Here $D^2$ denotes the closed unit disk in $\mathbb{R}^2$. I know that $D^2$ is not homeomorphic to $\mathbb{R}^2$ as $D^2$ is compact. Intuitively I believe that $D^2\setminus\{0\}$ is not homeomorphic to $\mathbb{R}^2\setminus\{0\}$. However, when I try to prove it, it is not so obvious because the elementary arguments (e.g. connectedness, compactness, fundamental group) do not work in this case. Is there a nice prove for this statement?
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1You could look at the Alexandrov compactifications. – Daniel Fischer Oct 25 '14 at 22:20
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2If you're more algebraic-topology minded, look at the local homology. That's different in the boundary points of $D^2$. Or the fundamental group of $X\setminus {p}$. – Daniel Fischer Oct 25 '14 at 22:21
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The space $X=\mathbb{R}^2\setminus\{0\}$ has the following property: For any $x\in X$, the fundemental group of $X\setminus x$ is not isomorphic to that of $X$. The space $Y=D^2\setminus\{0\}$ does not have this property.
Amitai Yuval
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Suppose we had a homeomorphism $f$ from $D^2\backslash\{0\}$ to $\mathbb{R}^2\backslash\{0\}$. Notice that, if we let $S$ be the boundary of the disk - that is, the circle of radius $1$ centered at the origin, we would have that the image $f[S]$ would be some non-self-intersecting loop in the plane. However, this would imply, from the Jordan curve theorem, that, by removing the curve from the plane, we get a set with two distinct regions - that is, the set $\mathbb{R}\backslash\{0\}\backslash f[S]$ is not connected. However, this is the image of $\mathbb{D^2}\backslash\{0\}\backslash S$ under $f$ - which is a connected region, and hence any given $f$ cannot be a homeomorphism, since any homeomorphism would preserve connectivity.
user84413
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Milo Brandt
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