From:greg@oreo.berkeley.edu (Greg)Newsgroups:sci.mathSubject:Resolution of my topology question.Date:28 Jan 89 07:07:27 GMTReply-To:greg@math.Berkeley.EDU (Greg)Organization:U.C. Berkeley

In a previous article I asked: Does a non-trivial knot in R^3 necessarily have four collinear points? I now know the status of this question. The answer is yes for knots in general position, where general position means everything but a closed set of measure zero in various reasonable spaces of knot embeddings. This was proved in a paper by Morton and Mond in 1980. As often happens, however, someone else had the same basic ideas in 1933. I can prove the statement for all C^1 knots, which I think suffices in this context in which this question arose. I'm still interested in the general case, especially for wild knots. Here is why a knot must have three collinear points: Suppose that for a point P there does not exist Q and R collinear with it on the knot. Then for every point Q, we may join P and Q by a chord, and the union of the chords is evidently an embedded disk with the knot as boundary. Therefore the knot is trivial. --- Greg