From: israel@unixg.ubc.ca (Robert B. Israel)
Newsgroups: sci.math
Subject: Re: another sphere decomposition
Date: 24 Nov 92 22:45:48 GMT

In <1992Nov23.204615.12917@mp.cs.niu.edu> rusin@mp.cs.niu.edu (David Rusin) writes:

>Prove or disprove (for a colleague's student): If the 2-sphere is
>written as the union of two compact pieces K and L having finitely
>many components, then (K intersect L) has finitely many components.

>Easily the answer is no if the union need not be all of the sphere,
>and it's yes if 'compact' is replaced with 'open'. If the sphere is
>replaced by a torus, say, then the number of components may be
>greater (e.g. K and L can be connected but not their intersection)
>but undoubtedly it's finite in the torus case if it must be so 
>in the sphere case.

>dave rusin@math.niu.edu 

  
In fact, if K_i and L_i are the components of K and L respectively, then
each K_i intersect L_j is connected.

Proof: Suppose not.  Let H = K_i intersect L_j.  
Then there are disjoint open sets U, V with H contained in U union V,
U intersect H and V intersect H nonempty.  Without loss of generality,
we may assume U is the union of a finite number of disks, none of them
tangent to each other, and is 
connected.  Let D be a component of U^c that intersects H.  Then D is 
topologically a disk, so its boundary B is connected.  Now B is contained
in H^c = K^c union L^c, so it is contained in K^c or L^c.  But if B is
in K^c, then Int(D) and Int(D^c) are disjoint open sets whose union contains
K_i, and both intersect K_i, contradicting connectedness of K_i.  Similarly
if B is in L^c.    
-- 
Robert Israel                            israel@math.ubc.ca
Department of Mathematics             or israel@unixg.ubc.ca
University of British Columbia
Vancouver, BC, Canada V6T 1Y4