This preview shows page 1. Sign up to view the full content.
Unformatted text preview: k n kk x k so that (as k n k is nonzero) n 6 k x k and k x k is an upper bound for F . Recalling (Archimedes) that the only subsets of N that are bounded above are nite, it follows that F is nite. Note : As noted in class, the existence of norms on X (indeed, on any real/complex vector space) follows easily from Zorns Lemma. Remark : As remarked in class, if F N is nite then X does indeed carry a norm kk F relative to which n is continuous whenever n F : simply let k k be any norm on X and for x X dene k x k F = k x k + X n F | x n | . 1...
View Full Document
This note was uploaded on 07/08/2011 for the course MHF 3202 taught by Professor Larson during the Spring '09 term at University of Florida.
- Spring '09