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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

1098.0. "topology problem" by HERON::BUCHANAN (Andrew @vbo DTN 828-5805) Mon Jul 10 1989 18:54

T.RTitleUserPersonal
Name
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1098.1HPSTEK::XIATue Jul 11 1989 04:109
    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
1098.2HPSTEK::XIATue Jul 11 1989 04:2110
    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
1098.3clarificationHERON::BUCHANANAndrew @vbo DTN 828-5805Tue Jul 11 1989 08:514
1098.4sorry for poor EnglishJRDV04::KOMATSUExistentialistFri Jul 14 1989 01:0559
1098.5HPSTEK::XIAFri Jul 14 1989 22:0634
    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
              
1098.6AITG::DERAMODaniel V. {AITG,ZFC}:: D'EramoSun Jul 16 1989 03:5367
1098.7cannot assume that F is path connectedAITG::DERAMODaniel V. {AITG,ZFC}:: D'EramoSun Jul 16 1989 14:1114
        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
1098.8AITG::DERAMODaniel V. {AITG,ZFC}:: D'EramoSun Jul 30 1989 03:0422
>>        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
1098.9next...HERON::BUCHANANAndrew @vbo DTN 828-5805Mon Jul 31 1989 10:154
	Well, how might this be false for n > 2?   Or given a proof for
n, can we get one for n+1?

Andrew.
1098.10AITG::DERAMODaniel V. {AITG,ZFC}:: D'EramoMon Jul 31 1989 22:414
	Gee, it seems like .0, or the theorem in .8, should be
	true for n > 2.  No counterexamples rush to mind.  :-)

	Dan
1098.11The wonder of Algebraic TopologyHPSTEK::XIAIn my beginning is my end.Thu Nov 22 1990 16:4753
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?  :-)

1098.12GUESS::DERAMODan D'EramoFri Nov 23 1990 18:5512
        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
1098.13HPSTEK::XIAIn my beginning is my end.Mon Nov 26 1990 19:3112
    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 
1098.14GUESS::DERAMODan D'EramoMon Nov 26 1990 19:518
        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
1098.15HPSTEK::XIAIn my beginning is my end.Mon Nov 26 1990 20:2212
    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
1098.16I say "path connected" when I mean "path connected".GUESS::DERAMODan D'EramoMon Nov 26 1990 22:147
        re .15,
        
>> Why is point set topology the only way to go?
        
        I like sets.
        
        Dan
1098.17HARLEY::DAVETue Nov 27 1990 13:449
    
    
    My fiancee successfully defended her doctoral thesis yesterday!
    
    Topic was something to do with semi metrizable spaces...
    
    Thank goodness it is finaly over...
    
    Dave
1098.18Always glad to hear good news from fellow mathematiciansHPSTEK::XIAIn my beginning is my end.Wed Nov 28 1990 15:576
    re .17
    
    Please extend my congratulations.  By the way, where did she get her
    degree?
    
    Eugene
1098.19HARLEY::DAVEFri Nov 30 1990 10:1310
    
    
    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