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By James R. Munkres

A readable creation to the topic of calculus on arbitrary surfaces or manifolds. obtainable to readers with wisdom of easy calculus and linear algebra. Sections comprise sequence of difficulties to augment concepts.

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A is compact}. We note that 2x = CL(X) when X is compact. We also note that when X is Hausdorff, 2x = {A c X : A is nonempty and compact}. 6 Definition. C(X) = {A E 2dY : A is connected}; in other words, C(X) = 2x n CLC(X). Starting with Chapter II, the book is almost exclusively about the two hyperspaces 2x and C(X) when X is a compact metric space. For the next definition, (Al denotes the cardinality of a set A. Also, recall that a Tl-space is a topological space, X, such that (2) is closed in A’ for each 2 E X.

Therefore, by (2), the base p is uncountable. 4 Theorem. Let (X, T) be a Ti-space. then (X, T) is a compact, metrizable space. If (CL(X), TV) is metrizable, Proof. 15), it is clear that (X,T) must be metrizable. We prove that (X,T) is compact. Suppose that (X, T) is not compact. Then, since (X, T) is metrizable, there is a countably infinite, closed (in X), and discrete subspace, (Y, TIY), of (X,T). 22 that CL(Y) with its own Vietoris topology (TIY) v is a subspace of (CL(X), TV). Thus, since we are assuming that (CL(X), 2’“) is metrizable, we have that (1) (CL(Y), (TIY)“) is metrizable.

11 trivial. 14 Exercise. Let (X, d) be a bounded metric space. If (X,d) is complete, then (CL(X), H d ) is complete. ) For each n = [Hint: Let {Ai}zl b e a Cauchy sequence in CL(X). 1,2,. , let Y, = d(UgO,,Ai). ] Remark. 5. 15 Exercise. If (X,d) is a complete metric space, then (2”,Hd) is complete. 14. Here, A, E 2x for each i; prove that E;, hence each Y,, is totally bounded and complete, therefore compact [5, P. , d(s, for Fz(I) First, R2. Next, that Exercise. Let I = [0, 11, and let d denote the usual metric for I t) = Js - tl for all s, t E I).

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