Re: Simply Connected Subsets of R^2



Maury Barbato wrote: 
Hello,
we have the following well known (see "Elements of the
Topology of Plane Sets of Points" by Newman)

Theorem. If S is an open subset of R^2,
then:
(I) S is simply connected if and only if it is
contractible.
Besides if S is supposed also bounded and connected,
then:
(II) S is simply connected if and only if Bd(S) is connected.
(II) S is simply connected if and only if R^2\S is connected.

This isn't true for non-domains. Take the (closed) unit
disk in the euclidean plane without the ponits (x,0),
with 0 <= x < 1. This is not simply connected (it is homotopy equivalent to the unit circle). But the
boundary is the unit circle together with the line {(x,0) ; 0<= x <=1}, hence connected.

However, I think the following statement remains true.

Conjecture. Let S be a subset of R^2. Then S is simply
connected if and only if it is contractible.
Besides, if S is a bounded simply connected subset of
R^2, then Bd(S) and R^2\S are connected.

If you have some opinion about the problem, I will be
glad to know it. Thank you a lot for your help.

Maury 

Your conjecture is incorrect.

Take the so-called "Warsaw circle" W:

	Let f(x) = sin(1/x) for x != 0.

	Let: C = {(x, f(x)) | 0 < |x| < pi},
	     D = {(0,y) | |y| <= 1}
	     K = arc joining (pi, sin(1/pi)) to (-pi, -(sin(1/pi)),
	         but missing all other points of the union of C and D.

	W = C U D U K

where I've used U to denote set-theoretic union.

Google "Warsaw Circle" or nose around this URL for a picture:
(glue the following strings together for a proper URL)

	http://www.informatics.bangor.ac.uk
	      /public/mathematics/research
	      /cathom/cathom2.html

1. W is simply-connected, since no path can cross over the tangled
   sin(1/x) part as x passes 0: the space is not even path-connected,
   having two path-components: a) C U K and b) D.
2. W is not contractible, since it has a non-nullhomotopic map into
   the ordinary circle. If you have seen the picture, the proof is
   so close to obvious that it's difficult.
3. R^2 \ W is not connected.

Dale
.

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