Are R^2 and R^n Homeomorphic If n is Not Equal to 2?

[gop]

Homework Statement

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).

Homework Equations

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

 

[CompuChip]

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

 

[HallsofIvy]

Homework Statement

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).

Homework Equations

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

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.