Is compactness needed in Birkhoff's theorem on dynamic systems?

Question

Let $G$ be a topological group acting on a space $X$. It is know that if $X$ is compact, then there exists a minimal set as follows:

Theorem (Birkhoff) If $X$ is compact and nonempty, there exists a nonempty, closed $G$-invariant subset of $X$ which is minimal with respect to these properties.

What is a nice example for a noncompact space $X$ where this goes wrong?

Edit: Sorry, I accidentally had the word "proper subset" in the theorem above, which would make it wrong.

Answer 1

As noted in the comments, the word "proper" should not be here : $S^1$ (or any compact group) acts on itself transitively.

Now for an example where this fails, even when removing "proper" is the following (there is probably a simpler one) :

Construct $X$ in the following way :at the first level you have a discrete $\mathbb{Z}$. To it, you attach $\mathbb{Z}$ points that behave like the infinite in the one-point compactification of $\mathbb{Z}$, i.e. a neighbourhood of such a point is this point + some cofinite set of the first $\mathbb{Z}$. The induced topology on this second $\mathbb{Z}$ is the discrete topology.

Then you add some more $\mathbb{Z}$'s on top of that.

Formally, our set is $X:=\displaystyle\bigcup_{n\in \mathbb{N}} \mathbb{Z}\times\{n\}$; and the topology is defined by induction : A neighbourhood of $(x,0)$ is any set containing it.

A neighbourhood of of $(x,n+1)$ is any set $U$ containing $(x,n+1)$ and such that there is a cofinite $A\subset \mathbb{Z}$ such that for each $a\in A$, there is a neighbourhood $U_a$ of $(a,n)$ such that $\displaystyle\bigcup_{a\in A} U_a \subset U$.

It is easily seen by induction on $m$ that $\displaystyle\bigcup_{n\geq m} \mathbb{Z}\times\{n\}$ is closed in $X$ and the induced topology makes it homeomorphic to $X$; and it is the closure of $\mathbb{Z}\times \{m\}$.

Then have $\mathbb{Z}$ act on it level by level : $n \cdot (x,m) = (n+x, m)$. This action is easily seen to be continuous by induction.

Now an invariant subset of $X$ contains whole levels, because $\mathbb{Z}$ acts transitively on each of them. So a closed invariant subset of $X$ must contain the closure of some $\mathbb{Z}\times \{n\}$, hence something of the form $\displaystyle\bigcup_{n\geq m} \mathbb{Z}\times\{n\}$, which shows that it is not minimal (since $\displaystyle\bigcup_{n\geq m+1} \mathbb{Z}\times \{n\}$ is also closed and invariant).

This space is moreover $T_1$ and locally compact. I believe with a similar spirit you can find $T_2$ locally compact spaces that work just as well; and I think you can even find some "non artificial" (whatever that means) ones

Answer 2

Here is an example with $X=\mathbf{R}^2$.

Let $D$ be the ordinal $\omega^\omega+1$ (here $\omega^\omega=\sup_n\omega^n$, and $+1$ means we take the 1-point compactification). This is a compact space. Its self-homeomorphism group acts on $D$, the orbits being given by the Cantor Bendixson rank (that is: by induction $D_n$ is the set of isolated points of $D\smallsetminus D_{n-1}$ and $D_\omega$ is a singleton).

Embed $D$ into the $2$-sphere $S$. Then each self-homeomorphism extends to a self-homeomorphism of $S$ (this is not trivial but not too hard). Let $G$ be the group of self-homeomorphisms of $S$ preserving $S$. Then the orbits of $G$ on $S$ are the $D_n$, the limit point $\infty$, and the open complement of $D$.

Now consider the action of $G$ on the plane $X=S\smallsetminus\{\infty\}$. Then its orbits are the same (except $\infty$), and hence the closed invariant subsets are $X$, the empty set and, for all $n$, the union $\bigcup_{k\ge n}D_n$. Hence there is no minimal nonempty closed invariant subset.