Thread: R^2 and R^2 - {(0,0)} are not homeomorphic

1. R^2 and R^2 - {(0,0)} are not homeomorphic

Let R denote the real line and R^2 the real plane.

How do I show that R^2 and R^2 - {(0,0)} are not homeomorphic? I don't even know where to start since I cannot think of a topological property that is in one but not in the other. Can anyone give me a clue on what that property is?

A similar problem is this:
Consider S = the union of the x and y- axes together with the subspace topology inherited from R^2. Show that S and R are not homeomorphic.

2. For the second problem. Let $\displaystyle f :S\rightarrow \mathbb R$ an homeomorphism. We denote by $\displaystyle g :S\setminus \left\{(0,0)\right\}\rightarrow \mathbb R\setminus \left\{f(0,0)\right\}$. $\displaystyle g$ is a homeomorphism. The first set has four connected components whereas the second has two.

3. Originally Posted by mysteriouspoet3000
Let R denote the real line and R^2 the real plane. How do I show that R^2 and R^2 - {(0,0)} are not homeomorphic?

Hint:

$\displaystyle \mathbb{R}^2$ is contractible and $\displaystyle \mathbb{R}^2-\{(0,0)\}$ it is not.

4. Thanks for your replies! It really helps. :-)

Sir fernandorevilla, I haven't encountered the concept of contractible (contractibility?) yet. Do you have any other suggestions?

5. Originally Posted by mysteriouspoet3000
Do you have any other suggestions?

Simply connected is a topological property, $\displaystyle \mathbb{R}^2$ is simply connected but $\displaystyle \mathbb{R}^2$ minus the origin is not.

6. consider the path p(t) = (cos(2πt), sin(2πt)) and the path q(t) = (cos(2πt) - 4, sin(2πt)). in $\displaystyle \mathbb{R}^2$ you can deform p to q continuously without ever leaving $\displaystyle \mathbb{R}^2$.

why can't you do this in $\displaystyle \mathbb{R}^2 - \{(0,0)\}$?

7. Originally Posted by girdav
For the second problem. Let $\displaystyle f :S\rightarrow \mathbb R$ an homeomorphism. We denote by $\displaystyle g :S\setminus \left\{(0,0)\right\}\rightarrow \mathbb R\setminus \left\{f(0,0)\right\}$. $\displaystyle g$ is a homeomorphism. The first set has four connected components whereas the second has two.
g is clearly bijective. But why is g and g^-1 continuous? Am I missing something obvious here?