From: greg@oreo.berkeley.edu (Greg)
Newsgroups: sci.math
Subject: Resolution of my topology question.
Date: 28 Jan 89 07:07:27 GMT
Reply-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