[proofplan] We prove that $\in$ is a well-order on $A$ by checking linearity and least-element existence. Linearity follows from the trichotomy property of ordinals. For a nonempty subset $B \subset A$, we choose an element $\beta \in B$; either $\beta$ is already least in $B$, or some element of $B$ lies below $\beta$, in which case the well-ordering of the ordinal $\beta$ supplies a least element of $B \cap \beta$. Trichotomy then shows that this element is least in all of $B$, not merely in $B \cap \beta$. [/proofplan] [step:Use ordinal trichotomy to make membership linear on $A$] Let $\alpha,\beta \in A$. By hypothesis, $\alpha$ and $\beta$ are ordinals. The trichotomy property of ordinals says that exactly one of \begin{align*} \alpha = \beta, \qquad \alpha \in \beta, \qquad \beta \in \alpha \end{align*} holds. Hence $\in$ is a strict linear order on $A$, with the associated non-strict order given by $\alpha \leq \beta$ iff $\alpha = \beta$ or $\alpha \in \beta$. [guided] We first need to show that any two elements of $A$ are comparable by membership. Let $\alpha,\beta \in A$. Since every element of $A$ is an ordinal, both $\alpha$ and $\beta$ are ordinals. Ordinals satisfy trichotomy: for any two ordinals, exactly one of equality or membership in one direction holds. Thus exactly one of \begin{align*} \alpha = \beta, \qquad \alpha \in \beta, \qquad \beta \in \alpha \end{align*} is true. This gives linearity of the order on $A$: whenever $\alpha,\beta \in A$, either they are equal, or one is a member of the other. The corresponding non-strict comparison is therefore $\alpha \leq \beta$ iff $\alpha = \beta$ or $\alpha \in \beta$. [/guided] [/step] [step:Reduce the least-element property to a subset of one ordinal] Let $B \subset A$ be nonempty. Choose $\beta \in B$. Suppose first that no element $\gamma \in B$ satisfies $\gamma \in \beta$. Then for every $\gamma \in B$, [ordinal trichotomy](/theorems/4812) applied to $\gamma$ and $\beta$ gives either $\gamma = \beta$ or $\beta \in \gamma$. Hence $\beta$ is an $\in$-least element of $B$. It remains to consider the case in which some $\gamma \in B$ satisfies $\gamma \in \beta$. Define \begin{align*} C := B \cap \beta. \end{align*} Then $C$ is a nonempty subset of $\beta$. [guided] Let $B \subset A$ be nonempty. To prove that $B$ has a least element, begin with any element $\beta \in B$. There are two cases. First suppose there is no $\gamma \in B$ with $\gamma \in \beta$. We claim that $\beta$ is already least in $B$. Indeed, take any $\gamma \in B$. Since $\gamma$ and $\beta$ are ordinals, trichotomy gives exactly one of \begin{align*} \gamma = \beta, \qquad \gamma \in \beta, \qquad \beta \in \gamma. \end{align*} The middle alternative is excluded by the case assumption. Hence either $\gamma = \beta$ or $\beta \in \gamma$, which is precisely the statement that $\beta \leq \gamma$ in the membership ordering. Thus $\beta$ is an $\in$-least element of $B$. Now suppose instead that there exists some $\gamma \in B$ with $\gamma \in \beta$. Define \begin{align*} C := B \cap \beta. \end{align*} This set is nonempty because the chosen $\gamma$ lies in both $B$ and $\beta$. Also $C \subset \beta$ by definition. [/guided] [/step] [step:Choose the least element inside the ordinal $\beta$] Since $\beta$ is an ordinal, $\beta$ is well-ordered by $\in$. Since $C \subset \beta$ is nonempty, there exists $\delta \in C$ such that for every $\eta \in C$ with $\eta \neq \delta$, one has $\delta \in \eta$. Because $C \subset B$, we have $\delta \in B$. [guided] The point of passing from $B$ to $C = B \cap \beta$ is that $C$ is now a subset of a single ordinal. Since $\beta$ is an ordinal, the relation $\in$ well-orders $\beta$. Therefore every nonempty subset of $\beta$ has an $\in$-least element. We have already checked that $C \subset \beta$ and $C \neq \varnothing$. Hence there exists $\delta \in C$ such that for every $\eta \in C$ with $\eta \neq \delta$, \begin{align*} \delta \in \eta. \end{align*} Since $C = B \cap \beta$, membership $\delta \in C$ also gives $\delta \in B$. Thus $\delta$ is a candidate for the least element of the original set $B$. [/guided] [/step] [step:Show the least element below $\beta$ is least in all of $B$] We prove that $\delta$ is an $\in$-least element of $B$. Let $\eta \in B$ with $\eta \neq \delta$. Since $\eta$ and $\delta$ are ordinals, trichotomy gives exactly one of \begin{align*} \eta = \delta, \qquad \eta \in \delta, \qquad \delta \in \eta. \end{align*} The first alternative is excluded. If $\eta \in \delta$, then $\delta \in C \subset \beta$, and transitivity of the ordinal $\beta$ gives $\eta \in \beta$. Hence $\eta \in B \cap \beta = C$, contradicting the choice of $\delta$ as the $\in$-least element of $C$. Therefore $\delta \in \eta$. Thus every $\eta \in B$ is either equal to $\delta$ or contains $\delta$ as an element. Hence $\delta$ is an $\in$-least element of $B$. Since every nonempty subset $B \subset A$ has an $\in$-least element, and since $\in$ is linear on $A$, the relation $\in$ well-orders $A$. [guided] It remains to prove that the element $\delta$ found inside $C = B \cap \beta$ is not merely least among the elements of $B$ below $\beta$, but least among all elements of $B$. Let $\eta \in B$ with $\eta \neq \delta$. Since $\eta$ and $\delta$ are ordinals, trichotomy gives exactly one of \begin{align*} \eta = \delta, \qquad \eta \in \delta, \qquad \delta \in \eta. \end{align*} The equality case is excluded by the choice of $\eta$. We rule out the possibility $\eta \in \delta$. Because $\delta \in C$, we have $\delta \in \beta$. Since $\beta$ is an ordinal, it is transitive: every element of an element of $\beta$ is itself an element of $\beta$. Therefore, if $\eta \in \delta$ and $\delta \in \beta$, then \begin{align*} \eta \in \beta. \end{align*} Together with $\eta \in B$, this implies $\eta \in B \cap \beta = C$. But $\delta$ was chosen to be the $\in$-least element of $C$, so no element of $C$ can lie in $\delta$. This contradicts $\eta \in \delta$. Thus the only remaining trichotomy possibility is $\delta \in \eta$. Since this holds for every $\eta \in B$ with $\eta \neq \delta$, the element $\delta$ is an $\in$-least element of $B$. We have shown that every nonempty subset of $A$ has an $\in$-least element, and the first step showed that membership is linear on $A$. Therefore $\in$ well-orders $A$. [/guided] [/step]