Homeomorphism between $\mathbb R^2 \setminus \left\{p \right\}$ and $\mathbb R^2 \setminus [p,q]$ where $[p,q]$ is an interval I have a problem with finding a homeomorphism between $\mathbb R^2 \setminus \left\{p \right\}$ and $\mathbb R^2 \setminus [p,q]$ where $[p,q]$ is an interval. Main problem is with continuity, of course. Thanks for help.
 A: Try finding a homeomorphism between each of these spaces with $R^2\setminus B(1)$ where $B(1)$ is a closed unit ball.
To obtain a homeomorphism between $R^2\setminus B(1)$ and $R^2\setminus [-1,1]$, just shift each point on the vertical ray above a point $x$ on the $x$-axis downward by $\sqrt{1-x^2}$ for all $x\in[-1,1]$ (and similarly for the negative ray: shift upward by the same amount). Otherwise the homeomorphism is the identity. 
A: Here's another construction.
Consider the set of all ellipses whose foci are the points $(-1,0)$ and $(1,0)$
Every point in the plane lies on exactly one of these ellipses. The geometric idea, now, is to continuously deform this collection of ellipses by moving the two foci to the origin. Each ellipse will ultimately be deformed into a circle centered on the origin.
This transformation compresses the interval $[-1, 1]$ of the $x$-axis to the origin, and is bijective everywhere else.
The most straightforward way to do this (without accidentally compressing everything to the origin) is to keep the $y$-intercepts of the ellipses fixed as they are deformed, and just translate each point horizontally.
I'm nearly certain there is something more beautiful you can do, but I'm failing to spot it.
