In my Algebra course the well-ordering property of $\mathbb{N}$ has been proven just by standard ("weak") induction, but I can't understand the proof given by the lecturer.
The proof proceeds as follows: let $S\subset \mathbb{N}$ a set with no minimum, we show that $S=\emptyset$. The claim is that if $S$ has no minimum and $ t \in \mathbb{N}$, then $S \cap t=\emptyset$; from this fact the conclusion would easily follow, since this proves that no $n \in \mathbb{N}$ can belong to $S$.
Now we suppose by contradiction that $S \cap t \ne \emptyset$; since $S \cap t \subset t$, $S \cap t$ is finite, since $S \cap t \subset \mathbb{N}$, $S \cap t$ is totally ordered. Hence there is $m=\min S \cap t$. We claim that this $m$ is in fact the minimum of the whole $S$, which is a contradiction.
We have to prove that $m \le x$ for every $x \in S$. If $x \in S \cap t$, this is obvious. If $x \notin S \cap t$, then $x \notin t$, hence either $x=t$ or $t \in x$ (by trichotomy property). Here begins the part I can't understand: the lecturer writes: " In this case [which one?] we have $m \in t$, so that $m \subsetneq t$, but also $ t \subsetneq x \implies m \subsetneq x \implies m <x$".
Can someone explain this last passage?
It seems to refer to the case where $t\in x$. But both cases can be treated the same way: in either case you have $t\subseteq x$ so since $m\in t$ you have $m\in x$, so $m<x$ by definition.