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 $A$.
(Halmos's Naive Set Theory, p.67)
I have difficulty parsing this sentence. Is he saying that $A$ is a continuation of $B$ if:
- $B$ is a subset of $A$
- $B$ is an initial segment of $A$ (??)
- Given the well orderings $(A,\leq)$ and $(B,\preceq)$, $\forall x,y\in B, (x\preceq y \iff x\leq y)$?
On point 2, I only know that $s(x)=\{y\in X\vert y\prec x\}$ is the initial segment in $X$ as determined by $x\in X$, but Halmos seems to mention the initial segment OF a set.
Let $(A, \leq)$ be a well-ordered set. An initial segment of $A$ is a subset $B \subseteq A$ that is closed downward, that is, it has the property that if $x \in A$, $y \in B$, and $x \leq y$, then $x \in B$. That is, $B$ starts from the "bottom" of $A$ and doesn't skip any elements - as you go up, once you find something not in $B$ you'll never find something higher that is in $B$.
The connection to the notion you've seen before is that, given $B$, we can define $x = \min\{a \in A : a \not\in B\}$. As long as $B \neq A$ this $x$ exists, and saying that $B$ is an initial segment of $A$ is the same as saying $B = s(x)$.
Your point (3) is correct, it just says that the ordering used on $B$ is the one it gets by virtue of being a subset of $A$.