Discussion:
Simply Connected Subsets of R^2
(too old to reply)
Maury Barbato
2005-10-18 08:21:29 UTC
Permalink
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
W. Dale Hall
2005-10-18 19:17:52 UTC
Permalink
Post by Maury Barbato
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,
(I) S is simply connected if and only if it is
contractible.
Besides if S is supposed also bounded and connected,
(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
Maury Barbato
2005-10-19 08:51:41 UTC
Permalink
Surely you're right: your counterexample is very
beautiful! I didn't know the Warsawe Circle: it's a very
interesting topological object!
However, what do you think of adding in my conjecture
the hypothesis that S is path-connected?

Thank you very much for your attention and your interest
in the subject.
My best regards.
Maury
William Elliot
2005-10-19 10:18:06 UTC
Permalink
Post by Maury Barbato
Surely you're right: your counterexample is very
beautiful! I didn't know the Warsawe Circle: it's a very
interesting topological object!
However, what do you think of adding in my conjecture
the hypothesis that S is path-connected?
The Warsawe Circle is path-connected.
Other than that, as you removed all context,
I'm unable to give additional cogent comments.
Post by Maury Barbato
Thank you very much for your attention and your interest in the subject.
Thank you were you to include context.

I've not the time nor the inclination for the clerical
chore of back tracking the thread to reconstruct the
line of thought. If you want intelligent and well thought
answers, then include the context pertinent to your reply
and to whom your are talking.

Please learn and use better math group manners as demonstrated
by other participants of this newsgroup and as described at
http://oakroadsystems.com/genl/unice.htm#quote

If you're using Google's quick reply, please don't, use quote feature
instead. Many of us have different news browsers than Google.
Lee Rudolph
2005-10-19 10:24:33 UTC
Permalink
Post by William Elliot
Post by Maury Barbato
Surely you're right: your counterexample is very
beautiful! I didn't know the Warsawe Circle: it's a very
interesting topological object!
However, what do you think of adding in my conjecture
the hypothesis that S is path-connected?
The Warsawe Circle is path-connected.
Okay, let's try "locally path-connected", which it ain't:
there's just no two ways about it.

Lee Rudolph
W. Dale Hall
2005-10-19 17:29:13 UTC
Permalink
Post by William Elliot
Post by Maury Barbato
Surely you're right: your counterexample is very
beautiful! I didn't know the Warsawe Circle: it's a very
interesting topological object!
However, what do you think of adding in my conjecture
the hypothesis that S is path-connected?
The Warsawe Circle is path-connected.
Other than that, as you removed all context,
I'm unable to give additional cogent comments.
But the Warsaw Circle is *not* path connected: note that
the "topologist's sine curve", defined as this:

S = {(x, sin(1/x)) | x > 0} U {0}x[-1 1]

isn't path-connected. The Warsaw circle is just the union
of two copies of S, identified along the vertical interval
(one copy to the left, the other to the right, of the y-
axis), joined by an arc connecting the tame ends. That
extra arc still doesn't get you a path to that vertical
line segment.
Post by William Elliot
Post by Maury Barbato
Thank you very much for your attention and your interest in the subject.
Thank you were you to include context.
... mini-rant on netiquette deleted ...

Dale
Maury Barbato
2005-10-19 18:16:28 UTC
Permalink
Post by W. Dale Hall
Post by William Elliot
Post by Maury Barbato
Surely you're right: your counterexample is very
beautiful! I didn't know the Warsawe Circle: it's a
very
Post by William Elliot
Post by Maury Barbato
interesting topological object!
However, what do you think of adding in my
conjecture
Post by William Elliot
Post by Maury Barbato
the hypothesis that S is path-connected?
The Warsawe Circle is path-connected.
Other than that, as you removed all context,
I'm unable to give additional cogent comments.
But the Warsaw Circle is *not* path connected: note
that
S = {(x, sin(1/x)) | x > 0} U {0}x[-1 1]
isn't path-connected. The Warsaw circle is just the
union
of two copies of S, identified along the vertical
interval
(one copy to the left, the other to the right, of the
y-
axis), joined by an arc connecting the tame ends.
That
extra arc still doesn't get you a path to that
vertical
line segment.
I trusted you, but now that I have thought a moment
about it, I have to conclude that you're wrong.

Surely, there's a misunderstanding!
What do you mean by "path-conneted"?

Definition. A space X is said to be path-connected if
for any two points x and y in X there exists a
continuous function f from the unit interval [0,1] to X
with f(0) = x and f(1) = y.
(This function is called a path from x to y.)

So W is obviously path-connected in this sense.
Surely it's not locally path-connected.

Definition. A topological space is said to be locally
path-connected if it has a base of path-connected sets.

So we can add the hypothesis of locally
path-connectedness to my conjecture, in the hope
that it becomes true.

Conjecture. Let S be a locally path-connected subspace
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.
Post by W. Dale Hall
Post by William Elliot
Post by Maury Barbato
Thank you very much for your attention and your
interest in the subject.
Post by William Elliot
Thank you were you to include context.
... mini-rant on netiquette deleted ...
Dale
My Best Regards
Maury
Robert Low
2005-10-19 18:26:04 UTC
Permalink
Post by Maury Barbato
Post by W. Dale Hall
But the Warsaw Circle is *not* path connected: note
that
S = {(x, sin(1/x)) | x > 0} U {0}x[-1 1]
isn't path-connected. The Warsaw circle is just the
union
of two copies of S, identified along the vertical
interval
(one copy to the left, the other to the right, of the
y-
axis), joined by an arc connecting the tame ends.
That
extra arc still doesn't get you a path to that
vertical
line segment.
Definition. A space X is said to be path-connected if
for any two points x and y in X there exists a
continuous function f from the unit interval [0,1] to X
with f(0) = x and f(1) = y.
(This function is called a path from x to y.)
So W is obviously path-connected in this sense.
So, describe the continuous function, f, with domain
[0,1] such that f(0) lies on the sin(1/x) part of the
curve, and f(1) lies on the {0}x[-1,1].
Maury Barbato
2005-10-19 20:03:24 UTC
Permalink
Post by Robert Low
Post by Maury Barbato
Post by W. Dale Hall
But the Warsaw Circle is *not* path connected: note
that
S = {(x, sin(1/x)) | x > 0} U {0}x[-1 1]
isn't path-connected. The Warsaw circle is just the
union
of two copies of S, identified along the vertical
interval
(one copy to the left, the other to the right, of
the
Post by Maury Barbato
Post by W. Dale Hall
y-
axis), joined by an arc connecting the tame ends.
That
extra arc still doesn't get you a path to that
vertical
line segment.
Definition. A space X is said to be path-connected
if
Post by Maury Barbato
for any two points x and y in X there exists a
continuous function f from the unit interval [0,1]
to X
Post by Maury Barbato
with f(0) = x and f(1) = y.
(This function is called a path from x to y.)
So W is obviously path-connected in this sense.
So, describe the continuous function, f, with domain
[0,1] such that f(0) lies on the sin(1/x) part of the
curve, and f(1) lies on the {0}x[-1,1].
Ok, that's a blunder! I had in my mind the graph of the
function f(x)=x*sin(1/x): I'm sorry ...

So W is not path-connected. What does W. Dale Hall mean
by the statement "W is simply connected"?
In any case, its meaning doesn't coincide with the
defintion which I refer to, in my conjecture.

Definition.A topological space X is called simply
connected if it is path-connected and any continuous map
f : S^1 -> X (where S^1 denotes the unit circle in
Euclidean 2-space) can be contracted to a point in the
following sense: there exists a continuous map
F : D^2 -> X (where D^2 denotes the unit disk in
Euclidean 2-space) such that F restricted to S1 is f.

I apologize to W. Dale Hall for my slip.
Thank you very much for your attention.

My Best Regards
Maury
W. Dale Hall
2005-10-20 00:45:00 UTC
Permalink
Post by Maury Barbato
Post by Robert Low
Post by Maury Barbato
Post by W. Dale Hall
But the Warsaw Circle is *not* path connected: note
that
S = {(x, sin(1/x)) | x > 0} U {0}x[-1 1]
isn't path-connected. The Warsaw circle is just the
union
of two copies of S, identified along the vertical
interval
(one copy to the left, the other to the right, of
the
Post by Maury Barbato
Post by W. Dale Hall
y-
axis), joined by an arc connecting the tame ends.
That
extra arc still doesn't get you a path to that
vertical
line segment.
Definition. A space X is said to be path-connected
if
Post by Maury Barbato
for any two points x and y in X there exists a
continuous function f from the unit interval [0,1]
to X
Post by Maury Barbato
with f(0) = x and f(1) = y.
(This function is called a path from x to y.)
So W is obviously path-connected in this sense.
So, describe the continuous function, f, with domain
[0,1] such that f(0) lies on the sin(1/x) part of the
curve, and f(1) lies on the {0}x[-1,1].
Ok, that's a blunder! I had in my mind the graph of the
function f(x)=x*sin(1/x): I'm sorry ...
So W is not path-connected. What does W. Dale Hall mean
by the statement "W is simply connected"?
In any case, its meaning doesn't coincide with the
defintion which I refer to, in my conjecture.
My meaning was only that every mapping of S^1 into W is
null-homotopic. There are surely problems with saying
W is simply-connected if one requires path-connectivity,
as W fails that requirement. In fact, I had forgotten
that people frequently do require this when I gave the
example.
Post by Maury Barbato
Definition.A topological space X is called simply
connected if it is path-connected and any continuous map
f : S^1 -> X (where S^1 denotes the unit circle in
Euclidean 2-space) can be contracted to a point in the
following sense: there exists a continuous map
F : D^2 -> X (where D^2 denotes the unit disk in
Euclidean 2-space) such that F restricted to S1 is f.
I apologize to W. Dale Hall for my slip.
Thank you very much for your attention.
No apology is necessary; everyone makes mistakes, and
everyone is capable of being in error. I could very well
have been mistaken, and having been called on that possibility
was not a problem.
Post by Maury Barbato
My Best Regards
Maury
Dale
W. Dale Hall
2005-10-20 01:13:16 UTC
Permalink
W. Dale Hall wrote:

... some stuff ...
Post by W. Dale Hall
My meaning was only that every mapping of S^1 into W is
null-homotopic. There are surely problems with saying
W is simply-connected if one requires path-connectivity,
as W fails that requirement. In fact, I had forgotten
that people frequently do require this when I gave the
example.
It just occurred to me that we can modify the Warsaw Circle
to obtain a path-connected space that is simply-connected
but not contractible. (An aside: in the ancient past, a
space that was path-connected + simply-connected was called
1-connected; my failure to recall the more restrictive usage
for the term "simply connected" was likely due to that
recollection)

Instead of taking two copies of the topologist's sine curve

S = {(x,sin(1/x)) | x > 0} U {(0,y) | |y| <= 1}

identified suitably, connected by an additional arc, just take
the one copy of S, and plug the tail end back into the vertical
interval. I'll call it the Warsaw semicircle: W/2.

Thanks to the tail coming back to meet that vertical line segment,
the space W/2 is path-connected. Due to the inability of a continuous
path to reach around the sin(1/x) part to join the vertical interval,
every map from S^1 to W/2 is null-homotopic.

Of course, W/2 is pathological in not being locally-connected nor
locally path-connected.

Dale
Lee Rudolph
2005-10-20 09:05:39 UTC
Permalink
Post by W. Dale Hall
Instead of taking two copies of the topologist's sine curve
S = {(x,sin(1/x)) | x > 0} U {(0,y) | |y| <= 1}
identified suitably, connected by an additional arc, just take
the one copy of S, and plug the tail end back into the vertical
interval. I'll call it the Warsaw semicircle: W/2.
Thanks to the tail coming back to meet that vertical line segment,
the space W/2 is path-connected. Due to the inability of a continuous
path to reach around the sin(1/x) part to join the vertical interval,
every map from S^1 to W/2 is null-homotopic.
Of course, W/2 is pathological in not being locally-connected nor
locally path-connected.
It was, in fact, this space that I thought (clearly wrongly) was
the "Warsaw circle"; hence my failed attempt to emulate quasi by
hinting "there's no two ways around it".

Lee Rudolph
W. Dale Hall
2005-10-19 19:43:16 UTC
Permalink
... stuff deleted ...
Post by Maury Barbato
Post by W. Dale Hall
Post by William Elliot
The Warsawe Circle is path-connected.
Other than that, as you removed all context,
I'm unable to give additional cogent comments.
But the Warsaw Circle is *not* path connected: note
that
S = {(x, sin(1/x)) | x > 0} U {0}x[-1 1]
isn't path-connected. The Warsaw circle is just the union of two
copies of S, identified along the vertical interval (one copy to
the left, the other to the right, of the y- axis), joined by an arc
connecting the tame ends. That extra arc still doesn't get you a
path to that vertical line segment.
I trusted you, but now that I have thought a moment
about it, I have to conclude that you're wrong.
You can conclude what you like, I'm sure. However, your
conclusion here is incorrect.
Post by Maury Barbato
Surely, there's a misunderstanding!
What do you mean by "path-conneted"?
You might be hard pressed to do so, but you can
take my word that I understand what it means for a
space to be path connected. However, I agree that
there's a misunderstanding; it's easy enough to
settle it.
Post by Maury Barbato
Definition. A space X is said to be path-connected if
for any two points x and y in X there exists a
continuous function f from the unit interval [0,1] to X
with f(0) = x and f(1) = y.
(This function is called a path from x to y.)
Yes, that's the definition of "path connected".
Post by Maury Barbato
So W is obviously path-connected in this sense.
Surely it's not locally path-connected.
Here's the problem. You've just buried the whole
concept of what's worth proving beneath the shroud
of what's "obvious". Maybe it's obvious that W is
path-connected. Let's see.

Recall

S = {(x, sin(1/x)) | x > 0} U {0}x[-1 1]

(the topologist's sine curve)

I'll show that the points (1,sin(1)) and (0,0) are
not in the same path-component of S.

Suppose

F: I --> S

with
F(0) = (1, sin(1))
F(1) = (0,0).

I'll let F(t) = (x(t),y(t)). As long as x is not zero,
y must be sin(1/x). Now, in order for F(t) to approach
(0,0) as t approaches 1 (which it must do, in order to
be continuous), x itself must approach 0. Similarly,
for y to approach 0 as t approaches 1, y must also
approach 0.

Now, your mistake is probably imagining that it is
sufficient for this pair of limits hold for *some*
sequence of values t_k ---> 1. It isn't. That pair
of limits (x(t) ---> 0, y(t) ---> 0) must hold
as t --> 1, i.e., for *every* sequence of values
t_k --> 1.

For instance, if t_k is defined as k*pi, then

x_k = 1/(k pi)
y_k = 0

and the set of points converges to (0,0) as we wanted.
However, let t_k = (2k+1)pi/2, and we get this:

x_k = 2/((2k+1)pi)
y_k = (-1)^k.

This sequence of points doesn't converge at all, and
0 is not even in the set of limiting values for the y_k.

Worse yet, you can take the set of integer values

t_k = k

and find that y_k = sin(1/k), which (as k ranges over
all integers) is dense in the interval [-1 1].

So, let's ask: where was the misunderstanding?

Way back in graduate school, it was common knowledge
that fully 80% (if not 95%) of errors in mathematics
proofs could be found in close proximity to one of the
weasel-words such as "clearly", "obviously", and so forth.

BTW, the Warsaw Circle is not only not locally path-connected,
it's not even locally connected.
Post by Maury Barbato
Definition. A topological space is said to be locally
path-connected if it has a base of path-connected sets.
So we can add the hypothesis of locally
path-connectedness to my conjecture, in the hope
that it becomes true.
Conjecture. Let S be a locally path-connected subspace
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.
I haven't given these much thought. I'll do so and
get back to you if I resolve the issue to my own
satisfaction.

... the rest deleted ...
Post by Maury Barbato
My Best Regards
Maury
Dale
William Elliot
2005-10-20 06:07:43 UTC
Permalink
Post by W. Dale Hall
Post by William Elliot
Post by Maury Barbato
Surely you're right: your counterexample is very
beautiful! I didn't know the Warsawe Circle: it's a very
interesting topological object!
However, what do you think of adding in my conjecture
the hypothesis that S is path-connected?
The Warsawe Circle is path-connected.
Other than that, as you removed all context,
I'm unable to give additional cogent comments.
But the Warsaw Circle is *not* path connected: note that
S = {(x, sin(1/x)) | x > 0} U {0}x[-1 1]
isn't path-connected. The Warsaw circle is just the union
of two copies of S, identified along the vertical interval
(one copy to the left, the other to the right, of the y-
axis), joined by an arc connecting the tame ends. That
extra arc still doesn't get you a path to that vertical
line segment.
Indeed, my first blunder was replying to a post without context included.
My second blunder was relying upon memory, thus the third blunder, my
posting a erroneous repose is the blunder of the poster who thinks
it's quick and slick to discard all previous discussion. He may
find it of interest to read

http://oakroadsystems.com/genl/unice.htm#quote
Maury Barbato
2005-10-19 17:53:52 UTC
Permalink
Post by W. Dale Hall
Post by Maury Barbato
Surely you're right: your counterexample is very
beautiful! I didn't know the Warsawe Circle: it's a
very
Post by Maury Barbato
interesting topological object!
However, what do you think of adding in my
conjecture
Post by Maury Barbato
the hypothesis that S is path-connected?
The Warsawe Circle is path-connected.
Other than that, as you removed all context,
I'm unable to give additional cogent comments.
Post by Maury Barbato
Thank you very much for your attention and your
interest in the subject.
Thank you were you to include context.
I've not the time nor the inclination for the
clerical
chore of back tracking the thread to reconstruct the
line of thought. If you want intelligent and well
thought
answers, then include the context pertinent to your
reply
and to whom your are talking.
Please learn and use better math group manners as
demonstrated
by other participants of this newsgroup and as
described at
http://oakroadsystems.com/genl/unice.htm#quote
If you're using Google's quick reply, please don't,
use quote feature
instead. Many of us have different news browsers
than Google.
I'm sorry ... I didn't know these principles about
quoting. I'll try to apply them from now on.
Thank you for your remarks: they will be useful!

My best regards
Maury
TheShrike
2005-10-19 20:04:30 UTC
Permalink
<snip>
Post by William Elliot
Post by William Elliot
Thank you were you to include context.
I've not the time nor the inclination for the
clerical
chore of back tracking the thread to reconstruct
the
Post by William Elliot
line of thought. If you want intelligent and well
thought
answers, then include the context pertinent to
your
Post by William Elliot
reply
and to whom your are talking.
Please learn and use better math group manners as
demonstrated
by other participants of this newsgroup and as
described at
http://oakroadsystems.com/genl/unice.htm#quote
If you're using Google's quick reply, please
don't,
Post by William Elliot
use quote feature
instead. Many of us have different news browsers
than Google.
I'm sorry ... I didn't know these principles about
quoting. I'll try to apply them from now on.
Thank you for your remarks: they will be useful!
My best regards
Maury
Don't apologize, your post was perfectly acceptable.
This William Elliot boob has posted the same snide
comment verbatim all over this forum. Seems like a
lot of malice from one who hasn't the time nor [sic]
the inclination. And to you, William: if you haven't
the time nor the inclination, please don't participate here--easier for you, easier for everyone else.
Maury Barbato
2005-10-19 20:37:07 UTC
Permalink
Post by TheShrike
<snip>
Post by William Elliot
Post by William Elliot
Thank you were you to include context.
I've not the time nor the inclination for the
clerical
chore of back tracking the thread to reconstruct
the
Post by William Elliot
line of thought. If you want intelligent and
well
Post by William Elliot
Post by William Elliot
thought
answers, then include the context pertinent to
your
Post by William Elliot
reply
and to whom your are talking.
Please learn and use better math group manners
as
Post by William Elliot
Post by William Elliot
demonstrated
by other participants of this newsgroup and as
described at
http://oakroadsystems.com/genl/unice.htm#quote
If you're using Google's quick reply, please
don't,
Post by William Elliot
use quote feature
instead. Many of us have different news
browsers
Post by William Elliot
Post by William Elliot
than Google.
I'm sorry ... I didn't know these principles about
quoting. I'll try to apply them from now on.
Thank you for your remarks: they will be useful!
My best regards
Maury
Don't apologize, your post was perfectly acceptable.
This William Elliot boob has posted the same snide
comment verbatim all over this forum. Seems like a
lot of malice from one who hasn't the time nor [sic]
the inclination. And to you, William: if you haven't
the time nor the inclination, please don't
participate here--easier for you, easier for everyone
else.
Dear TheShrike (I don't know your real name, so ...),
I thank you very much for your encouraging words.
If everyone were like you, ...
Thank you very very ... much again.

My Best Regards
Maury
William Elliot
2005-10-20 06:40:45 UTC
Permalink
Post by W. Dale Hall
Post by Maury Barbato
we have the following well known (see "Elements of the
Topology of Plane Sets of Points" by Newman)
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.
Set A = { (x, sin 1/x) | x in R\0 }

Let C = A \/ { (0,0) }
C is connected and as per discussion below, Dale's item 1,
simply connected. C however isn't contractible because
contractible spaces are path connected which C isn't.

Of interest is R^2 - C which is also connected,
ie C and R^2 - C are two disjoint connected sets
that pass thru each other. Of course R^2 - C also
isn't path connected.

Let B = R^2 - A
B is path connected. Is not B also simply connected?
Is B contractible? I think not.
Post by W. Dale Hall
Your conjecture is incorrect.
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.
Let W0 = C \/ { (0,0) } \/ K.
Surprise, R^2 - W0 is connected.
W0 is connected and simply connected as per Dale's item 1 below.
W0 isn't contractible as contractible spaces are path connected.
Post by W. Dale Hall
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.
W is not contractible for not being path connected.
Post by W. Dale Hall
3. R^2 \ W is not connected.
Maury Barbato
2005-10-24 10:35:57 UTC
Permalink
Post by Maury Barbato
Post by W. Dale Hall
Post by Maury Barbato
we have the following well known (see "Elements
of the
Post by W. Dale Hall
Post by Maury Barbato
Topology of Plane Sets of Points" by Newman)
Conjecture. Let S be a subset of R^2. Then S is
simply
Post by W. Dale Hall
Post by Maury Barbato
connected if and only if it is contractible.
Besides, if S is a bounded simply connected
subset of
Post by W. Dale Hall
Post by Maury Barbato
R^2, then Bd(S) and R^2\S are connected.
Set A = { (x, sin 1/x) | x in R\0 }
Let C = A \/ { (0,0) }
C is connected and as per discussion below, Dale's
item 1,
simply connected. C however isn't contractible
because
contractible spaces are path connected which C isn't.
Of interest is R^2 - C which is also connected,
ie C and R^2 - C are two disjoint connected sets
that pass thru each other. Of course R^2 - C also
isn't path connected.
Let B = R^2 - A
B is path connected. Is not B also simply connected?
Is B contractible? I think not.
These are very beautiful (and pathological) examples!
However, I think that the problem in all the
counterexamples found is that the existence of a simple
closed curve passing through an arbitrary point of the
subset S fails.
The hypothesis that S is locally path-connected assure
us this property holds?

Conjecture.Let S be a locally path-connected subspace
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.

However, I think that, indipendently of this conjecture,
the following part of my original conjecture holds.

Theorem. If S is a bounded simply connected subset of
R^2, then Bd(S) is connected.
Post by Maury Barbato
Post by W. Dale Hall
Your conjecture is incorrect.
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)),
Post by W. Dale Hall
but missing all other points of the union
of C and D.
Post by W. Dale Hall
W = C U D U K
where I've used U to denote set-theoretic union.
Let W0 = C \/ { (0,0) } \/ K.
Surprise, R^2 - W0 is connected.
W0 is connected and simply connected as per Dale's
item 1 below.
W0 isn't contractible as contractible spaces are path
connected.
Post by W. Dale Hall
1. W is simply-connected, since no path can cross
over the tangled
Post by W. Dale Hall
sin(1/x) part as x passes 0: the space is not
even path-connected,
Post by W. Dale Hall
having two path-components: a) C U K and b) D.
2. W is not contractible, since it has a
non-nullhomotopic map into
Post by W. Dale Hall
the ordinary circle. If you have seen the
picture, the proof is
Post by W. Dale Hall
so close to obvious that it's difficult.
W is not contractible for not being path connected.
Post by W. Dale Hall
3. R^2 \ W is not connected.
Thank you very very much for your attention and
your irreplaceable help.
Maury
José Carlos Santos
2005-10-24 10:50:51 UTC
Permalink
Post by Maury Barbato
However, I think that, indipendently of this conjecture,
the following part of my original conjecture holds.
Theorem. If S is a bounded simply connected subset of
R^2, then Bd(S) is connected.
I've asked once at this newsgroup whether or not this statemente was
true (except the I also added the hypothesis that S is an open set).
See:

http://groups.google.com/group/sci.math/browse_frm/thread/6efe9a6eaf537285/5c4546a6f6a28fbd#5c4546a6f6a28fbd

As you can see, there seems to be no simple answer.

Best egards,

Jose Carlos Santos
Maury Barbato
2005-10-24 13:19:12 UTC
Permalink
Dear Jose,
I thank you for your attention, but my question
is more general (see my first message

http://mathforum.org/kb/message.jspa?messageID=4027206&tstart=0 ).

My best regards.
Maury
William Elliot
2005-10-24 11:46:41 UTC
Permalink
Post by Maury Barbato
Post by William Elliot
Set A = { (x, sin 1/x) | x in R\0 }
Let C = A \/ { (0,0) }
C is connected and as per discussion below, Dale's item 1, simply
connected. C however isn't contractible because contractible spaces
are path connected which C isn't.
Of interest is R^2 - C which is also connected,
ie C and R^2 - C are two disjoint connected sets
that pass thru each other. Of course R^2 - C also
isn't path connected.
Let B = R^2 - A
B is path connected. Is not B also simply connected?
Is B contractible? I think not.
These are very beautiful (and pathological) examples!
Conjecture.Let S be a locally path-connected subspace
of R^2. Then S is simply connected if and only if it is
contractible.
Contractible ==> path-connected

Doesn't contractible ==> simply connected?

So your conjecture is actually,
locally path connected, simply connected ==> contractible?

The double rational slope 90 deg fan, one from (0,0) above the x-axis
and the other from (1,0) be low the x-axis is example of nowhere locally
connected, path-connected, simply connected, not contractible. This
belies the stronger conjecture
path connected, simply connected ==> contractible
Post by Maury Barbato
Besides, if S is a bounded simply connected subset of
R^2, then Bd(S) and R^2\S are connected.
However, I think that, indipendently of this conjecture,
the following part of my original conjecture holds.
Theorem. If S is a bounded simply connected subset of
R^2, then Bd(S) is connected.
Maury Barbato
2005-10-24 13:17:07 UTC
Permalink
Post by William Elliot
On Thu, Oct 20, 2005 2:40 AM, William Elliot
Post by William Elliot
Set A = { (x, sin 1/x) | x in R\0 }
Let C = A \/ { (0,0) }
C is connected and as per discussion below,
Dale's item 1, simply
Post by William Elliot
connected. C however isn't contractible because
contractible spaces
Post by William Elliot
are path connected which C isn't.
Of interest is R^2 - C which is also connected,
ie C and R^2 - C are two disjoint connected sets
that pass thru each other. Of course R^2 - C
also
Post by William Elliot
isn't path connected.
Let B = R^2 - A
B is path connected. Is not B also simply
connected?
Post by William Elliot
Is B contractible? I think not.
These are very beautiful (and pathological)
examples!
Conjecture.Let S be a locally path-connected
subspace
of R^2. Then S is simply connected if and only if
it is
contractible.
Contractible ==> path-connected
Doesn't contractible ==> simply connected?
So your conjecture is actually,
locally path connected, simply connected ==>
contractible?
The double rational slope 90 deg fan, one from (0,0)
above the x-axis
and the other from (1,0) be low the x-axis is example
of nowhere locally
connected, path-connected, simply connected, not
contractible.
A question: why isn't it contractible? (Maybe I
didn't understand exactly the definition of this set.)
Post by William Elliot
This
belies the stronger conjecture
path connected, simply connected ==> contractible
W.Dale Hall gave an other example of a subset which
contradicts this implication in one of his messages.

On Wed, Oct 19, 2005 9:13 PM, he wrote:

"<snip>
It just occurred to me that we can modify the Warsaw
Circle to obtain a path-connected space that is simply-
connected but not contractible.
<snip>
Instead of taking two copies of the topologist's sine
curve S = {(x,sin(1/x)) | x > 0} U {(0,y) | |y| <= 1}
identified suitably, connected by an additional arc,
just takethe one copy of S, and plug the tail end back
into the vertical interval. I'll call it the Warsaw
semicircle: W/2.
Thanks to the tail coming back to meet that vertical
line segment,the space W/2 is path-connected. Due to the
inability of a continuous path to reach around the
sin(1/x) part to join the vertical interval,every map
from S^1 to W/2 is null-homotopic.Of course, W/2 is
pathological in not being locally-connected nor locally
path-connected.
Dale"

However, if require S to be locally path-connected,
we can't use these pathological sets, and there's
some hopes that my conjecture is true.
Post by William Elliot
Besides, if S is a bounded simply connected subset
of
R^2, then Bd(S) and R^2\S are connected.
However, I think that, indipendently of this
conjecture,
the following part of my original conjecture holds.
Theorem. If S is a bounded simply connected subset
of
R^2, then Bd(S) is connected.
Thank you for your attention.
Maury
William Elliot
2005-10-25 10:56:55 UTC
Permalink
Post by Maury Barbato
Post by William Elliot
Post by Maury Barbato
Conjecture.Let S be a locally path-connected subspace
of R^2. Then S is simply connected if and only if
it is contractible.
Contractible ==> path-connected
Doesn't contractible ==> simply connected?
So your conjecture is actually,
locally path connected, simply connected ==> contractible?
The double rational slope 90 deg fan, one from (0,0)
above the x-axis
and the other from (1,0) be low the x-axis is example
of nowhere locally
connected, path-connected, simply connected, not contractible.
A question: why isn't it contractible? (Maybe I
didn't understand exactly the definition of this set.)
Whoops, it's not strongly contractible.
Post by Maury Barbato
However, if require S to be locally path-connected,
we can't use these pathological sets, and there's
some hopes that my conjecture is true.
Indeed, and it's strengthen as
connnected, locally path-connected ==> path connected

Is there path connected space that not's contractable?

What if the conjecture is broadened?
Is there counter example for R^3, R^n, other spaces?

----
Maury Barbato
2005-10-25 13:53:52 UTC
Permalink
Post by Maury Barbato
Post by Maury Barbato
Post by William Elliot
Post by Maury Barbato
Conjecture.Let S be a locally path-connected
subspace
Post by Maury Barbato
Post by William Elliot
Post by Maury Barbato
of R^2. Then S is simply connected if and only
if
Post by Maury Barbato
Post by William Elliot
Post by Maury Barbato
it is contractible.
Contractible ==> path-connected
Doesn't contractible ==> simply connected?
So your conjecture is actually,
locally path connected, simply connected ==>
contractible?
Post by Maury Barbato
Post by William Elliot
The double rational slope 90 deg fan, one from
(0,0)
Post by Maury Barbato
Post by William Elliot
above the x-axis
and the other from (1,0) be low the x-axis is
example
Post by Maury Barbato
Post by William Elliot
of nowhere locally
connected, path-connected, simply connected, not
contractible.
Post by Maury Barbato
A question: why isn't it contractible? (Maybe I
didn't understand exactly the definition of this
set.)
Whoops, it's not strongly contractible.
Now I understood, with some efforts, what set
you referred, and it's surely not contractible
A question: do "strongly contractible" have a special
meaning or do you mean "it's strongly non-contractible"?
Post by Maury Barbato
Post by Maury Barbato
However, if require S to be locally path-connected,
we can't use these pathological sets, and there's
some hopes that my conjecture is true.
Indeed, and it's strengthen as
connnected, locally path-connected ==> path connected
Is there path connected space that not's
contractable?
What do you mean by "not's"?
Please, can you use a less cryptical English?
Remember I'm Italian!
Post by Maury Barbato
What if the conjecture is broadened?
Is there counter example for R^3, R^n, other spaces?
----
Thank you for your attention.
Maury
William Elliot
2005-10-26 09:12:33 UTC
Permalink
From: Maury Barbato <***@aruba.it>
Newsgroups: sci.math
Subject: Re: Simply Connected Subsets of R^2
Post by Maury Barbato
Post by William Elliot
Post by Maury Barbato
Conjecture.Let S be a locally path-connected subspace
of R^2. Then S is simply connected if and only if
it is contractible.
Contractible ==> path-connected
Doesn't contractible ==> simply connected?
So your conjecture is actually,
locally path connected, simply connected ==> contractible?
The double rational slope 90 deg fan, one from (0,0)
above the x-axis
and the other from (1,0) be low the x-axis is example
of nowhere locally
connected, path-connected, simply connected, not contractible.
Now I understood, with some efforts, what set
you referred, and it's surely not contractible
path-connected nowhere locally connected space
{ (x, mx), (1-x, -mx) | 0 <= x, m in [0,1] /\ Q }
Post by Maury Barbato
A question: do "strongly contractible" have a special
meaning or do you mean "it's strongly non-contractible"?
The point of contraction is never moved.

S contractible when some h in C(SxI,S), a in S with
for all x in S, h(x,0) = x, h(x,1) = a

S strongly contractible when some h in C(SxI,S), a in S with
for all x in S, h(x,0) = x, h(x,1) = a, for all t, h(a,t) = a

A single fan, such as one of the above double fan, is contractible
to any point and strongly contractable at the vertex but at no other
point.
Post by Maury Barbato
Post by William Elliot
Post by Maury Barbato
However, if require S to be locally path-connected,
we can't use these pathological sets, and there's
some hopes that my conjecture is true.
Indeed, and it's strengthen as
connnected, locally path-connected ==> path connected
What do you mean by "not's"?
Please, can you use a less cryptical English?
Remember I'm Italian!
Whoops, my typo caused a derailment. Instead of correcting,
I remove question as it had vacuous significance.
Post by Maury Barbato
What if the conjecture is broadened?
Is there counter example for R^3, R^n, other spaces?
I'm puzzled, for what spaces are there counter example to your conjecture?
How general is it?

Here's concept related to simply connected.
S unicoherent when S connected,
for all closed connected K,L, (K\/L = S ==> K/\L connected)
sphere, line unicoherent; circle not unicoherent

----
Maury Barbato
2005-10-26 13:12:23 UTC
Permalink
Post by William Elliot
Newsgroups: sci.math
Subject: Re: Simply Connected Subsets of R^2
Post by Maury Barbato
Post by William Elliot
Post by Maury Barbato
Conjecture.Let S be a locally path-connected
subspace
Post by Maury Barbato
Post by William Elliot
Post by Maury Barbato
of R^2. Then S is simply connected if and only
if
Post by Maury Barbato
Post by William Elliot
Post by Maury Barbato
it is contractible.
Contractible ==> path-connected
Doesn't contractible ==> simply connected?
So your conjecture is actually,
locally path connected, simply connected ==>
contractible?
Post by Maury Barbato
Post by William Elliot
The double rational slope 90 deg fan, one from
(0,0)
Post by Maury Barbato
Post by William Elliot
above the x-axis
and the other from (1,0) be low the x-axis is
example
Post by Maury Barbato
Post by William Elliot
of nowhere locally
connected, path-connected, simply connected, not
contractible.
Post by Maury Barbato
Now I understood, with some efforts, what set
you referred, and it's surely not contractible
path-connected nowhere locally connected space
{ (x, mx), (1-x, -mx) | 0 <= x, m in [0,1] /\ Q }
Post by Maury Barbato
A question: do "strongly contractible" have a
special
Post by Maury Barbato
meaning or do you mean "it's strongly
non-contractible"?
The point of contraction is never moved.
S contractible when some h in C(SxI,S), a in S with
for all x in S, h(x,0) = x, h(x,1) = a
S strongly contractible when some h in C(SxI,S), a in
S with
for all x in S, h(x,0) = x, h(x,1) = a, for all t,
, h(a,t) = a
A single fan, such as one of the above double fan, is
contractible
to any point and strongly contractable at the vertex
but at no other
point.
Post by Maury Barbato
Post by William Elliot
Post by Maury Barbato
However, if require S to be locally
path-connected,
Post by Maury Barbato
Post by William Elliot
Post by Maury Barbato
we can't use these pathological sets, and
there's
Post by Maury Barbato
Post by William Elliot
Post by Maury Barbato
some hopes that my conjecture is true.
Indeed, and it's strengthen as
connnected, locally path-connected ==> path
connected
Post by Maury Barbato
What do you mean by "not's"?
Please, can you use a less cryptical English?
Remember I'm Italian!
Whoops, my typo caused a derailment. Instead of
correcting,
I remove question as it had vacuous significance.
OK, maybe it is insignificant, but I'm now curious
to know what "not's" stand for (I can't see the typo).
Post by William Elliot
Post by Maury Barbato
What if the conjecture is broadened?
Is there counter example for R^3, R^n, other
spaces?
I'm puzzled, for what spaces are there counter
example to your conjecture?
How general is it?
Surely it is not true for R^n, with n>2.
Take D={x in R^3: |x|<1}. Then D\{0} is
obviously connected, locally path-connected,
simply connected, but not contractible.

My best regards.
Maury
William Elliot
2005-10-27 10:08:50 UTC
Permalink
From: Maury Barbato <***@aruba.it>
Newsgroups: sci.math
Subject: Re: Simply Connected Subsets of R^2
Post by Maury Barbato
Conjecture.Let S be a locally path-connected subspace
of R^2. Then S is simply connected if and only
if is contractible.
OK, maybe it is insignificant, but I'm now curious to know
what "not's" stand for (I can't see the typo).
Nil, as it was suffix transposed incorrectly from another word to which it
was intended. However look out, you're asking about English which me
thinks is contender for world's most illogical language.
It's. It is. Compare with the phrases:
"its lid", "the pot's lid", "several pots".
That's it. That is it.

So proceed at your own risk. I used to converse with Russian even to the
point where we could discuss idioms and figures of speech. The Russian
one I remember best is "hanging noodles on one's ears," for example,
those weapons of mass destruction amassed in Iraq was just Bush hanging
noodles on our ears. Anyway have you wish to further discuss language,
culture, countries and/or politics do please remove "remove" from my email
address.

BTW, your mathematical English is proficient except for indulging the
recent fad of squeezing spaces frommathematicalexpressions.
Post by Maury Barbato
What if the conjecture is broadened?
Is there counter example for R^3, R^n, other spaces?
Surely it is not true for R^n, with n>2.
Take D={x in R^3: |x|<1}. Then D\{0} is
obviously connected, locally path-connected,
simply connected, but not contractible.
Oh, of course.

----

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