1
$\begingroup$

Let $G$ be a group of bijective transformations of a set $X$. For all $x\in X$, the orbit of $x$ is the set $$G(x)=\{gx:g\in G\},$$ where $gx\in X$ is the result of the action of $g$ on $x$.

A set $A\subseteq X$ is said invariant if $gA=A$ for all $g\in G$, where $gA=\{gx:a\in A\}$. Any orbit $G(x)$ is clearly invariant, and any invariant set is an arbitrary union of orbits. Additionally, the set $I$ of all invariant sets is a $\sigma$- algebra.

My question is the following. Consider the $\sigma$-algebra $O$ generated by all orbits $G(x)$. I found stated elsewhere (without proof) that $I=O$, but I can't seem to prove this. The inclusion $O\subseteq I$ is trivial, but I am not sure how to proceed with the reverse inclusion. Does anyone know if the equality $I=O$ is actually true, and if so how to prove it?

$\endgroup$
1
$\begingroup$

This is not true in general. Indeed, let $A$ be the collection of sets which are either a countable union of orbits or the complement of a countable union of orbits. It is not hard to see that $A$ is a $\sigma$-algebra. Since every orbit is an element of $A$, $O\subseteq A$. But if there are uncountably many orbits, then not every invariant set is in $A$ (split the orbits into two uncountable sets and take the union of one of them), so $O\neq I$.

(Conversely, if there are only countably many orbits, then every invariant set is a countable union of orbits so $O=I$. So $O=I$ iff there are only countably many orbits.)

$\endgroup$
2
$\begingroup$

$O$ and $I$ are not equal when there are uncountably many orbits.

In such a situation you can show that the set consisting of countable unions of orbits and complements of countable unions of orbits is a sigma algebra (in fact, it is $O$).

But since there are uncountably many orbits, you can take some uncountable collection of them that isn't all but countably many of them, and that gives you an invariant set not in $O$.

(This might be easier to work through if you first consider the case where $X$ is just some uncountable set and $G$ acts trivially.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.