I was thinking about this problem in trying to better get a feel for and understand well-orders (I'm trying to learn some set theory) - right now, I'm not really hoping for anybody to supply me with a proof, but I was wondering if I might be able to get a hint about how to approach this.
I am trying to show (without the axioms of choice or union - otherwise this is trivial) that for any set $W$ of well-orders, there is a well-order $(A, X)$ such that for all $(B, Y) \in W, \exists a \in A$ such that $(B, Y) \cong I_a$, where $I_a$ is the initial segment of $(A, X)$ up to $a$.
Here's what I'm thinking so far:
- $W$ is totally ordered under order-type-comparison
- if $W$ has a maximum $(M, N)$, then we take $(M, N)$ and enlarge $M$ to get the desired $(A, X)$
- if $W$ has no maximum, then we have some sort of infinitely ascending chain of longer and longer $(B, Y)$'s, so maybe we induct on them to show the existence of some kind of "limit" that we can use to construct the desired $(A, X)$?
- concerning the above point, maybe we make use of the axiom of infinity to produce this "limit"?
- wow, the thought just occurred to me in jotting down the above point: maybe we can just cheat with the axiom of infinity here? :-)
I'm not sure if these ideas are any good at all - they probably aren't - but, I'm just trying to get a better handle on this cool stuff, and I appreciate the wisdom from insightful users on here. Thanks in advance!
If you take any $W$, then you cannot assume that it is totally ordered. For example, if $W$ is made entirely of singletons? Then the order type comparison is not a total order. It's a total pre-order.
What you can do, however, is consider $W/\sim$ where $\sim$ is the order isomorphism relation. Now simply "fill in the gaps". So if $5$ is an order type of some element of $W$, but $2$ is not, add a subset of an element of $W$ of order type $2$.
So what do we want to really do? Replace $W$ by $W'$ the set of all initial segments (not necessarily proper initial segments) of the elements of $W$. Then consider $W'/\sim$. The result is indeed well-ordered with order-comparison. And you can easily prove that every order from $W$ correspond to some initial segment here.
As for the axiom of infinity, we haven't used it here. At all. But if $W$ is infinite, then either it is not a set (if you assume that Infinity fails) or that Infinity holds in the universe. Pick your pick.