# R^2 homeomorphic to R^n

1. Nov 9, 2008

### gop

1. The problem statement, all variables and given/known data

Prove that $$R^2$$ and $$R^n$$ are not homeomorphic if $$n\neq2$$ (Hint: Consider the complement of a point in $$R^2$$ or $$R^n$$).

2. Relevant equations

3. The attempt at a solution

The proof that $$R^n$$ is not homeomorphic to $$R$$ is done by considering that if they are homeomorphic i.e. there exists a continuous bijection with continuous inverse $$f:R\rightarrow R^n$$. Then the restriction $$f:R\backslash {0} \rightarrow R^n\backslash f(0)$$ is also a continuous bijection. Since $$R\backslash {0}$$ is not connected but $$R^n\backslash f(0)$$ is if $$n\neq 1$$ we have a contradiction.

However, to show that $$R^2$$ is not homeomorphic to $$R^n$$ this doesn't work. There is also no other topological invariant I could detect. Both $$R^2$$ and $$R^n$$ are contractible to a point and thus have the same fundamental group, for example.

I would appreciate any idea
thx

2. Nov 9, 2008

### CompuChip

Instead of taking out a point, try taking out something bigger :)

3. Nov 9, 2008

### HallsofIvy

Staff Emeritus
You can continue that idea. In R2\{0}, a closed loop containing a point p cannot be contracted to a point. In Rn, for n larger than 2, it can.