Show that 2 metric spaces aren't homeomorphic 
Show that $\mathbb{R}^2$ with railway metric and $\mathbb{R}^2$ with jungle river metric aren't homeomorphic.

In railway metric the distance between 2 points $A, B$ is measured by going from point $A$ to the origin and then to point $B$ (if $A, B$ are on the same ray from the origin then the distance is the Euclidean distance)
In jungle river metric the distance between $A, B$ is measured by travelling from point $A$ to the $x$-axis, travelling left/right along it and then moving up to the point $B$ (if $A, B$ are on the same vertical line then it's the Euclidean distance)
I missed the class on homeomorphisms. I need to show that there is no continuous bijection with continuous inverse between these 2 metric spaces, but why isn't there one? People have suggested using connectedness but we didn't cover that yet in class.
 A: HINT: Show that in the railway metric, every path-connected neighborhood $U$ of $0$ is such that $U \setminus \{0\}$ is path-connected. Proceed to show that $0$ is the only point with this property. In the jungle river metric, the whole $x$-axis has this property, so by a cardinality argument, the two spaces cannot be homeomorphic.

SOLUTION:
The balls in the railway metric look something like this for $x \neq 0$:

And balls in the jungle river metric look like this for $y' \neq 0$:

We will use that an interval and a disk are not homeomorphic (which can be seen by using path-connectedness - it is quite intuitive since paths get mapped to paths under continuous maps, so path-connectedness stays preserved under homeomorphisms). Maybe there may be other ways to prove this that I am not aware of.
Note that in the railway metric, $0$ is the only point such that every open set around zero contains an open set that is homeomorphic to the disk. This is because any other point has a ball around it that is homeomorphic to an interval (try to write down the homeomorphism if this is not clear).
However, in the jungle river metric, any point of the form $(x, 0)$ is such that every open set around $(x, 0)$ contains an open set that is homeomorphic to a square, which in turn is homeomorphic to the disk.
Since homeomorphisms preserve this property of points (check this if it is not clear), we have that there cannot possibly be a homeomorphism between these two metric spaces: There is only one point with the property in the railway metric, but uncountably many in the jungle river metric.
A: Basic obervations:


*

*If we remove the origin from the railway metric the resulting space has uncountably many connected components (all open half lines from the origin are connected, clopen and disjoint). 

*If we remove any point from the river metric plane, we get at most 4 components (depending on where the point lies, one on the "river" or not). 

*If $f: X \to Y$ is a homeomorphism, and $x \in X$ is arbitrary then $X\setminus \{x\}$ is (via the restriction of $f$) homeomorphic to $X\setminus \{f(x)\}$.
So assume there is a homeomorphism $f: X \to Y$ where $X$ is the railway-plane and $Y$ is the "river-plane".
Take $x$ the origin and apply $3$: $X\setminus \{x\}$ is homeomorphic to $Y\setminus\{f(x)\}$. But by 1 and 2. the former has more connected components than the latter, contradiction (homeomorphic spaces have the same number of connected components). 
So $f$ cannot exist. QED
