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?

Thank you in advance.

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

2 Answers 2


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 $\{x\in\Bbb Z\mid x\leq 100\}$).

In the context of well-orders, no well-ordered set is isomorphic to any of its proper initial segments. So a continuation adds no elements, or changes the structure's isomorphism type. And this is useful, and gets exploited later on (although in Halmos' book not as much as it should be, since it does not really talk about the ordinals).


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