Download Classical analysis on normed spaces by Tsoy-Wo Ma PDF

By Tsoy-Wo Ma

This e-book presents an trouble-free creation to the classical research of normed areas, paying targeted consciousness to non-linear issues similar to mounted issues, calculus and traditional differential equations. it's aimed toward novices who are looking to get throughout the uncomplicated fabric once attainable after which stream directly to do their very own examine instantly. It assumes basically common wisdom in finite-dimensional linear algebra, easy calculus and simple advanced research. because the remedy is self-contained with enough information, even an undergraduate with mathematical adulthood shouldn't have any challenge operating via it by myself. a number of chapters may be built-in into components of a grasp measure programme through direction paintings equipped via any neighborhood collage. constrained to finite-dimensional areas instead of normed areas, chosen chapters can be utilized for a direction in complex calculus. Engineers and physicists may possibly locate this booklet a convenient reference in classical research

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Example text

Next, since C = R , bounded sets are identical with precompact sets. Repeating the same argument for R " , we obtain the result for C " . 10. Exercise Show that the real line equipped with the discrete metric is bounded but not precompact. 11. Exercise Prove that the projection of a precompact set in a product metric space X x Y to X is precompact. 12. Exercise Is the set of all integers precompact? Is the set of all rationals i n (0,1) precompact? 1. In a precompact set A, every sequence has a Cauchy subsequence which need not converge in A.

A sequence is said to be bounded if its range is bounded. 2. Theorem diam'M) = diam(M). Proof. It is easy to prove that diam(M) < diam(M). If Af is unbounded, then both sides are oo. Without loss of generality, assume diam(M) < oo. Take any x, y e Af. For every e > 0, we select a e B ( x , e) n Af and b € B ( y , e) n Af. Hence, a\x,y) < a\x,a) + d(a,b) + d(f>,y) < diam{M) + 2e. Taking supremum over all x y e Af, we have diam(M) < diam(M) + 2e. Since e > 0 is arbitrary, t we have diam(M) < diam{M).

Let X,Y be metric spaces. Then a bijection / : X -* Y is called a homeomorphism if both / , f~ are continuous. 5. Corollary Let f : X -> Y be a continuous bijection. If A* is compact, then / is a homeomorphism. P r o °/- of X. 1 1 suffices to show that g = is continuous. Let A be a closed subset Since X is compact, so is A. Now f(A) is compact and hence closed. 2-7 Continuous Maps on Compact Spaces 31 Therefore the inverse image g~\A) = f(A) of a closed set under g is closed. Consequently, = g is continuous.

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