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4, F is a vector subspace of E and it is clear that it is the smallest closed vector subspace of E containing A. Assume now that A is countable and let B be the set of linear combinations of elements of A with coefficients in Q (resp. ~ + iQ). Then B is countable and m BcFcB. e. F is separable. 6 Let E be a normed (Banach) space and c~, fl, 7, 6 scalars such that ~- f17 # o. Then E x E > IR+, (z,y) , > sup{ II~x + ~YlI, 117x + @IF} is a (complete) norm. 1). > sup{ll*ll, Ilvll} n 40 1. 1 ( 0 ) An ordered set T is called upward (downward) directed if for every s, t 9 T , there is an r 9 T such that s<_r, t<_r (r<_s,r<_t).
It is obviuous that a sequence in I-I EL converges to a point x 9 11 E~ iff the projections of this LEI LCI sequence converge to the corresponding projections of x We deduce that the above norm generates the product topology of 1-[ EL. The last assertion is easy LEI to prove. 2 ( 0 ) Let (EL)~cI be a finite family of normed spaces. 1). These norms are denoten sometimes by I1" lb. Th~ 2-~o~m i~ also called the Euclidean n o r m of the product 11 EL. Unless otherwise specified, LEI we take the Euclidean norm on the product of normed space.
C) Every sequence in A has a subsequence, which converges in T . a => b => c is trival. c => a. Let (tn)nE~ be a sequence in A. ) < - n for every n E IN. ),E~ which converges in T . Let 20 1. Banach Spaces t'- lim sk~. n---+ (:XD m Then t E A and lim tkn = t . n---+ (x) m Hence A is compact. 12 t 0 ) Every relatively compact set of a metric space is precompact. The converse implication holds whenever the metric space is complete. m Take a relatively compact set A of the metric space T. 11 b =~ a).