# Homeomorphism between plane with different topologies

How would you show that spaces $(\mathbb{R^2},\cal{T}_r)$ and $(\mathbb{R}^2,\cal{T}_b)$, where $\cal{T}_r$ is a topology generated by jungle river metric (here) and equivalently $\cal{T}_b$ is generated by the British Rail metric (3.15 second one), are not homeomorphic?

I tried to think of any elementary topological property, with which I can characterize only one of those spaces, but they seem to be so alike... Yet not homeomorphic. Why?

No such points exists in the jungle. In short: The "main stream" is like $\mathbb R$ with standard metric, hence has at most two connected components after removing a point. But the corresponding open sets must reach into each of the vertical "arms", which look like $\mathbb R$ as well. To see this more explicitly: For any point $(x_0,y_0)$ the subsets $\{\,(x,y)\mid x<x_0\,\}$, $\{\,(x,y)\mid x>x_0\,\}$, $\{\,(x_0,y)\mid y<y_0\,\}$, $\{\,(x_0,y)\mid y>y_0\,\}$ are even path connected (along the obvious polyline paths that the metric suggests), hence $\mathbb R\setminus\{(x_0,y_0)\}$ has at most four connected components.
• Isn't the British rail metric the one in which an open set is an open interval on a line crossing $(0,0)$ point? Maybe I used the wrong name. If not, singletons are not open. – Jules Oct 20 '14 at 20:46
• Huh, apparently there are more than one rail metric. The one I'm used to is where two points are the usual Eucliden distance apart when they are linearly dependent, and $\|x\|+\|y\|$ otherwise. Definition 3.15 in these notes gives both. – Dan Rust Oct 20 '14 at 20:47
Removing the origin from the river metric gives us 4 path-components (the positive $y$-axis, the negative $y$-axis, and the sets $\{(x,y): x > 0\}$ and $\{(x,y): x < 0\}$.