Supremum in well orderings

Question

Problem *x7.27 from Moschovakis' textbook is asking to define a definite operation $\sup \mathscr E $, such that for every family $\mathscr E$ of well-ordered sets, $\sup \mathscr E$ has the following properties:

• $\sup \mathscr E$ is a well ordered set
• If $U\in \mathscr E$, then $U \leq_o \sup \mathscr E$
• If $W$ is a well ordered set and for each $U\in \mathscr E$, $U\leq_o W$ then $\sup \mathscr E\leq_o W$

($A\leq_o B$ means that $A$ is isomorphic (in the sense of well orders) to some initial segment of $B$)

Apparently, one way to do this is to use the fact that any well ordering is isomorphic to an ordinal, and then one can take $\sup \mathscr E$ to be the union of the corresponding ordinals in $\mathscr E$. But how to do this without invoking this fact?

(In case it matters -- the previous problem in the textbook was to define $\inf \mathscr E$, and the hint was to look for $\inf \mathscr E$ in the initial segments of $\chi(\bigcup \mathscr E)$ where $\chi(A)$ associates to each set $A$ a well ordered set $\chi(A)$ such that $\chi(A)\not\leq_c A$ (Hartogs' Theorem); $\leq_c$ means "less in size", i.e., "injects into". For this problem, there aren't hints.)

Answer 1

Essentially the same proof works without using ordinals:

Define a proper initial segment of a well-ordering to be an initial segment that isn't the entire well-ordering.

Let $\,\mathscr S\,$ be the collection of all proper initial segments of members of $\,\mathscr E.$

The members of $\,\mathscr S\,$ are well-ordered structures -- each one is an ordered pair $\langle W, \lt_W\rangle,$ with its components being the underlying set and its accompanying order relation.

Note: An alternative way of setting this up is to let $\,\mathscr S\,$ be the disjoint union $\;\bigsqcup \mathscr E = \{\langle a, W\rangle \mid a\in W \text{ and }W \in \mathscr E\}$ of all the members of $\mathscr E.$ We would then use $\langle a, W\rangle$ in the argument below instead of the proper initial segment $\{x \mid x \lt_W a\}$ of $W.$ But, even though that seems more natural, it ends up being a little bit messier notationally.

Define an equivalence relation on $\,\mathscr S\,$ by setting $$\langle W, \lt_W\rangle \sim \langle W', \lt_{W'}\rangle$$ iff $$\langle W, \lt_W\rangle \text{ and } \langle W', \lt_{W'}\rangle\text{ are isomorphic well-orderings.}$$

Now define $\,\mathsf{sup}\;\mathscr E\,$ to be the quotient $\,\mathscr S\, / \sim,\,$ and denote the equivalence class of $\langle W, \lt_W\rangle$ by $[ W, \lt_W].$

Finally, define an ordering $\prec$ on $\,\mathsf{sup}\;\mathscr E,$ by specifying that

$$[W, \lt_W]\prec [W', \lt_{W'}]$$

iff

$$\langle W, \lt_W \rangle\,\text{ is shorter than }\,\langle W', \lt_{W'}\rangle.$$

You can check that $\prec$ is well-defined and has the required properties.