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I thought this was easy until I realized that for a set to be well-ordered, EACH of it's non-empty subsets must have a least element.

Let $A$ be a non-empty totally ordered set. Let $x_0$ be some element of $A$. Define $S=\{x\in A|x\ge x_0\}$. $x_0 \in S$ so $S$ isn't empty.

Now, this seems like it would work. It certainly is totally ordered and cofinal in $A$, but it might not be well-ordered.

For example, if $A=\mathbb R$ then $(x_0 + 1, x_0 + 2)$ (where $()$ denotes an open interval) is a non-empty subset of $S$, but doesn't have a least element.

Luka Horvat
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2 Answers2

14

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 $\mathscr{W}$.
  • Prove that every chain in $\langle\mathscr{W},\preceq\rangle$ has an upper bound in $\mathscr{W}$.
  • Prove that a $\preceq$-maximal element of $\mathscr{W}$ is a well-ordered cofinal subset of $\langle X,\le\rangle$.
Brian M. Scott
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  • The last one seems a bit tricky but I'll ask for hints if I end up needing them. Thanks for the answer! – Luka Horvat Jan 20 '15 at 18:21
  • @LukaHorvat The last one is trivial. If it isn't cofinal, it can be extended. – user2345215 Jan 20 '15 at 18:22
  • @Luka: You’re welcome! And feel free to ask if you get stuck. – Brian M. Scott Jan 20 '15 at 18:25
  • So I've managed to do it, but I've got a question. Could the initial segment relation be replaced with a subset relation and still get the same result? – Luka Horvat Jan 20 '15 at 19:48
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    @Luka: If you do that, the union of a chain need not be well ordered. Consider, for instance, ${0},{-1,0},{-2,-1,0},\ldots$. – Brian M. Scott Jan 20 '15 at 19:51
  • How is it not well ordered? Every subset of the union has to be a subset of one of the elements of the chain, which are all well ordered. – Luka Horvat Jan 20 '15 at 19:52
  • @Luka: the union is the set of non-positive integers, which has no smallest element. – Brian M. Scott Jan 20 '15 at 19:53
  • @BrianM.Scott I'm curious why my argument doesn't hold. – Luka Horvat Jan 20 '15 at 19:54
  • @Luka: Because if you don’t require the shorter well-orders to be initial segments, the longer ones can keep extending to the left (or downward, if you think of your orders vertically), as in the example that I just gave. – Brian M. Scott Jan 20 '15 at 19:56
  • @BrianM.Scott I understand that and I see how what you wrote is a counterexample. But I don't see why I can't claim that any subset of the union has to be a subset of one of the elments of the chain. Those are all well ordered, so the subset has a least element. – Luka Horvat Jan 20 '15 at 19:58
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    @Luka: Look at the example: the union itself isn’t a subset of any of the elements of the chain. They’re all finite, and it’s infinite. You can only be sure that every finite subset of the union belongs to one of the elements of the chain. – Brian M. Scott Jan 20 '15 at 20:00
  • @BrianM.Scott I see. In that case, would you say that I could get away with the subset relation AND a restriction that every element in the set $W$ has to be greater than some fixed element of the original set $X$? – Luka Horvat Jan 20 '15 at 20:02
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    @Luka: I’m afraid not. Take $X=\Bbb R$, and let $0$ be the fixed element. Then you can build the chain ${1},\left{\frac12,1\right},\left{\frac13,\frac12,1\right},\ldots$. – Brian M. Scott Jan 20 '15 at 20:04
  • @BrianM.Scott Damn. The union fails for both conditions then... So did you figure this out beforehand or what? I'm interested in your thought process. Also, to avoid going overboard with the comments, this will be my last question. Thanks again for the help. – Luka Horvat Jan 20 '15 at 20:09
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    @Luka: You’re welcome. I’ve used similar arguments often enough that I don’t really have to think about that particular problem, and I’ve also used arguments in which I had to be able to extend to the left to get a non-well-order, so the whole cluster of ideas is simply very familiar to me. It’s been over $40$ years since I first encountered the ideas, and I honestly don’t remember my thought processes at the time. – Brian M. Scott Jan 20 '15 at 20:13
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    In fact it's simple. You just consider the "natural" order of your elements of W. If all you know about them is that they are subsets of some fixed set, subset is the natural order. If they are ordinals, \in is the natural order. For well-ordered sets, "being an initial segment of" is the natural order. I have yet to encounter an ordinary task (ie, not constructed specially for that purpose:) that requires you to think of some new ordering relation. – Veky Feb 16 '17 at 06:17
  • What is the definition of being an initial segment? I heard it is ${a\in W \mid a<x}$ for some $x\in W$. But then this relation is not reflexive. – CO2 Dec 22 '21 at 12:17
  • @CO2: That is one of several definitions of initial segment that are in use, but not, I think, the most common one. Here I’m using the following definition: if $\langle X,\le\rangle$ is a linear order, and $A\subseteq X$, then $A$ is an initial segment of $X$ iff ${x\in X:x\le a}\subseteq A$ for each $a\in A$. Equivalently, $A$ is an initial segment of $X$ iff $A=\bigcup_{a\in A}(\leftarrow,a]$, where $(\leftarrow,a]={x\in X:x\le a}$. – Brian M. Scott Dec 22 '21 at 19:31
  • How do you prove that the union of the sets in the chain is an upper bound? – jsmith Apr 07 '22 at 01:37
  • @jsmith: What part of that is giving you trouble? – Brian M. Scott Apr 07 '22 at 02:27
  • Ok, so I want to show $W_i \preceq (X = \cup \mathscr{C})$, where $\mathscr{C}$ is a chain and $W_i$ is an arbitrary element of $\mathscr{W}$. According to your definition of initial segment, proving $W_i \preceq X$ is equivalent to proving that $W_i$ is a subset of $X$ (easy, an immediate consequence of the definition of $X$) and proving that $\forall a \in W_i {x \in X: x \leq a } \subseteq W_i$. I assume $z$ is in this set, such that $z \in W_j$ for some $W_j \neq W_i$. Since $\mathscr{C}$ is a chain, I furthermore know that $W_i \preceq W_j$ or $W_j \preceq W_i$. (to be continued). – jsmith Apr 07 '22 at 16:51
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    Note that I'm assuming $z \in W_j$ for some $W_j \neq W_i$ since if $z \in W_i$, then for that case ${x \in X: x \leq a } \subseteq W_i$ comes immediately. So if $z \not \in W_i$, then $z$ is still in $X$, so it still belongs to some well-order. Anyways, if $W_i \preceq W_j$, then $\forall a \in W_i, {x \in W_j: x \leq a } \subseteq W_i$, which implies $z \in W_i$ since we assumed $z$ is ${x \in X: x \leq a }$ for some $a \in W_i$.

    In the 2nd case, well...$W_j \preceq W_i$ implies $W_j \subseteq W_i$ so $z \in W_i$ also..........

    welp, I figured it out. Thanks.

     – jsmith Apr 07 '22 at 16:57
  • My trouble was that I'd forgotten that in the second case, part of your initial segment definition (I'm using Halmos which has a different definition, so it took me a little getting used to) $W_j \preceq W_i$ also stated that $W_j \subseteq W_i$. And I couldn't deduce anything from $\forall b \in W_j {x \in W_i : x \leq b } \subseteq W_j$.

    I might still need help with proving that it's a well-order though. I'll get back to you on that, or simply type it up in the comments anyways as an addendum for future onlookers if I figure it out myself.

     – jsmith Apr 07 '22 at 17:01
  • For future reference, this question proves that the union of the chain is a well-order: https://math.stackexchange.com/questions/3158842/particular-union-of-well-ordered-sets-that-is-well-ordered – jsmith Apr 07 '22 at 17:53
  • And for completion, I'll include a proof that $A$ is cofinal. Since $X$ is a total order, you can assume $x \leq a$ or $a < x$ for arbitrary $a \in A$. Suppose $a < x$ for all $a \in A$. Then $A \preceq A \cup {x }$. Note in particular that $A \cup { x }$ is still a well-order. The maximality of $A$ implies $A = A \cup {x}$, so $x = a$ for some $a \in A$, a contradiction. Thus, there exists $a \in A$ such that $x \leq a$ – jsmith Apr 07 '22 at 18:48
  • @jsmith: Looks good! – Brian M. Scott Apr 07 '22 at 18:59
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By transfinite recursion, we can define a sequence $(a_\alpha)_{\alpha\in\mathrm{Ord}}$ of elements of $A$ (see ordinals) such that $a_\alpha$ satisfies $$\forall\beta,\ \beta<\alpha\implies a_\beta<a_\alpha$$ Each step is possible if and only if $(a_\beta)_{\beta<\alpha}$ isn't cofinal.

But this construction must stop somewhere for there isn't an unlimited number of elements of $A$. That's the point where you get a cofinal sequence, which is well ordered because ordinals are.

user2345215
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