3

There is an exercise in "Naive Set Theory" by Halmos given as (Sec. 17, p. 68):

A subset $A$ of a partially ordered set $X$ is cofinal in $X$ in case for each element $x$ of $X$ there exists an element $a$ of $A$ such that $x \leqq a$.
Prove that every totally ordered set has a cofinal well ordered subset.

I'm checking a sketch of the proof from the Answer by Brian M. Scott on Luka Horvat's question and it seems intuitively valid to me, but I stumble on trying to strictly prove the very first step from it:

Let $\langle X,\le\rangle$ be a non-empty linear order. If $X$ has a maximum element $x_0$, then $\{x_0\}$ is a well-ordered cofinal subset, so assume that $X$ has no largest element. Let $$\mathscr{W}=\{W\subseteq X:\langle W,\le\rangle\text{ is a well-order}\}\;,$$ and define a relation $\preceq$ on $\mathscr{W}$ by setting $W_0\preceq W_1$ if and only if $W_0$ is an initial segment of $W_1$.
- Prove that $\preceq$ partially orders $\mathscr{W}$.

To show that the relation $\preceq$ thus defined is a partial order, I have started from trying to show that it is reflexive, i.e.

For $w \in \mathscr W$, it is that $w \preceq w$ (or, $w$ is an initial segment of $w$)

However from Halmos' definition of an initial segment, I think it's impossible because $w$ being an initial segment of $w$ would mean that $\exists z \in w$ s.t. $w = \{ x \in w: x \le z \}$ (if we understand "initial segment" as a "weak initial segment"), and thus, that $w$ should have the largest element. Thus for all sets in $\mathscr W$ which don't have the largest element, the relation won't be reflexive. I think that examples of such could be constructed, i.e. a set $\{ 0, 1, 2, 3, 4, 5, ...\}$ is well ordered, but it won't be an initial segment of itself due to the lack of the greatest element.

To put it short, my question is: how can a partial ordering with respect to being an initial segment or a continuation exist if it seems to violate the reflexivity condition?

Addendum

As I see it, a similar problem would occur when we talk about partial orderings with respect to continuation as continuation is a special case of an initial segment according to Halmos. Halmos himself talks about some partial orderings with respect to continuation in Sec. 17 of the book, so the question arises not only in a context of the aforementioned answer

The definitions from the book

In "Naive Set Theory" by Halmos, continuation is defined as:

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 $A$

the definition of initial segment as given in one of the previous chapters of the book is:

if $X$ is a partially ordered set, and if $a \in X$, the set $ \{ x \in X: x < a \} $ is the initial segment determined by $a$

  • 1
    Accept the weak version or make the order A is a lower subset of B. – William Elliot Jul 13 '19 at 00:53
  • @WilliamElliot thank you, it seems to me that the definition of "Lower subset" from https://en.wikipedia.org/wiki/Upper_set should work. The whole chapter "Well Ordering" (that contains the Well Ordering Theorem) is filled with p.o. sets with respects to continuation. I will try to follow the arguments in the text reading it as a "p.o. set with respect to being a lower subset" according to that Wikipedia definition – Andrey Surovtsev Jul 13 '19 at 09:04

0 Answers0