How does one define a well-ordering for the set of Zermelo ordinals? For the von Neumann ordinals, $\in$ is the standard well-ordering.
I would like to know the standard way to define a well-ordering $<$ for the set of Zermelo ordinals.
In particular, I would like to know, for any two distinct Zermelo ordinals $u$ and $v$, how one determines, formally1, whether $u < v$ or $v < u$.
I assume that this definition of $<$ for Zermelo ordinals will depend only on Zermelo's axioms, with the possible addition of the Axiom of Foundation.  At any rate, I am interested only in such a definition.  (In particular, I would like to avoid any definition that requires the existence of the set $\omega$ of finite von Neumann ordinals.2)

In case the definition of Zermelo ordinals is not sufficiently standardized, below I give the one I know.
If $w$ is any set, let $\rho(w)$ be the formula that asserts $$\varnothing \in w \;\; \wedge \;\; \forall y\in w (\{y\} \in w).$$
Zermelo's original Axiom of Infinity postulates the existence of a set $w$ such that $\rho(w)$.
Let set $z$ be the smallest set $w$ such that $\rho(w)$ holds.  Formally, $$z := \bigcap \,\{w : \rho(w)\}.$$
A "Zermelo ordinal" is an element of the set $z$.

1 I use the word "formally" here to rule out the method of "counting the number of curly braces." :)
2 On p. 42 of his The Foundations of Mathematics Kenneth Kunen contrasts Zermelo's original Axiom of Infinity (Axiom des Unendlichen: $\exists w [\varnothing \in w \ \wedge \ \forall y\in w (\{y\} \in w)]$), which justifies the existence of the set of Zermelo ordinals, with the now more standard Axiom of Infinity ($\exists w [\varnothing \in w \ \wedge \ \forall y\in w (y \cup \{y\} \in w)]$), which justifies the existence of the set $\omega$ of finite von Neumman ordinals.  Kunen writes (my emphasis added):

...the two Axioms of Infinity are equivalent, although one needs to use [the Axiom of] Replacement (which Zermelo did not have) to prove the equivalence...".

As an exercise, I would like to prove, using the Axiom of Replacement, that the existence of the set $z$ of Zermelo ordinals implies the existence of the set $\omega$ of finite von Neumman ordinals, but all my attempts stumble for not having a proper definition of a well-ordering for $z$.
 A: First, we need to show that transitive closures (of relations) exist. This can be done using the axiom of the power set and $\Delta_0$ separation without reference to the axiom of infinity. It can also be done with full separation.
The transitive closure of $S \subseteq A^2$ is simply $\{(a, b) \in A^2 \mid \forall Q \subseteq A^2,$ if $Q$ is transitive and $S \subseteq Q$ then $(a, b) \in Q\}$. You can easily verify that the transitive closure is the smallest transitive relation containing $S$.
Then $< \subseteq z^2$ is the transitive closure of $\in_z = \{(a, b) \in z^2 \mid a \in b\}$. Since $\in_z$ is well-founded and the transitive closure of a well-founded relation is also well-founded, we see that $<$ is well-founded. The fact that $<$ is a strict total order can be shown using induction.
Edit: Here's a proof that the transitive closure of a well-founded relation is well-founded.
Consider any relation $\prec \subseteq S^2$, and let $<$ be its transitive closure. Prove that $x < y$ if and only if either $x \prec y$ or $\exists z \in S (z \prec y \land x < z)$. Dually, we see that $x < y$ if and only if either $x \prec y$ or $\exists z \in S (x \prec z \land z < y)$.
A set $Q \subseteq S$ is said to be "inductive" with respect to $\prec$ if $\forall x \in S (\{y \in S \mid y \prec x\} \subseteq Q \to x \in Q)$. Then $\prec$ is said to be a well-founded relation if for all $Q \subseteq S$ which are inductive with respect to $S$, $Q = S$. We are defining the term "well-founded" here using well-founded induction.
Now suppose that $\prec$ is well-founded. We wish to demonstrate that $<$ is also well-founded. Consider some $Q$ which is inductive with respect to $<$. Define $Q' = \{x \in S \mid \{y \in S \mid y < x\} \subseteq Q\}$. Note that $Q' \subseteq Q$.
I claim that $Q'$ is inductive with respect to $\prec$. For consider some arbitrary $x \in S$, and suppose $\{y \in S \mid y \prec x\} \subseteq Q'$. Now suppose we have some $y \in S$ with $y < x$. Then there are two cases. The first is that $y \prec x$; in this case, we have $y \in Q' \subseteq Q$. The second is that there is some $z \in S$ such that $z \prec x$ and $y < z$. In that case, we know that $z \in Q'$, and therefore $y \in Q$. So in either case, we know that $y \in Q$. Thus, we see that $\{y \in S \mid y < x\} \subseteq Q$. Therefore, $x \in Q'$.
Since $Q'$ is inductive with respect to $\prec$ and $\prec$ is well-founded, we have $Q' = S$. Then we have $S = Q' \subseteq Q \subseteq S$, so $Q = S$. $\square$
Now we show that any well-founded relation $\prec \subseteq S^2$ is irreflexive. Define $Q = \{x \in S \mid x \nprec x\}$; we will show that $Q$ is inductive. Now consider an arbitrary $x \in S$, and suppose $\{y \in S \mid y \prec x\} \subseteq Q$. Suppose $x \prec x$. Then $x \in Q$; that is, $x \nprec x$. This is a contradiction. Therefore, $x \nprec x$; that is, $x \in Q$. Since $Q$ is inductive, we have $Q = S$, so $S$ is irreflexive. $\square$
Let's return to our original topic, the relation $<$ on the Zermelo ordinals. We have shown that $<$ is well-founded and thus irreflexive. And of course it is transitive, being a transitive closure. All that remains is to show that $<$ is total.
We demonstrate $\forall a \in z \forall b \in z (a < b \lor a = b \lor b < a)$ by induction on $a$.
The base case is $\forall b \in z (0 < b \lor 0 = b \lor b < 0)$. We actually strengthen what must be proved to $\forall b \in z (0 < b \lor 0 = b)$; this follows immediately from induction on $b$.
For the inductive step, suppose $\forall b \in z (a < b \lor a = b \lor b < a)$. We wish to show that $\forall b \in z (Sa < b \lor Sa = b \lor b < Sa)$. Consider some $b \in z$. By the inductive hypothesis, we know that either $a < b$, $a = b$, or $b < a$. If $a = b$ or $b < a$, then clearly $b < Sa$. If $a < b$, then we have two possibilities. The first is that $a \in b$; in this case, we have $b = Sa$. The second is that there is some $c \in z$ such that $a \in c$ and $c < b$. In this case, we actually have $c = Sa$, and therefore $Sa < b$. This completes the proof. $\square$
