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If $\alpha : X \to X$ is a minimal homeomorphism on a compact Hausdorff space $X$, then if $X$ is connected, $\alpha$ is totally minimal, that is $\alpha^k$ is minimal for every $k \in \mathbb{Z}$.

I am interested in a minimal homeomorphisms on locally compact but not necessarily compact spaces.

Is there a similar result relating connected spaces to total minimality in the not necessarily compact case?

In general, it seems to be difficult to find results on minimal homeomorphisms on noncompact spaces, so in addition to the above question, any good reference here would be very useful.

Thank you, Nimh

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Assume $\alpha:X\to X$ is not totally minimal. Then there exists $k$ such that $\alpha^k:X \to X$ is not minimal. Consider the quotient space $X/\alpha^k$ endowed with the quotient topology and let $Y$ be the corresponding $T_0$ quotient, that is, the space obtained by identifying points not separated by the topology. The non-minimality of $\alpha^k$ is equivalent to $Y$ not being a singleton. $Y$ is acted upon by the finite group $\mathbb{Z}/(k)$ generated by the image of $\alpha$. If $\alpha$ was minimal on $X$ to begin with, then the $\mathbb{Z}/(k)$ action on $Y$ is minimal, thus transitive, and it follows that $Y$ is a finite discrete set of points. Thus the natural map $X\to Y$ gives a non-trivial discrete quotient of $X$, thus $X$ is not connected.

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    $\begingroup$ You use that a continuous minimal action of a finite group on a $T_0$ space $Y$ has to be transitive (and $Y$ is then finite discrete). Indeed, otherwise there are at least two orbits and we can then assume that $Y$ has two orbits, hence is finite. Then the union of minimal closed subsets is closed, invariant, and Hausdorff, and is all of $Y$ by minimality. (I'm writing this since it took me time to be convinced by this argument using non-Hausdorff topologies.) $\endgroup$ – YCor Nov 21 '17 at 10:40
  • $\begingroup$ This answer seems to work for any topological space $X$, with no restriction at all. The definition of minimality is that $X\neq\emptyset$ and the orbit $\{\alpha^nx,n\in\mathbf{Z}\}$ is dense for every $x\in X$. (For compact $X$ it's equivalent to the density of $\{\alpha^nx,n\in\mathbf{N}\}$ for all $x$.) $\endgroup$ – YCor Nov 21 '17 at 10:45

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