T.R | Title | User | Personal Name | Date | Lines |
---|
1098.1 | | HPSTEK::XIA | | Tue Jul 11 1989 04:10 | 9 |
| Andrew,
Thank you for thinking highly of me. However, I must admit my
ignorance here since I do not know what dF means. Could you (or
someone who knows) define it for us?
Thanks in advance.
Eugene
|
1098.2 | | HPSTEK::XIA | | Tue Jul 11 1989 04:21 | 10 |
| I guess that with n=1, if F is a subset of R and:
1. F is compact and connected.
2. R\F is connected.
Then F is the empty set.
I guess the above is obvious?
Eugene
|
1098.3 | clarification | HERON::BUCHANAN | Andrew @vbo DTN 828-5805 | Tue Jul 11 1989 08:51 | 4 |
1098.4 | sorry for poor English | JRDV04::KOMATSU | Existentialist | Fri Jul 14 1989 01:05 | 59 |
1098.5 | | HPSTEK::XIA | | Fri Jul 14 1989 22:06 | 34 |
| re -1,
Yes, I also had some difficulties when F is not path connected.
n
However, since F is compact, F is closed, so R \F is open, hence, path
connected. Now supposed dF is not connected, then you can pick two
disconnected pieces of dF say A and B. Pick x from A and y from B, and
take small open balls Nx and Ny around x and y respectively so that
Nx does not contain any point in B and Ny does not contain any point in
n
A. Since R \F is open and connected, there is a path that starts from
x and ends at y. Now all you have to do is to prove that this path
has to intesect F (which is intuitively obvious), but I haven't been
able to take that last step. I guess the last step will be somewhat
messy, but manageable. Base on the progress made above, I would say
that the problem shouldn't be very hard, but I cannot be absolutely
sure (hey, the Poincare conjecture looked easy too :-)).
I will think about it more when I have time.
Hope that helps.
Eugene
|
1098.6 | | AITG::DERAMO | Daniel V. {AITG,ZFC}:: D'Eramo | Sun Jul 16 1989 03:53 | 67 |
1098.7 | cannot assume that F is path connected | AITG::DERAMO | Daniel V. {AITG,ZFC}:: D'Eramo | Sun Jul 16 1989 14:11 | 14 |
| Let S be the subset of R^2 given by
S = {(x,sin 1/x) | 0 < x <= 1}
S is the image of the connected set (0,1] under a
continuous function, so S is connected. Let F be its
_
closure S in R^2. Then F is also connected. In fact F
satisfies all of the conditions of .0; F is compact and
connected and its complement in R^2 is connected, but F
is not path connected. (Here dF = F is connected so it
is not a counterexample to .0.)
Dan
|
1098.8 | | AITG::DERAMO | Daniel V. {AITG,ZFC}:: D'Eramo | Sun Jul 30 1989 03:04 | 22 |
| >> Herve said he thought he'd done it for n=2.
In my copy of _Counterexamples in Topology_, Second
Edition, by Lynn Arthur Steen and J. Authur Seebach, Jr.,
Springer-Verlag 1978, on pages 225-6 in the appendix, it
states:
"This follows directly from the theorem
(see, for instance, Newman [88], p. 124)
that every component of the complement of
a connected open subset of the plane has a
connected boundary."
The reference is to _Elements of the Topology of Plane
Sets of Points_ by M.H.A. Newman, Cambridge Univ. Press,
1939. What it means for the problem in .0, for the case
n=2, is that R^2 - F is a connected open subset of the
plane, and so every component of F = R^2 - (R^2 - F) has
a connected boundary. F is connected so it is its only
component, and therefore its boundary is connected.
Dan
|
1098.9 | next... | HERON::BUCHANAN | Andrew @vbo DTN 828-5805 | Mon Jul 31 1989 10:15 | 4 |
| Well, how might this be false for n > 2? Or given a proof for
n, can we get one for n+1?
Andrew.
|
1098.10 | | AITG::DERAMO | Daniel V. {AITG,ZFC}:: D'Eramo | Mon Jul 31 1989 22:41 | 4 |
| Gee, it seems like .0, or the theorem in .8, should be
true for n > 2. No counterexamples rush to mind. :-)
Dan
|
1098.11 | The wonder of Algebraic Topology | HPSTEK::XIA | In my beginning is my end. | Thu Nov 22 1990 16:47 | 53 |
| I know this is over a year late, but I finally come up with a proof (most
elegant, I might add). As a matter of fact, I think I deserve a MATH
notesfile prize for this. Hmm... Maybe I should get a free dinner at
next math party. Of course Andrew is free to send the solution to anyone,
provided credit is given where it is due. :-) I wander if I can
even get it published. Hmm...
The proof is very short, and it doesn't even require the whole space to
be R^n and neither does it require F to be compact. As a matter of fact,
it is suffice to require that the whole space itself is homotopically
equivalent to a single point (R^n certainly satisfies this condition for
all n).
Well, without much ado, I present:
Let X denote a space that is homotopically equivalent to a point (could be
R^n for example). F a subset of X. Denote F' to be the complement of F.
Let C(S) denotes the closure of S. By definition, dF (the boundry) is
equal to the intersection of C(F) and C(F'). Moreover, we have
C(F) U C(F') = X. Now we construct the reduced Mayer-Vietoris sequence
for dF, X, C(F) and C(F'):
~ ~ ~ ~ ~
... H (X) --> H (dF) --> H (C(F)) + H (C(F')) --> H (X) --> ...
1 0 0 0 0
~
Now the above sequece is exact. Moreover, it is immediate that H (X) = 0
1
since it is homotopically equivalent to a point. Moreover, since F and F'
are both connected, C(F)
~ ~
and C(F') are also connected. This implies that H (C(F)) + H (C(F'))
0 0
is also 0. Hence, we have an exact sequence that looks this:
~ ~
0 --> H (dF) --> 0 This implies that H (dF) is also 0.
0 0
Hence, dF is connected. Q.E.D.
I am afraid this proof does not offer much insight as to why it is so, but
on the other hand, this theorem is really quite clear intuitively any way.
Eugene
P.S. How is that for a proof Dan? :-)
|
1098.12 | | GUESS::DERAMO | Dan D'Eramo | Fri Nov 23 1990 18:55 | 12 |
| re .11,
>> P.S. How is that for a proof Dan? :-)
What's a reduced Mayer-Vietoris sequence? :-)
>> and neither does it require F to be compact.
I'll see if I can construct a counterexample when
F and R^n - F fail to be compact.
Dan
|
1098.13 | | HPSTEK::XIA | In my beginning is my end. | Mon Nov 26 1990 19:31 | 12 |
| re .12,
Dan, I think it would be great if you could post an expository essay on
on Homology and Mayer-Vietoris sequence. I tried to do that along with
the proof in .11, but I am no good at writing expository essays...
By the way, the word "connected" in .11 means "path connected", but
that is about all most people care about. Very few mathematician (all
of them point set topologist, and that is very few) care about the
pathological cases.
Eugene
|
1098.14 | | GUESS::DERAMO | Dan D'Eramo | Mon Nov 26 1990 19:51 | 8 |
| But .0 calls for connected, not path connected, and I
gave an example in an earlier reply that had the former
without the latter. And if I knew what a Mayer-Vietoris
sequence was, I wouldn't have asked.
Besides, point set topology is the only way to go.
Dan
|
1098.15 | | HPSTEK::XIA | In my beginning is my end. | Mon Nov 26 1990 20:22 | 12 |
| re .14,
Yea, I know Dan. It called for connected, but believe me, most people
say "connected" when they actually mean "path connected" especially
when dealing with R^n or manifolds like that. :-)
How about you write an expository essay on homological groups and I
then do the Mayer-Vietoris sequence part?
Why is point set topology the only way to go?
Eugene
|
1098.16 | I say "path connected" when I mean "path connected". | GUESS::DERAMO | Dan D'Eramo | Mon Nov 26 1990 22:14 | 7 |
| re .15,
>> Why is point set topology the only way to go?
I like sets.
Dan
|
1098.17 | | HARLEY::DAVE | | Tue Nov 27 1990 13:44 | 9 |
|
My fiancee successfully defended her doctoral thesis yesterday!
Topic was something to do with semi metrizable spaces...
Thank goodness it is finaly over...
Dave
|
1098.18 | Always glad to hear good news from fellow mathematicians | HPSTEK::XIA | In my beginning is my end. | Wed Nov 28 1990 15:57 | 6 |
| re .17
Please extend my congratulations. By the way, where did she get her
degree?
Eugene
|
1098.19 | | HARLEY::DAVE | | Fri Nov 30 1990 10:13 | 10 |
|
Laurie did her work at the university of new hampshire.
undergrade at mount holyoke
She seems to be much more relaxed lately, i wonder why....
dave
|