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