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

• Apr 1st 2011, 04:01 AM
mysteriouspoet3000
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.

• Apr 1st 2011, 05:20 AM
girdav
For the second problem. Let $f :S\rightarrow \mathbb R$ an homeomorphism. We denote by $g :S\setminus \left\{(0,0)\right\}\rightarrow \mathbb R\setminus \left\{f(0,0)\right\}$. $g$ is a homeomorphism. The first set has four connected components whereas the second has two.
• Apr 1st 2011, 05:41 AM
FernandoRevilla
Quote:

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:

$\mathbb{R}^2$ is contractible and $\mathbb{R}^2-\{(0,0)\}$ it is not.
• Apr 1st 2011, 08:27 AM
mysteriouspoet3000
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?
• Apr 1st 2011, 12:29 PM
FernandoRevilla
Quote:

Originally Posted by mysteriouspoet3000
Do you have any other suggestions?

Simply connected is a topological property, $\mathbb{R}^2$ is simply connected but $\mathbb{R}^2$ minus the origin is not.
• Apr 3rd 2011, 09:35 AM
Deveno
consider the path p(t) = (cos(2πt), sin(2πt)) and the path q(t) = (cos(2πt) - 4, sin(2πt)). in $\mathbb{R}^2$ you can deform p to q continuously without ever leaving $\mathbb{R}^2$.

why can't you do this in $\mathbb{R}^2 - \{(0,0)\}$?
• Apr 6th 2011, 10:13 AM
mysteriouspoet3000
Quote:

Originally Posted by girdav
For the second problem. Let $f :S\rightarrow \mathbb R$ an homeomorphism. We denote by $g :S\setminus \left\{(0,0)\right\}\rightarrow \mathbb R\setminus \left\{f(0,0)\right\}$. $g$ 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?