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I've already shown the existence of a homeomorphism between the open unit disc and $\mathbb{R}^2$ and now I'm trying to work out whether the closed unit disc is homeomorphic to $\mathbb{R}^2$ or not. Clearly the only real difference is the fact it includes the points actually on the circle, but because the continuous, invertible, bijective map I defined for the open disc was $\dfrac{x}{1-|x|}$ this clearly is not defined for the points on the line.

Intuitively I'm thinking it's not possible to form a homeomorphism with the closed disk precisely because of these points on the circle but can't really form a decent argument. I was thinking along the lines of showing that the non-open, closed subset $\bar{D}(0,1) \backslash D(0,1)$ (which is just the set of points on the actual circle i.e the boundary/frontier of the closed disk) can't possibly bijectively map to a closed subset of $\mathbb{R}^2$ - which clearly would have to be the case in a homeomorphism.

Any advice would be greatly appreciated.

Potato
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Noble.
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3 Answers3

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Perhaps the most immediate argument to show that the closed disk is not homeomorphic to $\mathbb R^2$ is that the former is compact while the latter is not. This argument is not quite in line with the intuition you were trying to formalize. You intuition can be formalized to some degree by using some concepts from homotopy theory. The fundamental group of $\mathbb R^2$ with a point removed is $\mathbb Z$, while the fundamental group of the closed disk with a boundary point removed is trivial. Thus the two spaces are not even homotopy equivalent, let alone homeomorphic.

Thinking in terms of paths (basically the argument above but without assuming you know the fundamental group is), any path (that is a continuous function from $[0,1]$) in the disk with a boundary point removed can be continuously deformed into a point (just shrink the path to the centre of the disk). But, not every path in the plane with a point removed can be so contracted. If the path is a circle around the missing point then it can't be deformed to a point, the missing point keeps you from doing that.

The property of deformability of paths is an invariant of homeomorphisms (and of homotopy equivalences) and thus the two spaces are not homeomorphic.

Ittay Weiss
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    @Ittay Weiss Thanks. We haven't covered homotopy yet, which might explain why I couldn't formalise my intuition. Thanks for taking the time to write that answer. – Noble. Nov 04 '13 at 23:22
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    @Noble. The disk and the plane are homotopy equivalent, but the spaces with points removed aren't, thus cannot be homeomorphic. – Stefan Hamcke Nov 04 '13 at 23:24
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No; the closed unit disk is closed and bounded in $\mathbb R^2$ , meaning it is compact, and $\mathbb R^2$ is not compact.

user99680
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  • Thanks a lot. I knew a homeomorphism preserved connectedness, but not compactness. Thanks for the simple solution! – Noble. Nov 04 '13 at 23:23
  • Careful: actually, homeomorphisms do preserve compactness, but $\mathbb R^2$ is not compact, but the closed disk is, and so the homeomorphic image of the closed disk (potentially $\mathbb R^2$ ) should be compact, but $\mathbb R^2$ is not. – user99680 Nov 05 '13 at 00:45
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    @user99680 I believe that Noble meant, "I knew a homeomorphism preserved connectedness, but did not know that it also preserved compactness." – lily Nov 05 '13 at 01:08
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No. $\Bbb R^2$ is not compact whereas the disk is by Heine Borel

Stefan Hamcke
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LASV
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