# Homeomorphism proof

$\mathbb{R}^2-(0 \times \mathbb{R}_+) \approx \mathbb{R}^2$

Now consider the map that sends the line $(-1 \times \mathbb{R}_+)$ to $(0 \times \mathbb{R}_+)$. And then continue this inductively. Every function is a map, because it is a translation. The composition of any finite number of maps is a map. The question I have is whether the countably infinite composition of these maps is a map. Thank you.

-
There's no meaningful definition of an infinite composition of maps. If $s(n)=n+1$ then what does the map $s^{\infty}$ map $n$ to? – Dan Rust Jun 22 '14 at 22:46
In the plane you can find simple homeomorphisms: first contract $R^2$ onto $R^*_+\times R$ via $(x,y)\mapsto(e^x,y)$, and then take the square by identifying the real plane with the complex numbers: $z\mapsto z^2$. This will give you a homeomorphism $R^2\to R^2\setminus R_-\times 0$ which you can then rotate to where you want it. – Olivier Bégassat Jun 22 '14 at 22:50
I guess, @Mike meant the map $(n,x)\mapsto (n+1,x)$ if $n\in\Bbb Z,\ n<0$ and $x>0$. – Berci Jun 22 '14 at 22:50
@Berci: yeah, that's basically what I meant. Additionally, all other points are mapped to themselves. – Mike Jun 22 '14 at 23:00
But this is not continuous on points of $\{n\}\times\Bbb R^{\ge 0}$ for $n<0,\,n\in\Bbb Z$. – Berci Jun 22 '14 at 23:02

Hint: Use polar coordinates, the angle measured starting from the given ray $\{0\}\times\Bbb R^+$ (i.e. the $y$-axis), then deform the angles from $(0,360^\circ)$ to $(90^\circ,270^\circ)$, thus you get the open half plane.
What is the image of the point $(0,-1)$ under this 'map'? If you can't answer this question then it is not a map. – Dan Rust Jun 22 '14 at 22:56
@Mike a small ball around $(0,0)$ has non-open preimage because the preimage has two connected components, one of which is the single point $\{(-1,0)\}$. – Dan Rust Jun 22 '14 at 23:07