By Kuczma M.E. (ed.)

ISBN-10: 0964095904

ISBN-13: 9780964095908

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**Additional resources for 144 problems of the Austrian-Polish Mathematics Competition, 1978-1993**

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Amer. Math. Soc. 71 (1951), p . 161. , p . 164. (3) Theorem of A. Pelczynski. See A remark on spaces 2X for zero-dimensional 3C, Bull. Polish Acad. Sei. 13 (1965), p . 85. (4) This problem was raised by V. Ponomarev. See A new space of closed subsets and multivalued mappings of bicompact spaces (Russian), Matem. Sb. 48 (1959), p . 195. § 42, il] THE SPACE 47 2X E e m a r k 3. Let us denote, for an arbitrary topological space X, by V(X) the set of all compact subsets of X. Since

E. that C = ~X. By Theorem 4, C—X is closed. Hence there exist, by Theorem 4 of Section III, a compact e^Yspace X*, a point peX*, and a continuous / : C -> X* such that f(C—X) = (p) and / is a homeomorphism of X onto X* — (p). This completes the proof of the first part of the theorem. If X is metric separable, then by the Urysohn Theorem, G can be assumed to be a closed subset of the Hubert cube. Since X* is a continuous image of the compact metric space (7, X* is metrizable by Theorem 3 of Section VI.

Tn the system of distinguished indices. It is easy to see that two disjoint members of the base must have a common distinguished index and different projections on the corresponding axis. For proving the theorem it suffices to show that, for a given w, every family of disjoint sets of length n is countable. We shall proceed by induction. For n■ = 1, our statement follows from the property just mentioned of disjoin members of the base. Suppose this is true for n — 1. Let G be the family of sets of length n.

### 144 problems of the Austrian-Polish Mathematics Competition, 1978-1993 by Kuczma M.E. (ed.)

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