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From: Timothy Little on 28 Feb 2005 01:16
Lasse wrote:
> How about the following: in R^2, take the union of the line segment
> from (0,0) to (1,0); the line segments {1/n} x [1,0] for all natural
> numbers n, and a half-circle (or any other curve) from (0,0) to
> (0,1) lying in the left half plane (i.e., x < 0) apart from these
> two endpoints. Then it would seem that this set is "star-like" by
> your definition, but it is clearly not contractible.

kI'm trying to follow this at home, but can't figure out what you mean
by "the line segments {1/n} x [1,0]". I haven't seen that notation
before.


- Tim
From: Lasse on 28 Feb 2005 03:15

Timothy Little wrote:
> Lasse wrote:
> > How about the following: in R^2, take the union of the line segment
> > from (0,0) to (1,0); the line segments {1/n} x [1,0] for all
natural
> > numbers n, and a half-circle (or any other curve) from (0,0) to
> > (0,1) lying in the left half plane (i.e., x < 0) apart from these
> > two endpoints. Then it would seem that this set is "star-like" by
> > your definition, but it is clearly not contractible.
>
> kI'm trying to follow this at home, but can't figure out what you
mean
> by "the line segments {1/n} x [1,0]". I haven't seen that notation
> before.
>

I meant 'the line segments from (1/n,0) to (1/n,1) for all n.

However, my example, as I wrote is, isn't really correct, since you
will have to change the metric on X as well to get a proper
counterexample: clearly any star-like set in R^2 is star-like in the
usual definition, so there can't be a counterexample with the induced
metric!

So I will come back to the abstract example I had in mind originally,
and be a little bit more careful. How about the following:

The first part of our set is indeed embedded into the plane; notation
is easier if we look at it in the complex plane. It is given by

Y := Union_n over [0,1]e^{i(pi/2 - pi/2n)},

where the union is over all natural numbers n >= 1. In other words, Y
is made up of countably many line segments emanating out from 0 and
accumulating on the line segment from 0 to i.

Now this set is star-like with the induced metric (and contractible).

Now we add an additional copy of the interval [0,1] to this set; let us
denote the elements of this copy by (t,*) when t\in [0,1].

We identify the element (t,0) with 0. On the copy of this interval, we
retain the usual euclidean metric of the interval. For any point z\in
Y, we define the distance from (t,*) to be

d( (t,*), z ) := min( t + |z| , 1 - t + |1-z| ).

Now this metric should satisfy the triangle inequality. Also, the space
X which we obtain is star-like by the above definition. However, it is
not contractible.

I hope this is correct now, or maybe I'm still making some silly
mistake.

Lasse

From: Stuart M Newberger on 28 Feb 2005 04:10

Lasse wrote:
> > That's what I had thought, but my problem is that although I can
show
> h_x:
> > [0,1] -> X, h_x(t) = h(x,t) is continuous (using this definition of
> > starlike), I cannot show h_t: X->X, h_t(x) = h(x,t) is continuous.
> Working
> > in R^2 for instance, and considering an arc (*p), with a point x on
> this
> > arc, and a neighborhood U of x, why couldn't U intersect (no matter
> how
> > "small" U is) another arc at point y so that for some fixed t, and
> for some
> > fixed e>0, abs(h_t(y)-h_t(x))>e (i.e h_t is not continuous) ?
That's
> what
> > bothers me.
>
> How about the following (it would seem that this can be done in R^2,
> but perhaps it is easier to define on the abstract level):
>
> How about the following: in R^2, take the union of the line segment
> from (0,0) to (1,0); the line segments {1/n} x [1,0] for all natural
> numbers n, and a half-circle (or any other curve) from (0,0) to (0,1)
> lying in the left half plane (i.e., x < 0) apart from these two
> endpoints. Then it would seem that this set is "star-like" by your
> definition, but it is clearly not contractible.
>
> You could assume that your space is compact and this problem would go
> away, but perhaps you don't want to do that.
>
> Hope this helps,
>
> Lasse
> ---
> (@remove.for.spam.maths.warwick.ac.uk)

Your example is not starlike with respect to any point using the
usual Euclidan metric on the plane.If you mean an arclength metric
then why isn't it contractible ? And how does compactness make any
problems go away. I suspect Hocking and Young meant to assume that the
map I called f_x(t) (=the point at distance t along the arc from the
star vertex p to x) is continuous in (x,t) together.But I have no
example so am not sure.
Regards Stuart M Newberger.

From: Stuart M Newberger on 28 Feb 2005 05:13

Lasse wrote:
> Timothy Little wrote:
> > Lasse wrote:
> > > How about the following: in R^2, take the union of the line
segment
> > > from (0,0) to (1,0); the line segments {1/n} x [1,0] for all
> natural
> > > numbers n, and a half-circle (or any other curve) from (0,0) to
> > > (0,1) lying in the left half plane (i.e., x < 0) apart from these
> > > two endpoints. Then it would seem that this set is "star-like" by
> > > your definition, but it is clearly not contractible.
> >
> > kI'm trying to follow this at home, but can't figure out what you
> mean
> > by "the line segments {1/n} x [1,0]". I haven't seen that notation
> > before.
> >
>
> I meant 'the line segments from (1/n,0) to (1/n,1) for all n.
>
> However, my example, as I wrote is, isn't really correct, since you
> will have to change the metric on X as well to get a proper
> counterexample: clearly any star-like set in R^2 is star-like in the
> usual definition, so there can't be a counterexample with the induced
> metric!
>
> So I will come back to the abstract example I had in mind originally,
> and be a little bit more careful. How about the following:
>
> The first part of our set is indeed embedded into the plane; notation
> is easier if we look at it in the complex plane. It is given by
>
> Y := Union_n over [0,1]e^{i(pi/2 - pi/2n)},
>
> where the union is over all natural numbers n >= 1. In other words, Y
> is made up of countably many line segments emanating out from 0 and
> accumulating on the line segment from 0 to i.
>
> Now this set is star-like with the induced metric (and contractible).
>
> Now we add an additional copy of the interval [0,1] to this set; let
us
> denote the elements of this copy by (t,*) when t\in [0,1].
>
> We identify the element (t,0) with 0. On the copy of this interval,
we
> retain the usual euclidean metric of the interval. For any point z\in
> Y, we define the distance from (t,*) to be
>
> d( (t,*), z ) := min( t + |z| , 1 - t + |1-z| ).
>
> Now this metric should satisfy the triangle inequality. Also, the
space
> X which we obtain is star-like by the above definition. However, it
is
> not contractible.
>
> I hope this is correct now, or maybe I'm still making some silly
> mistake.
>
> Lasse


Isn't f(t,x)= tx still continuous on [0,1]xM when your set M has the
new metric.It looks continuous to me.Also,why that min in your metric
definition,was that just to make all the distances at most 1 ? Also,why
don't you union over n>=2 so that the interval [0,1] is not in Y.That
way you dont need 2 [0,1] 's .But I think that your space is
contractible.
Tough problem. Regards ,Stuart M Newberger

From: Lasse on 28 Feb 2005 05:36
>
> Your example is not starlike with respect to any point using the
> usual Euclidan metric on the plane.If you mean an arclength metric
> then why isn't it contractible ? And how does compactness make any
> problems go away. I suspect Hocking and Young meant to assume that
the
> map I called f_x(t) (=the point at distance t along the arc from the
> star vertex p to x) is continuous in (x,t) together.But I have no
> example so am not sure.
> Regards Stuart M Newberger.

Of course you are right --- I posted the (hopefully) correct version
this morning; I'm not sure what I was thinking yesterday. (On the other
hand, the fact that the claim is in Hocking & Young makes me a bit
worried that there is some mistake also in the new example; I apologise
in advance if this is so.)

If the space X is compact, then I think the following argument works:

Suppose that x_n tends to x. Let us denote gamma_n(t) := h(x_n,t). Then
for each t, gamma_n(t) tends to some limit in X as n-> infinity; call
this limit gamma(t). Then

d(gamma(t_1),gamma(t_2)) <= d(gamma_n(t_1),gamma(t_1)) +
d(gamma_n(t_2),gamma(t_2)) +
d(gamma_n(t_1), gamma_n(t_1)) =
d(gamma_n(t_1),gamma(t_1)) +
d(gamma_n(t_2),gamma(t_2)) +
|t_1 - t_2| * d(x_n,p)
--> |t_1 - t_2| * d(x,p).
It follows that gamma is an arc connecting x to p which is congruent to
a line segment. So gamma(t) = h(x,t). Now it follows easily that h is
continuous: if x_n \to x and t_n \to t, then

d( h(x_n,t_n) , h(x,t) ) <=
d( h(x_n,t_n) , h(x_n, t) ) + d( h(x_n,t) , h(x,t) ) <=
|t_n - t| * d(x_n,p) + d(h(x_n,t) , h(x,t)) --> 0.

Or am I making another mistake?

In fact, this argument only seems to require local compactness.

Lasse

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