# Connectedness of $\mathbb{R}^2\setminus \lbrace 0\rbrace$

Is there a proof that $\mathbb{R}^2\setminus \lbrace 0\rbrace$ is connected without using the idea of path-connected sets ( i.e. using the definition of connected sets only ).

-
Otherwise you could take the mapping $f(x,y)=(e^x \sin y ,e^x \cos y)$ its range is $\frac{ \mathbb{R}^2 }{ \{ 0 \} }$ and it is continuous and we know that connected subsets through continuous mapping go to connected subsets and you are done.. – clark Jun 27 '12 at 7:05
@clark: +1. By far the best answer :) – t.b. Jun 27 '12 at 9:16
@clark, it deserves to be an answer. I was writing an answer along that line and then I saw your comment... – lhf Jun 28 '12 at 11:42

$\Bbb R$ is connected, isn't it ? So $\Bbb R^2$ is as well, as a product of connected spaces.

Consider the following lemma (proof below) :

Let $A$ and $B$ two subsets of a topological space $X$. If $A$ and $B$ are connected and if $A\cap B$ is not empty, then $A\cup B$ is connected.

And now consider $\Bbb R^2\setminus \{0\}$ as the union of four open halfplanes $\{ x>0 \}\cup\{x<0\}\cup\{y<0\}\cup\{y>0\}$. Each one is connected, being homeomorphic to $\Bbb R^2$. Using the lemma, you see that $\Bbb R^2\setminus \{0\}$ is connected.

Proof of the lemma — If $A\cup B \subset U\cup V$, with $U$ and $V$ open sets of $X$, with $U\cap V \cap (A\cup B) = \varnothing$. Then $A \subset U\cup V$ and since $A$ is connected, with have either $A\subset U$ or $A\subset V$. Similarly $B\subset U$ of $B\subset V$. If $A\subset U$ and $B\subset V$ — or $A\subset V$ and $B\subset V$ —, then $A\cap B \subset U\cap V$, which is a contradiction. So $A\cup B\subset U$ or $A\cup B\subset V$, what we wanted.

-
I started an answer along these lines, but your solution is far simpler and elegant. My solution was based on the same lemma, but using the observation that $\mathbb{R}^2 \setminus \{0\} = \cup_{p \in \mathbb{Z}^2 \setminus \{0 \}} B(p,1)$ (Euclidean norm). – copper.hat Jun 29 '12 at 5:53

HINT

Suppose that $A$ and $B$ are two disjoint open sets s.t. $A \cup B$ is $\mathbb{R}^2 \setminus \{0\}$, s.t. $A$ is nonempty. Then if $x \in A$, show that $B$ is empty, or alternatively that any $y$ is in $A$ as well. You might choose progresively larger open disks, or something similar.

-
The question is about $\mathbb R^2\setminus0$, not $\mathbb R\setminus0$. – Mariano Suárez-Alvarez Jun 27 '12 at 7:25