# Is there a reason Halmos defined continuation on well-ordered sets, but not total ordered sets?

In Naive Set Theory, Dr. Paul Halmos defines continuation for well-ordered sets as such (bullet points mine):

We shall say that a well ordered set $$A$$ is a continuation of a well ordered set $$B$$ if

• $$B \subset A$$,
• $$B$$ is an initial segment of $$A$$, and
• The ordering of elements in $$B$$ is the same as their ordering in $$A$$.

I'm not sure why we need well ordering for this property. For example, it seems like we should be able to describe $$\{ z \in \mathbb{Z} : z \leq 100 \}$$ as a continuation of $$\{ z \in \mathbb{Z} : z \leq 1 \}$$, under the usual ordering.

These are not well ordered sets, since they have subsets without a least element, but they seem to satisfy every other condition.

Is there a reason for defining this on well-ordered sets specifically?

• You could dial down the requirements even further - and end up with the boring subset relation Jun 30, 2018 at 11:07

We would like, in general, for properties of well-orders to be preserved under order-preserving maps - that is, we want it to be the order that matters, not the elements. The two sets you name, $\{z \in \mathbb{Z} : z \leq 100\}$ and $\{z \in \mathbb{Z} : z \leq 1\}$, are order-isomorphic; they differ only in elements, not in structure.
The property "$A$ is a continuation of $B$" is a property of the order - if $A$ is a continuation of $B$ and both are well-orders, then $A$ and $B$ are both isomorphic to ordinals and the ordinal of $B$ is greater than the ordinal of $A$. In your proposed change, "continuation" would be a property of the particular set; possibly an interesting notion anyway, but not what we usually care about when talking about well-orders.
Yes, you are correct. You can define the notion of "end-extension" in any sense of a linear order: $A$ is an end-extension of $B$ if $B$ is an initial segment of $A$.
And yes, that definition does merit some sort of interest. For example, two countable models of Peano arithmetic (the "standard theory of the natural numbers") are either order-isomorphic, or one of them is an initial segment of the other of type $\Bbb N$. In particular, you can have two models $M$ and $M'$ such that $M$ is an initial segment of $M'$, and they are isomorphic as linear orders, but they are not isomorphic as models of $\sf PA$.
However, outside the domain of well-orders, as Reese remarks in their answer, the notion of end-extension can become a bit trivialized: It could be that a linear order is isomorphic to one of its end-extensions (e.g. $(-\infty,0)$ and $\Bbb R$, or $\{x\in\Bbb Z\mid x\leq 1\}$ and $\{x\in\Bbb Z\mid x\leq 100\}$).