Membership-well-ordering and class of ordinal numbers I learnt how to construct a well ordered class of ordinals, ORD which satisfies 2 conditions.

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*If A is any well ordered set, there exists $a$$\in$ORD such that A is similar to $a$.

*If A is a well ordered set and $a$,$b$ $\in$ORD, then if A is similar to $a$ and A is similar to $b$ then $a$=$b$.

The book im studying constructed a class satisfying those 2 conditions, which is $\in$-well-ordered and each element of the class is also $\in$-well-ordered.
My question is that can any well ordered class satisfying those 2 conditions be $\in$-well-ordered?
Since the book keeps emphasizing that 'Which' class of ordinal numbers it is doesnt matter at all.
(i.e. 1meter can be measured by a ruler but can also be measured by 1meter-pencil)
I want to know this because I can prove that 'union of a family of ordinals is equal to supremum of the family of the ordinals' by using the property of the $\in$-well-ordered class of ordinals (class of ords, the one I constructed), but I do not know whether that holds for another class of ordinal numbers which satisfies above 2 conditions..
 A: It is not clear to me where $a, b, A$ are coming from so maybe I am missing something.
You have constructed some class. For reference say this class is defined by the predicate $P(x)$. Suppose that $X$ is a set that can not be well-ordered. Define a new predicate $R(x) \text{ iff } (P(x) \text{ or } x \in X)$. The predicate $R$ defines a class. If you could well-order this class then $X$ could be well-ordered.
Edit to answer
There are many classes that can be well-ordered by relations other than $\in$. Let $x$ be a set, none of whose elements is an ordinal. This is done so that I do not accidentally create an ordinal. Consider the class $C$ whose elements are ordered pairs of the form $(\alpha, x)$ where $\alpha$ is an ordinal. We can well-order this class by saying $(\alpha_{0}, x) \leq (\alpha_{1}, x)$ if and only if $\alpha_{0} \leq \alpha_{1}$. For ordinals the strict version of $\leq$ is $\in$.
Hopefully this is closer to what you are looking for.
A: No, you cannot assume that such a class is well-ordered by $\epsilon$. Let $\varphi(x)$ be a formula defining a class $\mathrm{ORD}$ satisfying the two conditions in question. Any $a$ satisfying $\varphi(a)$ is an ordered pair $\langle a_0,a_1\rangle$ such that $a_1$ is a well-ordering of $a_0$. It’s tedious but straightforward to write down a formula $\psi(x)$ such that any $b$ satisfying $\psi(b)$ is an ordered pair $\langle b_0,b_1\rangle$ such that there is a unique $a\in\mathrm{ORD}$ such that


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*$b_0=\big\{\langle a_0,x\rangle:x\in a_0\big\}$ and  

*$b_1=\Big\{\big\langle\langle a_0,x\rangle,\langle a_0,y\rangle\big\rangle: \langle x,y\rangle\in a_1\Big\}$.


Then $b_1$ is a well ordering of $b_0$ similar to the well-ordering $a_1$ of $a_0$, so the class $\mathrm{ORD}'$ of sets satisfying $\psi(x)$ also contains a unique $b=\langle b_0,b_1\rangle$ order-isomorphic to any given well-order. However, the construction ensures that for any $b=\langle b_0,b_1\rangle\in\mathrm{ORD}'$, the well-ordering $b_1$ of $b_0$ is not an $\epsilon$-ordering, and it also ensures that the natural ordering of $\mathrm{ORD}'$ is not an $\epsilon$-ordering.
