|
|
Simeon Stefanov wrote:
Herman Jurjus wrote:
Simeon Stefanov wrote:
There is another argument coming from the Corridors Lemma, which is a
2-dimensional analogue of the Mean Value Theorem.
Suppose that there is an embedding f of S^2 in R^2. It is easy to see
that f(S^2) may be enclosed in a rectangle P so that it intersects the
boundary of P in 4 (different) points A, B, X, Y where A, B lie in
opposite faces of P and X, Y lie in the other pair of opposite faces.
Consider their co-images A', B', X', Y' in S^2; it's clear
that we may connect A' with B' and X' with Y' by some
non-intersecting arcs p and q in S^2. But then f(p) and f(q) are arcs
in the rectangle P connecting the 2 pairs of its opposite faces, hence
by the Corridors Lemma (I'm not sure about the right name) they must
intersect; a contradiction with the assumption that f is an embedding.
Cool!
Note that the Corridors Lemma is implying the Jordan Curve Theorem, ...
[snip]
May we learn how?
I'll try to sketch the proof that a simple closed curve divides the
plane. Let C be such a curve, we may assume it enclosed in a rectangle
P so that it intersects two opposite sides in exactly 2 points and does
not intersect the other two sides. Then C is an union of two
topological segments C_1 and C_2 and by the corridors lemma each of
them is dividing P. Furthermore, we may assume that P\C_i have exactly
2 components, since otherwise C is dividing the plane and we are done.
It may be shown that a component of P\C_1 is intersecting a component
of P\C_2 and then each point of this intersection is lying in a bounded
component of P\C.
Of course, we don't get in such a manner that P\C has exactly 2
components.
Ok, so we get a weak version of the Jordan curve theorem. Fine.
What about the following sketch?
Embed R^2 into S^2 by stereographic projection. The curve is now a curve
in S^2. Take a point p on the curve, and map S^2 \ {p} homeomorphically
to the flat (open) disk-interior. If the image of the curve were a bit
well-behaved (especially around p), we'd now have a curve in the disk,
'from one end to the other', and we could then simply apply the
Corridors lemma.
But of course, the curve could be not-so-well-behaved, and then we're stuck.
Question: can the argument above be repaired so that it does work?
Also:
And is it also possible to have a similarly elementary proof of the
following:
"if p is in the bounded Jordan-component of a curve, then every
contraction of the curve hits p"
?
Any ideas on this one?
--
Cheers,
Herman Jurjus
|
|