Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Show that $\mathbb{R}^2$ is not homeomorphic to $\mathbb{R}^2 \setminus \{(0,0)\} $.

share|improve this question
12  
Consider the fundamental groups. It would certainly help to find the answer you're looking for if you gave a little hint at your background. –  t.b. Apr 4 '11 at 10:53

4 Answers 4

$\mathbb{R}^2\backslash\{(0,0)\}\cong S^1\times\mathbb{R}$. you can use co/homology or fundamental groups (if that's in your toolkit). you can note that one is contractible and the other is homotopy equivalent to a circle (and tell those apart). you can note that one has euler characteristic 0 and the other has euler characteristic 1 (if this is something you can calculate).

share|improve this answer

Let me give a solution to the problem that uses the idea of homotopy of loops but avoids that of the fundamental group or the co/homology groups. (This is actually a fleshing out of Christian Blatter's comment to Matt's answer. FWIW I thought of this independently a couple of hours ago but had to run off to proctor an exam for someone else's class.)

We say that two loops $\gamma_0, \gamma_1: S^1 \rightarrow X$ are homotopic if there exists a continuous function $G: S^1 \times [0,1] \rightarrow X$ such that for all $x \in S^1$, $G(x,0) = \gamma_0(x)$ and $G(x,1) = \gamma_1(x)$. (Note I am not fixing a basepoint here: it doesn't matter either way for what I am about to say.)

We say that a topological space $X$ is simply connected if it is connected and every loop in $X$ is homotopic to a constant loop. It is clear that if $X$ is simply connected and $Y$ is not, then $X$ cannot be homeomorphic to $Y$.

I claim that $\mathbb{R}^2$ is simply connected and $\mathbb{R}^2 \setminus \{0\}$ is not.

Step 1: We show directly from the definition that $\mathbb{R}^2$ is simply connected.
Indeed, if $\gamma: S^1 \rightarrow \mathbb{R}^2$ is any loop, then the map $G: S^1 \times [0,1]$, $G(x,t) = (1-t)\gamma(x)$ is a homotopy from $\gamma$ to the constant loop based at $0$.

Step 2: We show that $\mathbb{R}^2 \setminus \{0\}$ is not simply connected. For this we exploit the fact that $\mathbb{R}^2 \setminus \{0\}$ is an open subset in the complex plane and use complex analysis, specifically:

Theorem (Homotopy Form of Cauchy's Integral Theorem): Let $\Omega$ be an open subset of the complex plane, let $f$ be a holomorphic function on $\Omega$ and let $\gamma_1,\gamma_2$ be two homotopic paths in $\Omega$. Then $\int_{\gamma_1} f(z) dz = \int_{\gamma_2} f(z) dz$.

This is a nontrivial result, but it is a standard variant of the usual Cauchy Integral Formula that is found in most basic complex analysis texts. In particular, if $\Omega$ is simply connected, then every loop $\gamma: S^1 \rightarrow \Omega$ is homotopic to a constant loop and thus for every holomorphic function $f$ on $\Omega$ we have $\int_{\gamma} f(z)dz = 0$. In particular this applies to $\Omega = \mathbb{C}$ by Step 1.

The endgame is probably familiar: on $\mathbb{C} \setminus \{0\}$, if you integrate the holomorphic function $f(z) = \frac{1}{z}$ on the path $\gamma(t) = e^{2 \pi i t}$ then you get $2 \pi i$, which is not zero. Thus $\mathbb{C} \setminus \{0\}$ is not simply connected and hence not homeomorphic to $\mathbb{C}$.

share|improve this answer

Here's a slightly different way of looking at it that avoids fundamental groups (although has its own messy details to check). One of the spaces, upon removing a compact set, can be separated into two connected components with noncompact closure. The other can't.

share|improve this answer
2  
Nice. It's not even very messy: about all you need is that subsets of $\mathbb{R}^n$ are compact iff closed and bounded. –  Chris Eagle Apr 4 '11 at 14:21
4  
I guess another way to phrase this is that $\mathbb{R}^2$ has only one "end" whereas $\mathbb{R}^2 \setminus \{0\}$ has at least two. (There is a formal definition of "ends" of a topological space: see e.g. en.wikipedia.org/wiki/End_%28topology%29.) –  Pete L. Clark Apr 4 '11 at 14:35

conceptually its because your mapping a continuous space to one with a hole in it. you cant have a continuous map from $\mathbb{R}^2$ to $\mathbb{R}^2\setminus (0,0)$ since to do so you'd have to tear a hole in it.

share|improve this answer
    
am I wrong about this? please feel free to correct me if I am. –  Kate Apr 4 '11 at 11:03
3  
First of all, I'm not the one who voted this down, although I was tempted. Second, you probably mean homeomorphism where you say continuous map ($(x,y) \mapsto (e^{x}\cos{(y)},e^{x}\sin{(y)})$ is very continuous and even onto, isn't it?). Third, I'd rather say intuitively than conceptually but still, this would be more of a comment than an answer, I think. –  t.b. Apr 4 '11 at 11:08
    
fair enough, Thanks for clarification. I'll keep this in mind for future reference. –  Kate Apr 4 '11 at 11:13

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.