Conference rusure::math

    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

    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

	dF is the boundary of F.
	I would presume that dF = d(F�).

Andrew.

>	Say R^n has a normed topology.
>	F is a subthing of R^n, compact and connected
>	F� connected.   (That's R^n \ F, I think)
>
>	Show dF is connected.

Re .1

Because R^n has a normed topology, the definition of dF is as follows,

dF = { p ; take A includes p, shch that A is an open set
           (open ball in this case),
                 -> intersection of A and F is not empty,
                    intersection of A and R^n \ F is not empty. }

Re .2

>    I guess that with n=1, if F is a subset of R and: 
>    Then F is the empty set.

I think it's the right answer.

Note : If I can assume F and R^n\F is arcwise connected, this excersize must
      not be so difficult. The outline of the proof will be as follows,

      Choose 2 points from dF whose name is P and Q.
      I will show you there is a path from P to Q.

      w(t) is a path, Continuous Mapping from [0,1] to R^n, with follwing
      features.
	w(0) = P, w(1) = Q, w(t) included in F (t is in [0,1])
      The reason why you can choose this path is that F is arcwise connected.

      w'(t) is also a path, such that,
	w'(0) = P, w'(1) = Q, w'(t) included in R^n\F (t is in [0,1])
      Same reason.

      We can choose w as, if t is not s then w(t) is not w(s).
      Also w'.

      W(s,t) : continuous mapping from [0,1]x[0,1] to R^n
               W(0,t) = w(t)
               W(1,t) = w'(t)
               W(s,0) = P    ! constant
               W(s,1) = Q    ! constant
      cf. W(s,t) is a homotopy between w and w'.

      You can't choose a path from a point in F to a point in R^n\F that never
      be a point in dF. (Jordan?'s theorem) This means we can find a path in
      [0,1]x[0,1] with mapping image of the path is from P to Q and included in
      dF.

Above must be almost right answer in arcwise connected condition. But I think
I need more explanations in Last Part. But this is a 2-dimensional problem.
I think this must be related to Jordan's closed curve theorem ? or something.

I think case of R^n with only connected condition should be very difficult.
This kind of things reminds me Poincare's conjectur.

    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
              

	re .0

>>	Say R^n has a normed topology.
>>	F is a subthing of R^n, compact and connected
>>	F� connected.   (That's R^n \ F, I think)
>>
>>	Show dF is connected.
>>
>>	Herve said he thought he'd done it for n=2.   Even *I* can do it for
>>	n=1.   So what's the story?

	In note 1038 we show that all norms over R^n for finite n
	are equivalent, so the problem is dealing with R^n with
	the usual Euclidean topology.

	re .3

>>	dF is the boundary of F.
>>	I would presume that dF = d(F�).

	The "dF" notation looks like a "terminal" representation
	of the partial derivative "delta" symbol that is usually
	(or is it just often?) used for boundaries.

	The boundary of a subset A of a space X is the
	intersection of the closure of A with the closure of X-A.
	This is symmetric with complementation, so dF = d(R^n - F).

	Compact subsets of a Hausdorff space are closed.  R^n is
	definitely Hausdorff and F is given as compact, so F is
	closed.  The closure of F is therefore F, and so dF is
	the intersection of F with the closure of (R^n - F).
	This will be a closed subset of F, and since closed
	subsets of compact sets are compact, dF is also compact.

	re .4

>>	Re .1
>>
>>	Because R^n has a normed topology, the definition of dF is as follows,
>>
>>	dF = { p ; take A includes p, shch that A is an open set
>>	           (open ball in this case),
>>	                 -> intersection of A and F is not empty,
>>	                    intersection of A and R^n \ F is not empty. }

	This works out the same as the above, that dF is the set
	of points p that are in both closures.

	re .5

>>         Yes, I also had some difficulties when F is not path connected.
>>    However, since F is compact, F is closed, so R \F is open, hence, path
>>    connected.

	Yes, R^n is locally path connected, so its open
	connected subsets are also path connected, and R^n - F is
	given as open and connected.

	An "obvious" visualization is something like F a solid,
	filled in sphere in R^3.  F is compact, it and its
	complement are both connected, and dF is a sphere, which
	is connected.  But I don't see how to prove the general
	case or how to prove the desire in earlier replies that F
	be path connected.

	Dan

        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

>>        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

	Well, how might this be false for n > 2?   Or given a proof for
n, can we get one for n+1?

Andrew.

	Gee, it seems like .0, or the theorem in .8, should be
	true for n > 2.  No counterexamples rush to mind.  :-)

	Dan

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?  :-)

        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

    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 

        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

    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

        re .15,
        
>> Why is point set topology the only way to go?
        
        I like sets.
        
        Dan

    
    
    My fiancee successfully defended her doctoral thesis yesterday!
    
    Topic was something to do with semi metrizable spaces...
    
    Thank goodness it is finaly over...
    
    Dave

    re .17
    
    Please extend my congratulations.  By the way, where did she get her
    degree?
    
    Eugene

    
    
    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