Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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..

share|improve this question
    
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
add comment

2 Answers

up vote 0 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.

share|improve this answer
add comment

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.