If $n$ is a finite ordinal, either $n=\emptyset$, or there exists a finite ordinal $m$ such that $n=m \cup\{m\}$

Question

I have a difficulty with this statement :

If $n$ is a finite ordinal, either $n=\emptyset$, or there exists a finite ordinal $m$ such that $n=m \cup\{m\}$.

My professor's proof is totally unclear. I did the following, using the following definitions.

A transitive set is a set a such that $\forall x, y$, if $x \in y$ and $y \in a$, then $x \in a$.

An ordinal is a transitive set which is well-ordered for the relation $\in$.

We define $m=\bigcup n=\bigcup_{z \in n}z$, using the axiom of reunion.

First step : $m$ is transitive : Let $x,y$ such that $x \in y$ and $y \in m$ ; $y \in m$ implies $\exists z\in n, y\in z$. Since $n$ is transitive, $y\in z$ and $z\in n$ implies $y \in n.$ Then, $x\in y$ and $y \in n$ implies $x \in \bigcup n =m$. Therefore m is transitive.

Then I suppose step 2 would be to show that $m$ is finite, and well-ordered by $\in$, and step 3 would be to finally show the double inclusion $n\subset m \cup\{m\}$ and $m \cup\{m\}\subset n$.

My questions are : how to show that $m$ is finite ? (The rest is ok I think). Is there a quick way to show that $m$ is well-ordered ? I have a proof the fact that $\in$ is a total order (cf bottom), but the well-ordering seems long to prove. Last question : my professor seems to say that we can use $\subset$ or $\in$ indifferently for the proofs, is that so, and if so, why ?

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Proof for the total ordering : Let $x,y \in m$. $\exists z_x,z_y \in n, x\in z_x, y\in z_y$. Since $n$ is an ordinal, it is totally ordered by $\in$, therefore either $z_x\in z_y$, or $z_y\in z_x$. Without loss of generality, suppose $z_x\in z_y$. Then $x\in z_x$ and $z_x\in z_y$ imply $x\in z_y$ by transitivity of relation $\in$ on $n$. So $x\in z_y$ and $y\in z_y$. By transitivity of $n$, $x\in n$ and $y\in n$. Since $n$ is totally ordered by $\in$, we have $x\in y$ or $y\in x$. Therefore $m$ is totally ordered by $\in$

Answer 1

There are only two ways to build new ordinals: successor or limit. You start from the base case $\emptyset$. To be a limit ordinal, you have to have infinitely many ordinals smaller than you, so you are not finite. It means that if you are finite, either you are $\emptyset$, or you are a successor ordinal of the form $m\cup\{m\}$ (all successor ordinals have this form). Moreover $m$ needs also to be finite, since if $m$ is infinite $n$ is also infinite.

Answer 2

You can note that since $n$ is transitive, $m\in n$ and therefore $m\subseteq n$. Therefore $m$ is a subset of a finite set, so it is finite, and it is a subset of a well-ordered set so it is well-ordered.

Also, note that for ordinals, $\alpha<\beta\iff\alpha\in\beta$, but since those are transitive sets $\alpha\subseteq\beta$, with inclusion being proper if and only if $\alpha<\beta$, so we have that $\alpha\in\beta\iff\alpha\subsetneq\beta$.