# Bijection between closed and open disks (of the same radius)?

I am contemplating ways of defining a bijection between closed and open disks of the same radius. By these terms I mean http://en.wikipedia.org/wiki/Disk_(mathematics)

I am aware that the technique of perceiving disk as a set of circles and then mapping circles of the given radius as follows: $$\forall n \in \mathbb{N}: \quad \frac{R}{n} \rightarrow \frac{R}{n+1}$$ appears to work, but how would you personally do it? I mean do you happen to know other elegant and fancy ways?

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Well, there is a pretty obvious bijection between the interior of the closed disc and the open disc, so keeping that bijection as untouched as possible would be a good thing when trying to do it elegantly. Thus the question remains: what to do with the circle at the boundary of the closed disc? That is where the solution you mention comes up, and to me it seems the most elegant option. Because of open/closedness, there is no "nice", continuous way of doing it. – Arthur Feb 12 '13 at 21:54
This would be my approach. I think it is elegant, but not fancy. – Ross Millikan Feb 12 '13 at 21:54
Why the downvote? This is good question. – 1015 Feb 12 '13 at 21:57
I feel stupid since everybody seems to understand your construction...and I don't. Is there any other way to explain it? – 1015 Feb 12 '13 at 22:50

I would say first that the open disk is homeomorphic to the whole plane and the closed disk is homeomorphic to the standard square. Then finding the bijection is the same thing as finding a bijection between $\mathbb R$ and $[0,1]$.
An alternative is to find a bijection between $^[0,1)$ and $[0,1]$ and then use polar coordinates.