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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
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
@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
@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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