Choosing elements of linear orders

Question

Is it consistent with ZF that there can be a countable family of linear orders, each isomorphic to $\mathbb Z$ (that is, every element has a unique predecessor and successor, and any two elements have finitely many elements between them), such that one cannot choose an element from each of them?

(Random musing inspired by this question.)

Let ${P_n: n \geq 1}$ be a countable family of pairs of socks - So there is no set that contains exactly one sock from each $P_n$. Then ${P_n \times Z: n \geq 1}$ should work. — hot_queen, Jun 29 '14 at 16:52
@hot_queen: But you can't choose the ordering for these sets. If you could, you can choose from $P_n$ the element for which $(a,0)$ was smaller than $(b,0)$. — Asaf Karagila, Jun 29 '14 at 17:58
@Tom: Heavens no. It follows easily from the axiom of countable choice for families of countable sets. It is even weaker than "countable unions of countable sets are countable". By which I mean, of course the negation of the statement given here. — Asaf Karagila, Jun 30 '14 at 02:12
Ok, so it is weaker than CC$(\omega)$. Interesting that this is still independent. — Forever Mozart, Jun 30 '14 at 02:17

Asaf Karagila · Accepted Answer · 2014-06-30T01:35:19.417

6

Yes, it is consistent. Let me sketch an argument using atoms, then the Jech-Sochor transfer theorem can do its magic (since it's a bounded statement).

Start with a countable set of atoms which is the disjoint union of $A_n$'s each countably infinite. Fix $\leq_n$ on $A_n$ which is a linear order of type $\omega^*+\omega$ (or $\Bbb Z$, in other words).

The automorphism group is $\mathscr G$, of permutations of the atoms which preserve both the partition into the $A_n$'s and their linear orders. In other words, we take a sequence of automorphisms of $\leq_n$ (which are finite shifts), and they each act on the relevant $A_n$. Let us denote a permutation by $\pi$ and $\pi_n$ is the permutation on the $n$-th coordinate.

We define a filter of subgroups in the following way. Given $H$ a subgroup of $\mathscr G$, $H$ is in the filter if and only if there exists a finite $E\subseteq\omega$, such that whenever $\pi\in\mathscr G$ for which $\forall n\in E$, $\pi_n=\operatorname{id}$, then $\pi\in H$. (For example taking $E=\varnothing$ we have that $\{\pi\in\mathscr G\mid\forall n\in\varnothing(\pi_n=\operatorname{id})\}=\mathscr G$ itself.)

Some easy propositions:

Each $A_n$ is symmetric, and each $\leq_n$ is symmetric.
The enumerated partition $\{A_n\mid n\in\omega\}$ is symmetric, and the enumeration of the orders is also symmetric.

And the important point: There is no choice function from the $A_n$'s.

Sketch for a proof: If $f$ was such a choice function, it would have a finite support $E$ as above, now taking any $k\notin E$ we can apply any $\pi$ which fixes $f$ but $\pi_k\neq\operatorname{id}$ to get a contradiction.

(This is the usual argument.)

Now we can use the Jech-Sochor theorem. Or we could have used forcing to begin with, it would just make things harder to write down properly in this case.

edited Jun 30 '14 at 01:35

answered Jun 29 '14 at 18:08

Asaf Karagila

393,674

I'm afraid you lost me at "The filter of subgroups ..." Are you sure that sentence says what you meant it to? As written it looks like every subgroup $H$ meets the condition because we can take $E=\varnothing$ and then "for all $n\in E$" is vacuously true. (And where is $\pi$ quantified in this definition?) – hmakholm left over Monica Jun 30 '14 at 01:17
Also, did I miss a definition of "symmetric", which you seem to apply to a lot of different things in the subsequent text? – hmakholm left over Monica Jun 30 '14 at 01:17
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@Henning: I sort of skipped the most of it; I'd thought that you know a bit more about the topic. In any case, here are two different threads with the definitions. As for the empty case, note that if $E=\varnothing$ then it is necessarily the case that $H=\mathscr G$. – Asaf Karagila Jun 30 '14 at 01:23
x @Asaf: Really? Take any odd $H$. Then there exists $E$ (namely $\varnothing$) such that $E$ is a finite subset of $\omega$ (yes, $\varnothing$ is a finite subset of $\omega$), and such that $\forall n\in E$ such-and-such holds (trivially because $E$ is empty). – hmakholm left over Monica Jun 30 '14 at 01:27
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Henning, I do not appreciate the tone of your text. I'm trying to answer and be helpful. And since we require that ALL the permutations $\pi$ such that $\pi_n=\operatorname{id}$, for all $n\in E$ this is necessarily means that EVERY $\pi\in\mathscr G$ satisfies this condition. So all in all, $H=\mathscr G$ (yes, double inclusion and whatnot). – Asaf Karagila Jun 30 '14 at 01:29
a @Asaf: I'm not intending any tone. I'm trying to understand your helpful answer. I don't see how the choice $E=\varnothing$ fails make every $H$ satisfy your condition. Which of the two parts of it does it fail? – hmakholm left over Monica Jun 30 '14 at 01:31
Let me rewrite this condition in the answer in a way which will be clearer. – Asaf Karagila Jun 30 '14 at 01:32
x @Asaf: Okay, I read you original wording as "... and $\forall n\in E;(\pi_n=\operatorname{id}\Rightarrow \pi\in H)$" which confused me. – hmakholm left over Monica Jun 30 '14 at 01:51
@Henning: Fair enough. At least it's clear now. – Asaf Karagila Jun 30 '14 at 01:59

Choosing elements of linear orders

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