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."
 A: 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.
A: 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.
A: This survey paper on the Bing-Borsuk conjecture may be useful.
A: 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".
A: 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: 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).
