Basic closure properties of ordinals

Claim (a): let $x\in \alpha$; then $x\subseteq \alpha$ by transitivity of $\alpha$, so $\in$ strictly well-orders $x$ by [L1], and $x$ is transitive, because $z\in y\in x$ gives $y\in \alpha$ and $z\in \alpha$ by transitivity of $\alpha$, whence $z\in x$ by transitivity of the relation $\in$ on $\alpha$; so $x$ is an ordinal.

Claim (b): if $\alpha \in \alpha$ then $\alpha$ is an element of $\alpha$ satisfying $\alpha \in \alpha$, contradicting irreflexivity of $\in$ on $\alpha$; hence $\alpha \mathrm{\notin }\alpha$.

Claim (f), the easy direction: $\alpha \in \beta$ gives $\alpha \subseteq \beta$ by transitivity of $\beta$, and $\alpha =\beta$ gives $\alpha \subseteq \beta$ trivially.

Claim (f), the substantial direction: assume $\alpha \subseteq \beta$ and $\alpha \mathrm{\ne }\beta$, let $\gamma$ be the $\in$-least element of the nonempty set $\beta \setminus \alpha \subseteq \beta$, and check $\gamma =\alpha$; indeed $\delta \in \gamma$ forces $\delta \in \beta$ by transitivity of $\beta$ and then $\delta \in \alpha$, since otherwise $\delta \in \beta \setminus \alpha$ with $\delta \in \gamma$ contradicts minimality, so $\gamma \subseteq \alpha$; conversely $\delta \in \alpha \subseteq \beta$ compared with $\gamma$ by trichotomy in $\beta$ cannot satisfy $\delta =\gamma$ or $\gamma \in \delta$, since each would put $\gamma \in \alpha$, using transitivity of $\alpha$ in the second case, so $\delta \in \gamma$ and $\alpha \subseteq \gamma$; hence $\alpha =\gamma \in \beta$.

Claim (c): ${\alpha }^{+}$ is transitive, because $x\in {\alpha }^{+}$ means $x\in \alpha$, whence $x\subseteq \alpha \subseteq {\alpha }^{+}$, or $x=\alpha \subseteq {\alpha }^{+}$; the relation $\in$ is irreflexive on ${\alpha }^{+}$ by [A1] and step 1.2, transitive there because $x\in y\in z$ with $z=\alpha$ gives $x\in y\subseteq \alpha =z$ and with $z\in \alpha$ reduces to transitivity in $\alpha$, and trichotomous there because two elements of $\alpha$ are comparable in $\alpha$ while $x\in \alpha$ satisfies $x\in \alpha$ and neither $\alpha \in x$ nor $x=\alpha$, both of which would give $\alpha \in \alpha$; finally a nonempty $S\subseteq {\alpha }^{+}$ has an $\in$-least element, namely the $\in$-least element of $S\cap \alpha$ when that is nonempty and $\alpha$ otherwise.

Claim (d): $\bigcap A$ is transitive, since $x\in \bigcap A$ gives $x\in \delta$ and hence $x\subseteq \delta$ for every $\delta \in A$, so $x\subseteq \bigcap A$; and $\bigcap A$ is a subset of any fixed member of the nonempty set $A$, so $\in$ strictly well-orders it by [L1].

Claim (g): $\gamma =\alpha \cap \beta$ is an ordinal by claim (d) applied to $\{\alpha ,\beta \}$, and $\gamma \subseteq \alpha$ and $\gamma \subseteq \beta$, so claim (f) gives $\gamma \in \alpha$ or $\gamma =\alpha$, and likewise for $\beta$; both memberships at once would give $\gamma \in \alpha \cap \beta =\gamma$, contradicting claim (b), so $\gamma =\alpha$ or $\gamma =\beta$, that is $\alpha \subseteq \beta$ or $\beta \subseteq \alpha$.

Claim (e): $\bigcup A$ is transitive, since $x\in \delta \in A$ gives $x\subseteq \delta \subseteq \bigcup A$; its elements are ordinals by claim (a), so $\in$ is irreflexive on it by claim (b) and transitive on it because $x\in y\in z$ with $z\in \delta \in A$ puts $x,y,z$ all in the ordinal $\delta$; any two of its elements lie in a common member of $A$ by claim (g) and are therefore comparable, which gives trichotomy by claim (b) and claim (f); and a nonempty $S\subseteq \bigcup A$ has an $\in$-least element, namely the $\in$-least element of $S\cap \delta$ for any $\delta \in A$ meeting $S$, since an element of $S$ lying $\in$-below it would lie in $\delta$ by transitivity and contradict minimality.

Claims (a) to (g) are established.