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From: Lasse on 28 Feb 2005 05:45
The idea of the metric is to ensure that points in the complex plane
which converge to i will converge to the endpoint (1,*) in the new
space, while the points (t,*) with t\in (0,1) cannot be accumulated on
by points in the original set Y.

So the point is that the endpoints of the line segments in Y, i.e., the
points
x_n := e^{\pi/2 - \pi/2n}
converge to (1,*). However, the points
x_n(t) := t x_n
do not converge to (t,*). Am I missing something?

Lasse

From: Stuart M Newberger on 28 Feb 2005 06:42

Lasse wrote:
> The idea of the metric is to ensure that points in the complex plane
> which converge to i will converge to the endpoint (1,*) in the new
> space, while the points (t,*) with t\in (0,1) cannot be accumulated
on
> by points in the original set Y.
>
> So the point is that the endpoints of the line segments in Y, i.e.,
the
> points
> x_n := e^{\pi/2 - \pi/2n}
> converge to (1,*). However, the points
> x_n(t) := t x_n
> do not converge to (t,*). Am I missing something?
>
> Lasse

The distance from d(x_n,(1,*)) converges to Min(1+|i|,|1-i|)=2^1/2>0 so
the sequence does not converge to (1,*) .Regards,Stuart M Newberger

From: Lasse on 28 Feb 2005 07:17
OK, I see that I made a typo: it should have been |i-z| rather than
|1-z| in the definition of the metric. Whoops. Does this make more
sense now?

Lasse

From: Zbigniew Fiedorowicz on 28 Feb 2005 12:23

Julien Santini wrote:
> The theorem I mentioned is theorem 4.8 (Hocking & Young) and the proof is
> "left to the reader" (it is clearly implied it must be more or less
> trivial). However, it seems clear to me that, using this interpretation of
> starlike, a counterexample can be found easily. I really can't tell.
>

I think the following would be a counterexample. First consider the
subset X of the Cartesian plane defined as follows. Let P=(0,1),
Q=(0,-1). Let X be the union of all line segments joining points
(x,0), x rational 0<=x<=1, to P together with all line segments joining
points (x,0), x irrational 0<=x<=1, to Q. Endow X with the subspace
metric of the Cartesian plane. Let Y be the quotient topological space
obtained by identifying P and Q. Endow Y with the following metric
d'([u],[v]) = min( d(u,v), d(u,P)+d(v,Q), d(u,Q)+d(v,P) )
where d denotes the (Euclidean) metric on X and [u], [v] denote the
equivalence classes of u and v. Then the line segments in X are still
geodesics (congruent arcs) in Y and Y is "starlike" with respect to
[P]=[Q]. But clearly Y is not contractible.
From: Lasse on 28 Feb 2005 16:24

Zbigniew Fiedorowicz wrote:
> Julien Santini wrote:
> > The theorem I mentioned is theorem 4.8 (Hocking & Young) and the
proof is
> > "left to the reader" (it is clearly implied it must be more or less
> > trivial). However, it seems clear to me that, using this
interpretation of
> > starlike, a counterexample can be found easily. I really can't
tell.
> >
>
> I think the following would be a counterexample. First consider the
> subset X of the Cartesian plane defined as follows. Let P=(0,1),
> Q=(0,-1). Let X be the union of all line segments joining points
> (x,0), x rational 0<=x<=1, to P together with all line segments
joining
> points (x,0), x irrational 0<=x<=1, to Q. Endow X with the subspace
> metric of the Cartesian plane. Let Y be the quotient topological
space
> obtained by identifying P and Q. Endow Y with the following metric
> d'([u],[v]) = min( d(u,v), d(u,P)+d(v,Q), d(u,Q)+d(v,P) )
> where d denotes the (Euclidean) metric on X and [u], [v] denote the
> equivalence classes of u and v. Then the line segments in X are still
> geodesics (congruent arcs) in Y and Y is "starlike" with respect to
> [P]=[Q]. But clearly Y is not contractible.

Yes --- this is essentially the same idea as in my previous example,
except that I only had one line segment in the "lower" area and
countably many line segments "above".

(In other words, in my example X would have been the union of all line
segments joining points (1/n,0) to P together with the single line
segment joining (0,0) to Q. However, though we clearly have the same
example in mind, your description is a lot clearer than what I wrote
(what I really wanted to do was draw a picture on a blackboard, but
that still seems to be difficult in newsgroups ...))

Of course my example is a lot tamer, since there is only one point
where we get trouble, whereas in your example there is trouble
everywhere.

Lasse

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