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