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

set-theorywell-orders

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.

Best Answer

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.

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