Let $X$ be a totally disconnected $G$-space, where $G$ is a locally compact Hausdorff group. Is the orbit space X/G also totally disconnected?

The same question for locally compact, Hausdorff, totally disconnected topological group $G$.


2 Answers 2


The orbit space is not necessarily disconnected. For a simple example, let $X$ be the irrational reals, with the order topology and let $G$ be the rationals with the discrete topology. $G$ is then locally compact, metrizable and totally disconnected, and it acts continuously on $X$ by the addition of real numbers. It is easy to see though, that every orbit is dense in $X$, therefore the orbit space is indiscrete.

Some additional notes:

  • Any subgroup of $\text{Homeo}(X)$ with the discrete topology has all the required properties, including being totally disconnected.
  • Adding the additional requirement that $G$ is not discrete makes no difference, because you can e.g. multiply $X$ and $G$ by a disconnected local field and let that act on itself by addition. This produces essentially the same orbit space.
  • Requiring that the group is totally disconnected makes no real difference anyway, since a continuous map into a totally disconnected space must be constant on on every connected subset of its domain. In this case that means that the action of the connected component of $G$ is always trivial.
  • $\begingroup$ Thanks for the answer. If we also assume that the orbit space X/G is Hausdorff, can we conclude that X/G is totally disconnected? $\endgroup$
    – m07kl
    Apr 28, 2014 at 6:08
  • $\begingroup$ I'm not sure. If you also add the condition that $X$ is compact Hausdorff, then it is zero-dimensional and the quotient map is closed as well as open. In that case $X/G$ is also zero-dimensional. $\endgroup$ Apr 30, 2014 at 0:13
  • $\begingroup$ Maybe $X$ only has to be locally compact Hausdorff, since then the homeomorphisms of $X$ can be extended to homeomorphisms of its one point compactification which fix the added point. $\endgroup$ Apr 30, 2014 at 0:35

Well, let's see. To show that $X/G$ is totally disconnected, (by which I assume you mean that every point is open), we need to show that every $y \in X/G$ is open.

Now, the open sets in a quotient space are by definition, the images under the projection map of the open sets of $X$. So, a set $S$ is open, if it is $\pi(U)$ for some open set $U$ in $X$ where $\pi: X \rightarrow X/G$ is the projection.

However, if $X$ is totally disconnected, then every subset of $X$ is open. Hence, $$y = \pi(\pi^{-1}(y))$$ is open and thus $X/G$ is totally disconnected.

This is actually an example of the more general fact that the images of totally disconnected spaces under open maps are totally disconnected.

  • $\begingroup$ Every point is open means space is discrete. Totally disconnected spaces are not necessarily discrete. A topological space X is totally disconnected if the connected components in X are the one-point sets. $\endgroup$
    – m07kl
    Apr 21, 2014 at 17:54
  • $\begingroup$ Is $X$ not necessarily locally path connected? In locally path connected spaces connected components are open. I think that was in my assumption sorry. $\endgroup$ Apr 25, 2014 at 1:15
  • $\begingroup$ X is not necessarily locally path connected. At least, I don't know the reason for it. $\endgroup$
    – m07kl
    Apr 25, 2014 at 6:58

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