You have to re-read all the passage from page 67-on :
We shall say that a well ordered set $A$ is a continuation of a well ordered set $B$, if, in the first place, $B$ is a subset of $A$, if, in fact, $B$ is an initial segment of $A$, and if, finally, the ordering of the elements in $B$ is the same as their ordering in $A$.
Thus if $X$ is a well ordered set and if $a$ and $b$ are elements of $X$ with $b < a$, then $s(a)$ is a continuation of $s(b)$, and, of course, $X$ is a continuation of both $s(a)$ and $s(b)$.
If $\mathcal C$ is an arbitrary collection of initial segments of a well ordered set,
then $\mathcal C$ is a chain [see page 54 : "a totally ordered set"] with respect to continuation; this means that $\mathcal C$ is a collection of well ordered sets with the property that of any two distinct members of the collection one is a continuation of the other.
A sort of converse of this comment is also true and is frequently useful. If a collection $\mathcal C$ of well ordered sets is a chain with respect to continuation, and if $U$ is the union of the sets of $\mathcal C$, then there is a unique well ordering of $U$ such that $U$ is a continuation of each set (distinct from $U$ itself) in the collection $\mathcal C$.
Roughly speaking, the union of a chain of well ordered sets is well ordered. This abbreviated formulation is dangerous because it does not explain that "chain" is meant with respect to continuation. If the ordering implied by the word "chain" is taken to be simply order-preserving inclusion, then the conclusion is not valid.
The relevant fact is : the collection $\mathcal C$ of well-ordered set is a chain w.r.t continuation.
A collection $\mathcal C$ is a chain when, for all $A,B \in \mathcal C$ : $A \subseteq B$ or $B \subseteq A$.
If a collection $\mathcal C$ is a chain w.r.t continuation, it has "something more" : in addition to the property (common to all chains) that for all $A,B \in \mathcal C$ : $A \subseteq B$ or $B \subseteq A$, we have also that (supposing : $B \subseteq A$) $B$ is an initial segment of $A$, and the ordering of the elements in $B$ is the same as their ordering in $A$.
Thus, when we "merge" all the members of the collection $\mathcal C$ into the "mega-set" $U$ every subset "preserve" its "original" minimal-element.
I hope it may help ...
I'm not able to "manufacture" an example different from the "trivial" one built from $\mathbb N$.
If you consider a collection $\mathcal C = \{ X_n \}$ where all $X_i$ form a chain w.r.t continuation, we have that $X_n = \{ 0,1,2, \ldots n \} = n+1$.
Thus the union $U$ of the collection is $\omega$ itself.
Halmos mentions that for any $a$ in an ordered set $(A, \le)$ we can construct $\mathcal{O}(a) = \{\,x \in A \mid x \le a\,\}$, and thus the set $\mathcal{O}(A) = \{\,\mathcal{O}(a)\mid a \in A\,\}$. He then describes a process by which we can recover the original order on $A$ by taking the smallest set (by inclusion) of $\mathcal{O}(A)$, extracting the element of $A$ it represents, then taking the next smallest set, which has added one element of $A$, etc. This description of the reverse process implies to me that the order $\le$ he's referring to is actually a well-ordering, where $A$ has a least element and we can remove elements of $A$ and still have a least element.
By a characterization, I think he means a set of properties that is necessary and sufficient for a set $\mathcal{O}$ to have if it is $\mathcal{O}(A)$ for some $A$. By "intrinsic", I think he means this characterization shouldn't reference the order operation $\le$.
I thought about this yesterday, and have got it down to 3 properties:
- $\bigcup \mathcal{O} = A$.
- Any nonempty $\mathcal{S} \subseteq \mathcal{O}$ has a minimal element $M$ such that $M \subseteq X$ for all $X \in \mathcal{S}$. (In other words, $\mathcal{O}$ is well-ordered by inclusion).
- For any two elements $x$ and $y$ of $\bigcup \mathcal{O}$ (where $x \neq y$) there is a set $X \in \mathcal{O}$ that contains either $x$ or $y$, but not both.
The first property assures that $\mathcal{O}$ isn't missing any elements of $A$. It's not really necessary if you consider $\bigcup \mathcal{O}$ as an implicit definition of $A$, but you need it if you're matching a given $A$. The second guarantees we can follow the process of successively picking and removing the smallest element of $\mathcal{O}$. And the third guarantees distinct elements of $A$ can be separated by the derived order.
To show that an $\mathcal{O}$ with these properties defines a well-ordering of $A$, I found it easier to work with a strict order $\lt$ on $\bigcup \mathcal{O}$ defined by $x \lt y$ iff there is some set $X \in \mathcal{O}$ that contains $x$ but not $y$. Irreflexivity follows directly from the definition of $\lt$. Asymmetry can be derived by contradiction from property 2 by considering the minimal set between a set in $\mathcal{O}$ that contains $x$ but not $y$ and one that contains $y$ but not $x$. Transitivity is derived by using property 2 to show that a set in $\mathcal{O}$ containing $x$ but not $y$ must be a subset of one containing $y$ but not $z$. Trichotomy ($x \lt y$ or $x = y$ or $y \lt x$) can be derived directly from property 3. And well-ordering requires both 2 and 3. For well-ordering, we get the minimal element of some nonempty $A' \subseteq A$ by finding the minimal $X \in \mathcal{O}$ that contains at least one element of $A'$. If it contains more than one element of $A'$, then property 3 shows it's not minimal, which contradicts property 2, so it must have exactly one element of $A'$, which is then shown to be the minimal element of $A'$.
Whether or not this is the most concise or understandable characterization I don't know (it's why I searched for this question). I'm answering it long after it was asked, since it does come up in searches, and I would have found a different answer useful.
Best Answer
Yes, $A\times B$ is the (new) set of all ordered pairs $(x,y)$ with $x\in A$ and $y\in B$. There will be no problem if $A$ and $B$ share common elements since an ordered pair is precisely such that it can contain duplicates (an unordered pair, i.e. a set $\{a,b\}$ can not). So, for example, $A\times A$ is just the set of all ordered pairs of elements of $A$. For example, $\{1,2\}\times\{1,2\}=\{(1,1),(1,2),(2,1),(2,2)\}$.
EDIT: Ordered means that the order of the elements matters, that is, $(1,2)\ne(2,1)$ whereas $\{1,2\}=\{2,1\}$.