Help defining a topological property After an introductory course in General Topology I started to look up many examples of topologies (mainly on $\mathbb{R}$) just to get a feel for how different they can be and it seems to me that they all can be fit in one of the following classes:


*

*Topologies with "special points" like the $K$-topology or the collapse $\mathbb{R}/\mathbb{Z}$ where the neighborhoods of the point $0$ are substantially different from those of the other points.

*Topologies in which all points are "alike" such as the standard topology on $\mathbb{R}$ or the cofinite topology.
I attempted to formalize this property for topologies of type (2):

if $U$ is a neighborhood of any point $x$ then every other point $y$ have a neighborhood $V_y$ homeomorphic to $U$.

My question is: is this intuition justified? 
 A: There is a topological property called "homogeneity":
A space $X$ is homogeneous iff for every $x,y \in X$ there exists some homeomorphism $f: X \to X$ such that $f(x) = y$.
This implies your local version, as for any neighbourhood $U_x$ of $x$, $f[U_x]$ is a (homeomorphic) neighbourhood of $y$. Your version is weaker (as all manifolds are locally the same, but need not be homogeneous).
All topological groups (like $\mathbb{R}$) are homogeneous, because translations are all homeomorphism (if we have a group structure $1, x\cdot y,  x^{-1}$ on $X$, with all operations continuous, we can use $f(t) = y \cdot x^{-1} \cdot t $ as such a map, mapping $x$ to $y$..)
There are also homogeneous spaces that do not admit a topological group structure, it's a more general idea.
A: I'll expand on what yanko mentioned above. Let $\mathcal{T}$ be any topology on $\mathbb{R}$ where $+ : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ is continuous, and has continuous inverse 
Fix $x \in \mathbb{R}$, and pick $y \neq x \in \mathbb{R}$ . Let $U$ be a neighbourhood of $x$.
If $y=x$, $U$ is also a neighbourhood of $x$ and the identity map is the desired homeomorphism.
If $x < y$, then let $f : U \to \mathbb{R}$ be defined by $f(t) = t + |x-y|$, $f$ is clearly continuous, and bijective onto its image $f[U]$, and it has continuous inverse given by  $f^{-1}(t) = t - |x+y|$, hence $V = f[U]$, since $f$ is a homeomorphism onto its image and since $f(x) = x+ |x-y| = y$, we can see that $V$ is a neighbourhood of $y$.
A similar argument proves the existence of a neighbourhood $V$ of $y$, in the case of $y < x$.
So your property is correct, and your intuition justified.
