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Homeomorphism

Siron

Active member
Jan 28, 2012
150
Hi,

How can I prove there's no homeomorphism between $\mathbb{R}$ and $\mathbb{R}^2$. I thought they are topological almost the same.

Thanks in advance.
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
they almost are, there's just "one point of difference".

what happens to the connectedness of $\Bbb R$ when you remove one point? is this same true for $\Bbb R^2$?

(surely if they were homeomorphic this wouldn't happen, right?)
 

Siron

Active member
Jan 28, 2012
150
they almost are, there's just "one point of difference".

what happens to the connectedness of $\Bbb R$ when you remove one point? is this same true for $\Bbb R^2$?

(surely if they were homeomorphic this wouldn't happen, right?)
I guess if I remove one point $x \in \mathbb{R}$ then $\mathbb{R}$ is not connected anymore because it wouldn't be an interval then. An interval $X$ is defined as $\forall x,y \in X: x \leq y \leq z \Rightarrow y \in X$, which is not true in the case of $\mathbb{R}\setminus \{0\}$ for example, because if we take $x=-1, y=0 $ and $z=1$ then if $\mathbb{R}$ would be an interval we would have $0 \in \mathbb{R}$ which is not true.

Thus I guess that $\mathbb{R}^2$ is still connected if we remove one point. But how do I prove that? Maybe with path connectedness?