Uncategorized – Page 12 – Random Permutations

I don’t claim that I’m the first person to have come up with this example; I only claim that yesterday is the first time I thought of it. This is a rather sorry state of affairs, because it’s an obvious and nice example.

First, some background, the example of a nonmeasurable set that I was given when I first learnt measure theory went basically as follows. Start with ${[0,1]}$ , and call ${x,y\in[0,1]}$ equivalent if ${x-y\in{\bf Q}}$ . This defines an equivalence relation, and thus we can define a set ${Z}$ to consist of exactly one representative from each equivalence class. Then the interval ${[0,1]}$ is just the disjoint union of translates mod ${1}$ of ${Z}$ by rational numbers mod ${1}$ (or something). Because the rationals in ${[0,1]}$ are countably infinite this is incompatible with ${Z}$ being measurable: if ${\mu(Z)>0}$ then then ${\mu([0,1])=\infty}$ , and if ${\mu(Z)=0}$ then ${\mu([0,1])=0}$ .

This example is fine and beautiful and all that, but I find it rather hard to remember and especially difficult to visualize. Thus, yesterday, when I was explaining to my girlfriend in an intuitive way why not all sets are measurable, the example I produced was, by accident, not the above example.

From an intuitive point of view, if we’re talking about ${[0,1]}$ and we’re generally thinking about mod ${1}$ then we’re probably better off just talking about the circle ${S^1}$ and not confusing the issue. Moreover, now translation mod ${1}$ is just rotation, which is much easier to visualize and, importantly, much easier to believe is measure-preserving. Now we just need to decompose ${S^1}$ into a countably infinite collection of copies of some set ${Z}$ . Towards this end, let ${T}$ be any irrational rotation of ${S^1}$ . (When I was talking yesterday, I started with the concrete example of rotation by ${1}$ degree, and then hastily changed this to rotation by ${1}$ radian.) Then for any point ${x}$ we can consider the orbit ${\{T^n(x)\}_{n\in\mathbb{Z}}}$ , and we thus decompose ${S^1}$ into a union of orbits under the action of ${T}$ . Let ${Z}$ be a set consisting of exactly one representative of each of these orbits. Since every orbit is infinite, ${S^1}$ is the disjoint union of

$\displaystyle \ldots,T^{-2}(Z), T^{-1}(Z), Z, T(Z), T^2(Z),\ldots.$

Since ${T}$ is measure-preserving, we have a contradiction as before.

Recall the following theorem: Given a set ${X}$ with two topologies ${\cal U}$ and ${\cal V}$ , with ${\cal U}$ weaker than ${\cal V}$ , if ${\cal U}$ is Hausdorff and ${\cal V}$ is compact then in fact ${\cal U = \cal V}$ .

In general, is there a “right” topology, in this sense? Specifically, given a Hausdorff topology ${\cal U}$ can you weaken it to a compact topology which is still Hausdorff, and given a compact topology ${\cal V}$ can you strengthen it to a Hausdorff topology which is still compact?

For example, consider the cofinite topology on ${\omega}$ . The cofinite topology is always compact. Can we extend this topology to a Hausdorff topology? Yes we can: ${\omega}$ could just as well have been any countably infinite set, say ${\{0\}\cup\{1/n: n\in{\bf Z}^+\}}$ , and now one checks that the usual topology (induced by ${\mathbf{R}}$ , say) is compact, Hausdorff, and stronger than the cofinite topology (because finite sets are closed).

EDIT: A more or less complete answer can be found in this MO answer.

Category: Uncategorized

A simpler example of a nonmeasurable set

A question in general topology