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 ).
$\begingroup$
$\endgroup$
4
-
5$\begingroup$ In order to get the best possible answers, it is helpful if you say what your thoughts on the problem are so far; this will prevent people from telling you things you already know, and help them write their answers at an appropriate level. Also, people are much happier to help those who demonstrate that they've tried the problem themselves first. Many see giving commands as rude when asking for help; please consider rewriting your post. $\endgroup$– Zev ChonolesJun 27, 2012 at 7:04
-
11$\begingroup$ 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.. $\endgroup$– clarkJun 27, 2012 at 7:05
-
3$\begingroup$ @clark: +1. By far the best answer :) $\endgroup$– t.b.Jun 27, 2012 at 9:16
-
1$\begingroup$ @clark, it deserves to be an answer. I was writing an answer along that line and then I saw your comment... $\endgroup$– lhfJun 28, 2012 at 11:42
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
1
$\Bbb R$ 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.
-
$\begingroup$ 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). $\endgroup$ Jun 29, 2012 at 5:53
$\begingroup$
$\endgroup$
1
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.
answered Jun 27, 2012 at 6:46
-
4$\begingroup$ The question is about $\mathbb R^2\setminus0$, not $\mathbb R\setminus0$. $\endgroup$ Jun 27, 2012 at 7:25