Abstract

PROOF. Suppose not. Let {A, } be an uncountable collection of different types. We may regard each AO as a subset of the Hilbert cube I@. For each : let c be chosen so that AO is a retract of its en-neighborhood UB in Iw. Let r,: UO-*A, be a retraction map. Since {AO is uncountable we may assume (by choosing a subcollection) that there is a positive e so that E> e for each ,B. Note that if f and g are any two maps of a space X into an AO so that the distance in 1X between f(x) and g(x) is less than E, for all x in X, thenf and g are homotopic. In fact F(x, t) -r#((1 -t)f(x) +tg(x)) is the desired homotopy, where we use the linear structure in I@'. For each ,B there is a 8> 0 so that r, restricted to the 8-neighborhood of AO moves no point as far as e/2. Again we may assume that there is a positive a so that 8! ?8 for each S. The hyperspace of all closed subsets of IX with the Hausdorff metric is separable (e.g. if D is any countable dense set in I@1, the collection of finite subsets of D is countable and dense in the hyperspace). Since every uncountable set, in a space satisfying the Second Axiom of Countability, contains a limit point, we see that for some : and y, AO is in the 8-neighborhood of A, and vice versa. Then rj IA and r,J A, are homotopy inverses of each other, since their compositions in either direction move points a distance less than E, hence, are homotopic to the corresponding identity maps. This shows A3 and A. have the same homotopy type, a contradiction.

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