in particular $n\in m\cup\{m\}$ and we're done.

By the way, I suggest you rewrite the definition of $W$ as $W=\{n\in\omega:(n\in m)\vee (m\in n)\vee (n=m)\}$ which is logically equivalent and shows the trichotomy clearer.
