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I learnt how to construct a well ordered class of oridinals, ORD which satisfies 2 conditions. 1. If A is any well ordered set, there exists $a$$\in$ORD such that A is similar to $a$. 2. 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..

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Are $a$ and $b$ elements of your class? – Jay May 28 '12 at 21:57
Yes. Those are elements of the class of ordinals, so are ordinals. – Katlus May 28 '12 at 22:01
up vote 1 down vote accepted

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

  1. $b_0=\big\{\langle a_0,x\rangle:x\in a_0\big\}$ and
  2. $b_1=\Big\{\big\langle\langle a_0,x\rangle,\langle a_0,y\rangle\big\rangle: \langle x,y\rangle\in a_1\Big\}$.

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.

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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.

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Hmm.. My question is 'any class satisfying above two consitions can be $\in$-well-ordered?' Maybe im not getting your answer.. – Katlus May 28 '12 at 22:00
Your conditions say "If $A$ is a well-ordered set then ...." Is $A$ a subset of your class? If a set is not well-ordered your conditions say nothing about it. As a result I can add my set $X$ to your class $P$ and the statements will still hold. Please state explicitly where the set $A$ resides. Is it any set what-so-ever or is it a subset of your class? This is not clear to me. – Jay May 28 '12 at 22:08
No. $A$ doesnt need to be a subset of the class. $A$ is just an arbitrary set and the class has an element $a$ such that $a$ is similar to $A$. – Katlus May 28 '12 at 22:11
@Jay: Does the class $C$ from your edit have the property that for every well-ordered set $A$ there is a unique $(a,x)\in C$ such that $A$ and $(a,x)$ are similar? It seems to me that it doesn't, so that isn't quite what the OP is looking for. – Cameron Buie May 28 '12 at 23:39
@Cameron You are correct. – Jay May 29 '12 at 21:27

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