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I want to know the example which satisfies the following.

X is topological space.

for every point x,y in X, there exist open nbhd Ux,Uy of x,y which are homeomorphic.

X has some kind of good conditions, i.e hausdorff,locally connected,locally compact, 2nd countable.. etc..

X is not locally Euclidean .

I think.... the thing like topologically homogeneous space which is not manifold. but I can't find the good example.

additional supplement

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

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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
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5 Answers

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.

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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
Yes, I wondered about that, but I figured this answer would at least force you to clarify :) Indeed, I think you should try to state precisely what you mean by "infinite-dimensional manifold" since there are several notions. –  Nate Eldredge Aug 24 '10 at 12:26
@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
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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.

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

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Very good survey! –  Victor Protsak Aug 25 '10 at 8:11
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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 $\{0,1\}^{\mathbb{N}}$.

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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. 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".

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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
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