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

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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
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5 Answers 5

$\mathbb{R}^2\backslash\{(0,0)\}\cong S^1\times\mathbb{R}$. 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).

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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 $\mathbb{C}$.

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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.

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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
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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
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To expand on Theo Buehler's comment:

The fundamental group of a topological space $X$ at a point $x_0 \in X$ is the set of equivalence classes of loops at $x_0$ where two paths are equivalent iff they're homotopic.

$\mathbb{R}^2$ and $\mathbb{R}^2 - \{ (0,0)\}$ are both path-connected so the fundamental group is independent of the point $x_0$ you pick.

In $\mathbb{R}^2$, you see that any loop at any $x_0$ is homotopic to the constant map $x_0$. (the homotopy is easy to write down, if you can't do that I'll provide it in a second edit). This means that the fundamental group $\pi_1(\mathbb{R}^2) = \{ [c] \} $ where $c$ is the constant map which is the neutral element of the group, i.e. the fundamental group of $\mathbb{R}^2$ is trivial.

In $\mathbb{R}^2 - \{ (0,0)\}$ on the other hand you can pick a loop around $(0, 0)$ and you will see that it is not homotopic to the constant map. Why? Because if you had a homotopy $h$ that contracts the loop to a point, the loop will go through $(0,0)$ at some point in time. But $(0,0)$ is not in $\mathbb{R}^2$, so $h$ is not a valid homotopy.

Hope this helps.

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In order to actually prove that in the case of $\dot{\mathbb R}^2$ there is no such homotopy for the loop around $(0,0)$ one needs stronger means, e.g. the fact that the integral ${1\over 2\pi}\int_\gamma{dz\over z}\in{\mathbb Z}$ is invariant. –  Christian Blatter Apr 4 '11 at 13:09
    
@Christian Blatter: Many thanks for pointing this out. I'm a beginner in this subject, so may I ask: why is it not enough to argue that any homotopy would go through $(0,0)$? –  Matt N. Apr 4 '11 at 13:35
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Intuitively it is clear that any homotopy "would go through $(0,0)$", but there are things between heaven and earth... That's where invariants come in: They furnish the irrefutable impossibility argument. –  Christian Blatter Apr 4 '11 at 13:49
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@Christian Blatter: I'm sorry but I think this doesn't answer my question satisfactorily. Could you give me a mathematically rigorous argument why it is not enough? To me arguing like I argued in my answer above is a proof by contradiction, so it seems rigorous to me. –  Matt N. Apr 4 '11 at 14:06
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@Matt: it doesn't really work that way: the burden of rigor is on you in giving your argument. For instance, the result that you claim for $\mathbb{R}^2$ is false for $\mathbb{R}^3$, but where in your argument did you call attention to some distinction between the two spaces that makes your argument work? –  Pete L. Clark Apr 4 '11 at 14:19
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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.

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am I wrong about this? please feel free to correct me if I am. –  Kate Apr 4 '11 at 11:03
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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
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