Re: The Gordian knot


John Jones wrote:
Rupert wrote:
John Jones wrote:
Alexander stumbled across another Gordian knot. This one consisted of
an infinite length of a single string coiled randomly into a ball.
Alexander cut the knot with his his shining sword of discrimination.
Could the two halves have an unequal number of ends?

We need to make this problem a bit more precise, I think. The string
could be taken to be an embedding f:R->R^3, presumably we want the
image to be closed, so it can't be bounded. I have difficulty
reconciling this with the image of "being coiled into a ball". Perhaps
you could take it to be an embedding f:[0,1]->R^3, but that's hard to
reconcile with "infinite length".

Then we could take the cut to be a subset S of R^3 such that there is a
topological automorphism of R^3 mapping S to {(x,y,z)|z=0}. But what
happens if S contains subsets of the image of f whose preimage under f
is open? Or if there's a point f(x) in S such that there's a
neighbourhood of x that is mapped by S into the union of S with just
one side of S?

If we assume that for every point f(x) in S, there's a neighbourhood
(a,b) of x such that (a,x) gets mapped into one side of S, (x,b) into
the other, then the problem is trivial. But you presumably want to
explore what other sorts of cases there can be.

A random coiling of an infinite string would mean that in both halves
of Alexanders cut knot there would be an uncountable number of ends.

Well, not necessarily. That wouldn't be the case on the assumption I
gave above,for example.

What do you mean by a "random coiling"?

You need to make this problem more precise.

I don't see how you coudl reasonably call a point f(x) in S an "end"
unless there were points y arbitrarily close to x such that f(y) was on
one side of S, and also points y arbitrarily close to x such that f(y)
was on the other side of S. And in that case it is trivial that every
point which is an end from one side is also an end from the other side.

I
need to give the number of these ends in order to say that they are
numerically equal or not.

On any reasonable precise statement of the problem, I would say they
are equal.

The problem is this: adding two to infinity
simply results in infinity, and that as far as infinity of ends is
concerned both halves are equal, but if they are equal then I cannot
distinguish between a linear string with two ends and a circular string
with no ends.

I don't follow you. I thought you said your string was an infinite
string to start with. I assumed that meant a linear string with no ends.

.

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