Are there non-homogeneous topological spaces which are "almost homogeneous" in Rudin's sense?

Question

Let $X$ be a nonempty topological space and $G$ the group of all homeomorphisms $g:X\to X$. Recall that $X$ is homogeneous if $G$ acts transitively on $X$. In

"Homogeneity Problems in the Theory of Čech Compactifications", The Mathematical Legacy of Eduard Čech (1993), pages 81-92,

Walter Rudin considers the following condition on $X$: for all point $x\in X$ and all nonempty open subset $U\subset X$ there is a $g\in G$ such that $g(x)\in U$. He calls such a space almost homogeneous.

I suspect that there are simple examples of almost homogeneous non-homogeneous spaces, but I haven't been able to find any. Let me ask the implicit question formally, together with an obvious follow-up question:

Question 1. Are there almost homogeneous non-homogeneous spaces?

Question 2. Assuming the answer to Question 1 is Yes, what is the least possible cardinality of an almost homogeneous non-homogeneous space?

[I added the quotient-spaces tag because, in the above notation, $X$ is almost homogeneous if and only if the quotient topology of $G\backslash X$ is coarse.]

Answer 1

Let $X_n$ be the Cayley graph of the free product $G_n=\frac{1}{2^n}\mathbb{Z}*\frac{1}{2^n}\mathbb{Z}$ (with respect to the two obvious generators). Call edges of $X_n$ that come from the first generator horizontal edges and edges that come from the second generator vertical edges. The inclusion map $G_n\to G_{n+1}$ induces an inclusion map $X_n\to X_{n+1}$. Let $G=\mathbb{Z}[1/2]*\mathbb{Z}[1/2]$ be the colimit of the groups $G_n$, and let $X$ be the colimit of the spaces $X_n$. Note that $G_n$ acts on $X_n$ for each $n$, and these actions glue to give an action of $G$ on $X$. There is also an automorphism of $G_n$ that swaps the two generators which induces an automorphism of $X_n$ that swaps the horizontal and vertical edges, and these glue to give an automorphism $\sigma$ of $X$.

Now I claim $X$ is almost homogeneous. Let $x\in X$ and let $U\subseteq X$ be a nonempty open subset. Then there is some $n$ such that $x\in X_n$ and $U$ contains an open interval of some edge of $X_n$. Applying the automorphism $\sigma$ if necessary, we may assume that $x$ is in a horizontal edge of $X_n$ and that $U$ also contains an open interval of a horizontal edge of $X_n$.

Now note that each edge of $X_n$ gets split in half to become two edges in $X_{n+1}$. So for some $N\geq n$, the open interval that $U$ contains in $X_n$ actually contains an entire horizontal edge of $X_N$. But now the action of $G_N$ is transitive on the horizontal edges of $X_N$, so some element of $G_N$ maps $x$ into the horizontal edge of $X_N$ that $U$ contains. This gives an automorphism of $X$ that maps $x$ into $U$.

On the other hand, $X$ is not homogeneous. To see this, note that if $x\in X$ is a vertex of $X_n$ for some $n$, then $X\setminus \{x\}$ has $4$ connected components (the four "branches" of the tree coming out from the vertex), whereas if $x\in X$ is not a vertex in any $X_n$, then $X\setminus\{x\}$ has only $2$ connected components (the two sides of the tree coming off of the edge that $x$ is in the middle of in each $X_n$).

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This example has cardinality $2^{\aleph_0}$; I don't know whether you can get a smaller example but I suspect it is possible to get a countably infinite one. It is easy to see that an almost homogeneous finite space must be homogeneous (consider the minimal nonempty open sets; by almost homogeneity every point must be in one and they all must have the same cardinality, which means the space is a disjoint union of indiscrete spaces of the same cardinality).

Answer 2

Yes, there are such spaces, as showed by Eric Wofsey. Another way to produce examples of almost homogeneous but not homogeneous spaces is to look at any minimal $\Bbb Z$-flow (meaning a compact $\Bbb Z$-space such that every orbit is dense) which is not homogeneous, here is a way to build one:

Given a rectangle $B=[a,a+h]\times[b,b+k]$ let $\lambda(B)$ denote the disjoint union of three rectangles obtained by removing the second and fourth vertical fifth of $B$, and then removing the top half and the bottom half of the left and right rectangles thus obtained (see the two figures below for the first two stages). More formally $\lambda(B)=B_0\cup B_1\cup B_2$ where \begin{align*} B_0&=[a,a+h/5]\times[b,b+k/2]\\ B_1&=[a+2h/5,a+3h/5]\times[b,b+k]\\ B_2&=[a+4h/5,a+h]\times[b+k/2,b+k]. \end{align*}
Now let $B^{(0)}=[0,1]\times[0,1]$ and define $B^{(n+1)}=\lambda(B^{(n)})$, so that $B^{(n)}$ consists of $3^n$ disjoint rectangles, numbered as in the two figures below, that is $$\lambda(B^{(n)}_j)=B^{(n+1)}_{j}\cup B^{(n+1)}_{j+3^n}\cup B^{(n+1)}_{j+2\cdot 3^n}.$$

Let $X=\bigcap_n B^{(n)}$, which is a compact metric space made of vertical line segments (which are all degenerate to points except for countably many segments). There is one central segment of length $1$ and for each $n\in\Bbb N$ there are infinitely many segments of length $1/2^n$. This space is not homogeneous, because there are points with different local dimensions, namely $0$ and $1$ depending on whether they belong to a degenerate segment or not. The permutations $B^{(n)}_j\mapsto B^{(n)}_{j+1}$ induce a homeomorphism $\phi$ of $X$ (note that $\bigcap_n B^{(n)}_{j_n}\neq\varnothing$ iff $j_{n+1}\equiv j_n\;\;(\mathrm{mod }\;\;3^n)$, so the permutations induce a map $\bigcap_n B^{(n)}_{j_n}\to \bigcap_n B^{(n)}_{{j_n+1}}$, if both are nontrivial segments map one onto the other linearly). Then the $\Bbb Z$-flow coming form $\phi$ is minimal, since every point has dense orbit (the orbit of every point intersects every $B^{(n)}_j$), which means that this space is almost homogeneous, since $\phi\in\mathrm{Homeo}(X)$.

Stage 1 of the construction.

Stage 2 of the construction.

Regarding the cardinality question here are some observations. First of all note that $X$ also has cardinality $2^{\aleph_0}$. In general you won't find any countable example among compact hausdorff spaces:

Lemma: Let $G$ be a group and $X$ a minimal $G$-flow. Then either $X$ is finite or $X$ is perfect.
Proof: Suppose that there is an isolated point $x_0$. Since every orbit is dense, $x_0$ belongs to every orbit, which implies that there is a single orbit and every point is isolated. By compactness of $X$, it must be a finite space.

together with the standard result that perfect compact hausdorff spaces are uncountable.

Answer 3

A countable example is given by the prime spectrum of the ring $\mathbb Z$.

For the reader not familiar with prime spectra, here is a self-contained definition of a topological space $X$ homeomorphic to the prime spectrum of $\mathbb Z$.

As a set $X$ is the union of $\mathbb N$ and $\{\infty\}$ with $\infty\notin\mathbb N$. The topology of $X$ is characterized by the condition that the proper closed subsets of $X$ are precisely the finite subsets of $\mathbb N$.

If $G$ is the group of all self-homeomorphisms of $X$, then a bijection $g:X\to X$ is in $G$ if and only if $g(\infty)=\infty$. The orbits of $G$ in $X$ are $\mathbb N$ and $\{\infty\}$, each of them being dense in $X$.