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Say that a preordering $\le$ on a set of subsets of some space preserves strict inclusion provided that $A\lt B$ whenever $A\subset B$ (where $A\lt B$ iff $A\le B$ and $B \not\le A$).

Let the space be the unit circle $T$. Then it's easy to see that there is no preorder on all countable subsets of $T$ that preserves strict inclusion and that satisfies the condition that $A \le B$ iff $r[A] \le B$ for every rotation $r$. (Just let $A = \{ e^{2\pi i n u} : n \in \mathbb N \}$ for an irrational number $u$, and let $r$ be a rotation by $-2\pi u$.)

Is there a preordering of all Borel-measurable subsets of $T$ that (a) preserves strict inclusion, (b) satisfies the condition that $A \le B$ iff $r[A] \le r[B]$ for every rotation $r$ and (c) is total, i.e., all Borel subsets are comparable? (Without (c), we could let $\le$ be $\subseteq$.)

(One can think of a total preorder as a very general generalization of measure or probability, so the question is whether there is any measure-like thing that preserves strict inclusion, i.e., is "regular" in Bayesian terminology.)

Update: The answer is negative if we require $r[A] \le r[B]$ for all isometries $r$ of the circle (i.e., we include reflections). That's because (b)-for-isometries and (c) in that case imply that $A \le B$ iff $r[A] \le B$ for every isometry $r$, and we already saw that in that case we have a problem.

To see that $A \le B$ iff $r[A] \le B$, note that all rotations are compositions of reflections, and so all we need to show is that if $A \le B$, then $\sigma[A] \le B$ for any reflection $\sigma$. If $\sigma$ is a reflection and $A \le B$, suppose for a contradiction that we don't have $\sigma[A] \le B$. Then by (c) we have $\sigma[A] > B$. By the analogue of (b), we have $A = \sigma^2[A] > \sigma[B]$. Since $B \ge A$, we have $B > \sigma[B]$. Thus, by (c) $\sigma[B] > \sigma^2[B]$ and so $\sigma[B] > B$, and we have a contradiction.

But I still don't know what happens if instead of all isometries we just have all rotations.

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What if instead of T we ask the same question about Z? The integers are a subgroup of the circle, so it should be easier to give a preorder on them, if there is one. –  domotorp Sep 8 '13 at 8:23
    
Indeed: a negative answer for Z will give a negative answer for T. –  Alexander Pruss Sep 9 '13 at 2:21
    
But in fact we have a positive answer for T, hence also for Z. :-) –  Alexander Pruss Sep 16 '13 at 4:39
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1 Answer

up vote 1 down vote accepted

The answer is positive, given Choice. It turns out that whenever $G$ is an abelian group acting on a set $X$, then there is a $G$-invariant preorder $\le$ on $2^X$ such that if $A$ is a proper subset of $B$, then $A<B$.

The proof uses the fact that $G$-invariant partial orders on $\Omega$ extend to $G$-invariant linear orders, provided that $G$ is abelian and no element of $G$ has a finite orbit of length greater than one. While $2^X$ in general won't satisfy the orbit condition, one can instead take $\Omega$ to be a quotient of $2^X$ by the relation $A\sim B$ iff there is a $g\in G$ and $n\in\mathbb N$ such that $gB=A$ and $g^nB=B$.

A writeup is here.

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