Question

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.

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additional supplement

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

Answer 1

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.

Answer 2

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".

Answer 3

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 $\{0,1\}^{\mathbb{N}}$.

Answer 4

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.

Answer 5

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