Just by cardinality alone, there are 2^aleph_0 circles in the plane with any fixed center and any real radius. But there are only aleph_0 rational points. So there are lots of circles that miss the set of rational points, or for that matter any countable subset, dense or not. (Also, every countable dense subset of the plane is homeomorphic to any other, and every complement of such a subset is homeomorphic to any other as well.) --Dan On 2013-09-16, at 8:39 AM, Cordwell, William R wrote:

Hello,

There was some interesting discussion about rational points being dense in the unit circle (corresponding to all the Pythagorean triples). It is straightforward to show that there are many circles that contain no rational points.

So, is there any theory about (smooth, closed, whatever modifier you like) curves in the plane that completely miss a dense subset?

Thanks, Bill

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There is the theory of elliptic curves over Q. Such have to have at least one rational point, sometimes have finitely many, sometimes have infinitely many. When there are infinitely many the component containing the identity is dense with rationals, but the other component (if any) may have no rationals. Not your question, but closely related. On Mon, Sep 16, 2013 at 10:39 AM, Cordwell, William R <wrcordw@sandia.gov>wrote:

Hello,

There was some interesting discussion about rational points being dense in the unit circle (corresponding to all the Pythagorean triples). It is straightforward to show that there are many circles that contain no rational points.

So, is there any theory about (smooth, closed, whatever modifier you like) curves in the plane that completely miss a dense subset?

Thanks, Bill

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