# Topologically homogeneous space?

I want to know an example of a topological space $$X$$ which satisfies the following.

• for all points $$x,y\in X$$ and neighborhoods $$U_x$$, $$U_y$$, there exist neighborhoods $$U'_x\subset U_x$$, $$U'_y\subset U_y$$ of $$x$$ and $$y$$ that are homeomorphic (the homeomorphism does not have to map $$x$$ to $$y$$).

• $$X$$ has some kind of good conditions, i.e Hausdorff, locally connected, locally compact, second countable, etc.

• **X is not locally Euclidean ** (i.e., not a topological manifold)

I can't find the good example.

In fact, one goal of my question is this: "How can I make locally euclidean property from other topological properties."

• Reminds me of uniform spaces: en.wikipedia.org/wiki/Uniform_space – Will Jagy Aug 24 '10 at 5:13
• Your edit "How can I make locally euclidean property from the other topological properties" should properly be posted as a new question (fleshed out a little). – Nate Eldredge Aug 24 '10 at 12:27
• @Kim, in the literature what you're asking for would be called a manifold in the literature. Take a google scholar search of "fractal manifold". Things that look locally like a Sierpinski gasket would be Sierpinski gasket manifolds. FYI it doesn't take much to make the Sierpinski space into a locally homogeneous space I believe removing the three extremal vertices should do the job. – Ryan Budney Aug 24 '10 at 17:53
• Ryan, the factoid about Sierpinski gasket doesn't sound right: isn't it the case that it has points of ramification index 2,3,4? – Victor Protsak Aug 24 '10 at 20:57
• In Kim's definition the homeomorphism between the neighbourhood U of x and V of y does not have to send x to y. At least, it's not written that way. – Ryan Budney Aug 25 '10 at 4:33

An infinite dimensional torus $X = \prod_{n=1}^\infty S^1$ has all these properties. It's a topological group, so certainly homogeneous. It is compact, metrizable, connected, locally connected, and second countable. But it is not locally Euclidean.

• The infinithe dimensional torus is an infinite dimensional manifold, i.e locally R^infinity. In a certain sense, It is also locally euclidan. – Seonhwa Kim Aug 24 '10 at 6:17
• @Seonhwa Kim: X is not locally Euclidean if endowed with the product topology, see e.g. en.wikipedia.org/wiki/Product_topology . The point is that a basic open set is a product of open sets $U_i$, only a finite number of which are different from $S^1$. The topology you have in mind is too fine and would not be compact. – Benoît Kloeckner Aug 24 '10 at 7:06
• @ Benoît Kloeckner: thank you for good point. but I think that it depends on what is the meaning of infinite dimensional "locally euclidean". I think it means usually infinite dimensional vector space or function space. and that has product topology or compact-open topology. I have known that the infinite dimensional torus with product topology is an an infinite dimensional manifold. I'd like to find the example which is not manifold. – Seonhwa Kim Aug 24 '10 at 9:05
• @Seonhwa Kim: Nate Elredge is right to ask you for clarification, it seems to me that even infinite-dimensional, a manifold should be locally homeomorphic to some topological vector space. I hardly want to call a space, each open set of which has a non-finitely generated fundamental group, a manifold. – Benoît Kloeckner Aug 24 '10 at 12:32
• Also, the infinite dimensional torus is not locally $\mathbb{R}^\infty$ or any other infinite dimensional topological vector space, since it is locally compact. – Nate Eldredge Aug 24 '10 at 13:13

The $\mathbb R^2\setminus \mathbb Q^2$ provides an example that is locally connected but not locally compact.

Proof. We need to show that the group of automorphisms of $\mathbb R^2\setminus \mathbb Q^2$ is acting transitively on $\mathbb R^2\setminus \mathbb Q^2$. The idea is to conisder piecwise linear automorphisms of $\mathbb R^2\setminus \mathbb Q^2$ with rational coefficients with infinately many breaks.

In order to explain how this works we will conisder the $\mathbb R\setminus \mathbb Q$ and prove the statement here. Let $x$ and y be two points in $\mathbb R\setminus \mathbb Q$. Then conisder two monotonly decresing (for $i\ne 0$) seuqences of rational numbers $x_i$, $y_i$, $i\in \mathbb Z$ with $x_0=x$, $y_0=y$, with $lim_{i\to + \infty} x_i=x$, $lim_{i\to + \infty} y_i=y$ and $lim_{i\to - \infty} x_i=x$, $lim_{i\to - \infty} y_i=y$. Finally take the piecwise linear map from $\mathbb R\setminus \mathbb Q$ to itself that sends $x_i$ to $y_i$. This is the automorphism we a looking for. In the case of $\mathbb R^2\setminus \mathbb Q^2$ the same thing can be done by chosing anappropriate triangulations of $\mathbb R^2$.

I don't see how to make a locally compact set.

This survey paper on the Bing-Borsuk conjecture may be useful.

• Very good survey! – Victor Protsak Aug 25 '10 at 8:11

If you drop the locally connected assumption, the middle third Cantor set satisfies the desired properties. The proof can be given as in Nate Eldredge's answer, since it is homeomorphic to $\{0,1\}^{\mathbb{N}}$.

A homogeneous continuum is a compact connected metric space X such that for any two points x,y there is a homeomorphism of X taking x to y. This obviously implies that X is locally the same everywhere (a priori, it is a stronger condition). There are plenty of examples in books on general topology. My favorite one is a solenoid, which is not a manifold because, for example, it is not locally connected.

ADDENDUM The Menger curve C (also known as the Menger sponge, Menger universal curve, and Sierpinski universal curve) is a one-dimensional locally connected continuum. R.D. Anderson proved a characterization which implies that C is n-point homogeneous and that, moreover, up to a homeomorphism, the circle and C are the only one-dimensional homogeneous locally connected continua.

Anderson, R. D. A characterization of the universal curve and a proof of its homogeneity. Ann. of Math. (2) 67 1958 313-324 MR

Anderson, R. D. One-dimensional continuous curves and a homogeneity theorem. Ann. of Math. (2) 68 1958 1-16 MR

By the way, I am not a general topologist: all information can be easily found using web searches starting with "homogeneous continuum".

• I have pointed that "locally connected" property. The examples of homogeneous continuum in wikipedia "en.wikipedia.org/wiki/Continuum_(topology)"; are locally euclidean or not locally connected. and the examples like topologist's sine curve what I know is also not locally connected. As wikipedia, Peano continuum is locally connected homogeneous continuum, but I can not make the concrete example too. – Seonhwa Kim 28 secs ago – Seonhwa Kim Aug 24 '10 at 6:34
• Sorry, I didn't notice the "locally connected" assumption (you may want to make your requirements on X more explicit, instead of including them as an afterthought). I can think of two negative results:  (1) Masurkiewicz proved that a locally connected homogeneous planar continuum is a Jordan curve;  (2) Bing and Borsuk proved that an n-dimensional homogeneous continuum that is an absolute neighborhood retract (ANR) is a topological manifold for n=1,2. Apparently, it's unknown whether this remains true for higher n. – Victor Protsak Aug 24 '10 at 7:32
• The Menger curve is a very good example indeed. – Benoît Kloeckner Aug 24 '10 at 12:37

The Hilbert cube $$\,\ \mathcal H\ :=\ [0;1]^\mathbb N\,\$$ is a homogeneous space and it seems beautiful to me (it's compact, connected, locally connected, contractible, it's an absolute retract, ...).

The Hilbert cube decomposes into a cartesian product in a great variety of ways.

Oh, by the way, $$\ \mathcal H\$$ is not locally Euclidean nor locally Hilbert (no infinite-dimensional compact is).

• Is it obvious that it's homogeneous? – YCor Feb 21 at 13:15
• @YCor Homogeneity of HQ is not obvious, it is a theorem, see here for references. – Moishe Kohan Feb 21 at 14:24
• Thanks! Note that the fact it satisfies the weaker homogeneity (apparently) asked by the OP is much easier: indeed it's immediate that every point of $\mathcal{H}$ has a basis of neighborhoods all homeomorphic to $\mathcal{H}$. – YCor Feb 21 at 17:21
• @YCor, construct a homeomorphism which maps $p$ to pseudo-interior (the rest is obvious). Let $\ k\$ be the first index for which $\ p_k\in\{0\ 1\}\$ (or else we're done). Rotate sq. $[0;1]^{\{k\ k\!+\!1\}}\$ so that the $k$-th coordinate $\notin \{0\ 1\}$ (the two coord's of the image of $p$ move within the border of the sq. away from $\ \{0\ 1\}\times[0;1].\$ Now, repeat it for the first border coord. of the image, say $p_n\$ (by necessity, $k<n$). Etc. This procedure ends in or converges to a homeomorphism -- the image of $p$ will belong to the pseudo-interior. – Wlod AA Feb 22 at 2:34