I would like to show that $\Bbb{R}^2$ and $\Bbb{R}^2\setminus \{0\}$ are not homeomorphic without using Algebraic Topology. Is there an elementary way to do this?
Asked
Active
Viewed 158 times
2
-
http://en.wikipedia.org/wiki/End_%28topology%29 might be helpful? – Blah Nov 25 '14 at 07:48
-
1Your question was answered here: http://math.stackexchange.com/a/30888/5798 – Rudy the Reindeer Nov 25 '14 at 08:15
-
@RudytheReindeer Sorry I couldn't understand that solution , do you know any basic argument? – Mathronaut Nov 26 '14 at 14:06
2 Answers
0
I'm not sure one could prove it completely without AT, however we can prove it in a way that doesn't require a lot of deep AT. Using covering spaces one could show that the fundamental group $\mathbb R^2\setminus \{0\}$ is $\mathbb Z$, whereas that of $\mathbb R^2$ is the trivial group (every two paths are homo topic to each other). This is a contradiction since homeomorphisms induce an isomorphism between the fundemental groups
0
This doesn't fully answer the question, but on $\mathbb{R}^2\setminus\{0\}$, there exists a vector field $(x_2/(x_1^2+x_2^2), -x_1/(x_1^2+x_2^2))$ that has curl zero but is not a gradient. This shows that they are at least not diffeomorphic.
Peter Franek
- 11,522