Monday, 14 November 2011

A simpler example of a nonmeasurable set

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
$\ldots,T^{-2}(Z), T^{-1}(Z), Z, T(Z), T^2(Z),\ldots.$
Since $T$ is measure-preserving, we have a contradiction as before.

Wednesday, 26 October 2011

A question in general topology

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 $\mathbb{R}$, 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 on http://mathoverflow.net/questions/15841/how-do-the-compact-hausdorff-topologies-sit-in-the-lattice-of-all-topologies-on-a.