On the proof that the successor of an ordinal is also an ordinal

Question

Definition. A set $\alpha$ is an ordinal if it is transitive and strictly well-ordered by $\in$.

The proofs I have seen (e.g. Lemma 2.5 in Jech and Hrbacek or p.32 in these notes) go like this:

First they show that $\alpha^+:=\alpha\cup \{\alpha\}$ is transitive.

Second they show that every non-empty subset $S\subset \alpha^+$ contains an $\in$-least element.

I have no problem with these arguments, but it seems to me we are missing a piece, namely the proof that $\in$ is a strict linear order on $\alpha^+$.

I think asymmetry and transitivity both follow from the fact that $\in$ is a strict order on $\alpha$ and that $\alpha \notin \alpha$ for any ordinal $\alpha$, which is proved after as Lemma 2.7 in Jech and Hrbacek. The fact that $\in$ is total on $\alpha^+$ follows from the fact that $\in$ is total on $\alpha$.

Am I missing something?

Answer 1

To show that a set $\alpha$ is an ordinal is suffices to show that

1. $\alpha$ is transitive.
2. Every non-empty subset of $\alpha$ has an $\in$-minimal member.
3. Trichotomy: For every $x,y\in\alpha$ either $x\in y$ or $x=y$ or $y\in x$.

Here's how to get the rest

• Irreflexivity: If $x\in\alpha$, the subset $\{x\}$ has an $\in$-minimal member, which must be $x$, so $x\not\in x$.
• Antisymmetry: If $x,y\in\alpha$ and $x\ne y$, the subset $\{x,y\}$ has an $\in$-minimal member, so either $x\not\in y$ or $y\not\in x$.
• Transitivity of $\in$: Let $x,y,z\in\alpha$ with $x\in y$ and $y\in z$. The subset $\{x,y,z\}$ has an $\in$-minimal member, which must be $x$. So $z\not\in x$. Also, $z\ne x$ because otherwise $y\in x=z$ contradicting the $\in$-minimality of $x$. By the trichotomy, $x\in z$.
• Totality of order - by the trichotomy.
• Every non-empty subset has an $\in$-least member - since $\alpha$ is totally ordered by $\in$, an $\in$-minimal member is an $\in$-least member.

Edit: If $S$ is a set, $R$ a relation and $x\in S$

• $x$ is called $R$-minimal if $\forall z\in S\; \neg zRx$.
• $x$ is called $R$-least if $\forall z\in S\; (x=z\lor xRz)$.

The definition of $R$-least does not prohibit the possibility that $xRx$ or $zRx$ for some $z\in S$.

Edit 2: To complete the proof that if $\alpha$ is an ordinal so is $\alpha^+$:

• $\alpha^+$ is transitive is in the notes.
• Trichotomy: If $x,y\in\alpha^+$, three cases. Either $x,y\in\alpha$ in which case they're $\in$-comparable since $\alpha$ is an ordinal. Or exactly one of $x,y$ is in $\alpha$, say $x\in\alpha$ and $y=\alpha$. Then $x\in y$. Third case is $x=y=\alpha$.
• Let $S$ be a non-empty subset of $\alpha^+$. First, $\alpha\not\in\alpha$ because otherwise $\{\alpha\}$ would be a non-empty subset of $\alpha$ with no $\in$-minimal member. Also, if $x\in\alpha$ then $\alpha\not\in x$ because otherwise by transitivity of $\alpha$, $\alpha\in\alpha$. Now, if $S\cap\alpha\ne\emptyset$, let $y$ be an $\in$-minimal member of $S\cap\alpha$. Since $\alpha\not\in y$, $y$ is also an $\in$-minimal member of $S$. Second case, if $S\cap\alpha=\emptyset$ then necessarily $S=\{\alpha\}$ so $\alpha$ is the $\in$-minimal member of $S$.