This preview shows page 1. Sign up to view the full content.
Unformatted text preview: R such that K 1 K 2 K 3 ... . Prove that there exists at least one point x R such that x K n for all n N ; that is, the intersection T n =1 K n is not empty. Suppose that T K n is empty. Since K n is compact, it is closed. Thus, K c n is open. Since T K n is empty, we have that S K c n = R . In particular, { K c n } n =2 is an open cover for K 1 . Thus, there are a nite number of indices 1 ,..., N so that K 1 K c 1 K c N = ( K 1 K N ) c . This means that K 1 K 1 K N is empty. But by the nexting property, if 1 N , then K 1 K 1 K N = K N which was nonempty by assumption. 1...
View
Full
Document
This note was uploaded on 07/15/2010 for the course MATH 521 taught by Professor Metcalfe during the Spring '10 term at University of North Carolina Wilmington.
 Spring '10
 MetCalfe
 Calculus

Click to edit the document details