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Class of ordinals is defined as follows:

  1. If $A$ is 'any' well-ordered set, there exists an element $a$ in the class such that $A$ is similar to $a$.
  2. If $A$ is a well-ordered set and $a$,$b$ are elements in the class, then [$A$ is similar to $a$ and $A$ is similar to $b$ ⇒ $a=b$].

I have constructed a well-ordered class, $C$, which satisfies two conditions and is $\in$-well-ordered.

First, I have proved that if there exists another class, $D$, satisfying above two conditions, $D$ is well ordered by -< (isomorphism).

Second, I have proved that there exists an isomophism $f$ between $C$ and $D$.

My questions is if $\{r_i\}$ is a family of ordinals in $C$, $f(\bigcup r_i$) = $\bigcup f(r_i)$?

If this is false, how come $sup\{r_i|i \in I\}$ = $\bigcup r_i$ is generally true?

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up vote 1 down vote accepted

My questions is if $\{r_i\}$ is a family of ordinals in $C$, $f(\bigcup r_i) = \bigcup f(r_i)$?

No. It may not even be true that if $r,s\in C$ with $r<s$, then $f(r)\subseteq f(s)$; it depends entirely on the nature of $D$ and its ordering.

The fact that $\sup\{r_i:i \in I\}=\bigcup_{i\in I}r_i$ for ordinary von Neumann ordinals is specific to that choice of representatives for well-orderings: it comes from the fact that these ordinals are transitive sets of smaller ordinals.

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