From:  (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.