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A Note on Completely Metrizable Spaces of Continuous Injections Volume 36, 2010 Pages 141{143 http://topology.auburn.edu/tp/ A Note on Completely Metrizable Spaces of Continuous Injections by Pratibha Garg Electronically published on February 19, 2010 Topology Proceedings Web: http://topology.auburn.edu/tp/ Mail: Topology Proceedings Department of Mathematics & Statistics Auburn University, Alabama 36849, USA E-mail: [email protected] ISSN: 0146-4124 COPYRIGHT ⃝c by Topology Proceedings. All rights reserved. http://topology.auburn.edu/tp/ TOPOLOGY PROCEEDINGS Volume 36 (2010) Pages 141-143 E-Published on February 19, 2010 A NOTE ON COMPLETELY METRIZABLE SPACES OF CONTINUOUS INJECTIONS PRATIBHA GARG Abstract. In this paper, we present a stronger version of R. A. McCoy's theorem given in Completely metrizable spaces of embeddings [Proc. Amer. Math. Soc. 84 (1982), no. 3, 437{442] on complete metrizability of the space I(X; Y ), of continuous injections from a Tychonoff space X to a complete metric space Y equipped with the compact-open topology. The space Ck(X; Y ) of all continuous functions from a Tychonoff space X to a complete metric space (Y; d) equipped with the compact- open topology is completely metrizable if X is a hemicompact kR- space. For each f 2 C(X; Y ), compact set A in X, and 휖 > 0, define hf; A; 휖i = fg 2 C(X; Y ): d(f(x); g(x)) < 휖 for all x 2 Ag: Then for each f 2 Ck(X; Y ), the collection fhf; A; 휖i : A is a com- pact set in X; 휖 > 0g forms a neighborhood base at f in Ck(X; Y ). Since complete metrizability is not a hereditary property, some nat- ural subspaces of Ck(X; Y ) may not be completely metrizable even if Ck(X; Y ) is completely metrizable. In the theorem given in [2], a sufficient condition has been given on X for the subspace I(X; Y ) of all injections in Ck(X; Y ) to be completely metrizable. This theorem says that if X is a hemicompact metric space and Y is a complete metric space, then I(X; Y ) is completely metrizable. In 2010 Mathematics Subject Classification. Primary 54C35; Secondary 54E50. Key words and phrases. completely metrizable space, function space, kR- space, submetrizable. ⃝c 2010 Topology Proceedings. 141 142 P. GARG this paper, we show that the condition of metrizability on X in R. A. McCoy's theorem can be removed. The idea of the proof lies in the fact that, if I(X; Y ) is non-empty, then X is submetrizable. A space X is said to be submetrizable if it has a weaker metrizable topology. If I(X; Y ) is nonempty, then there exists a continuous injection from X to the complete metric space Y . This existence shows that X is submetrizable. Also it can be easily shown that every submetrizable space is a kR-space. Theorem. Suppose X is a hemicompact space and Y is completely metrizable, then I(X; Y ) is completely metrizable. Proof: Without loss of generality, we may assume that I(X; Y ) 6= ;. So there exists a one-to-one continuous function f : X −! Y . So X is submetrizable. Consequently, X is a hemicompact kR-space. Hence, Ck(X; Y ) is completely metrizable. Since X is hemicompact, then there exists a sequence fAn : n 2 Ng of compact subsets of X such that X = [An and whenever A is compact in X, A ⊆ An for some n. Since each An is compact submetrizable, each An is separable. Hence, X is a countable union of separable spaces. Consequently, X is also separable. Let f 2 I(X; Y ). Then f : X −! f(X) is a continuous bijec- tion. Since X is separable, f(X) is separable. So X has a sepa- rable metrizable compression. Hence, Ck(X) is separable. (Here Ck(X) = Ck(X; R), where R is the real line with the usual topol- ogy.) Let D = f휙n : n 2 Ng be a dense subset of Ck(X). Then D separates the points of X. For each compact set A in X, 휙 2 Ck(X), and 휖 > 0, define [휙, A; 휖] = ff 2 C(X; Y ): j휙(x) − f(x)j < 휖 8 x 2 A; for some continuous function : Y −! Rg Claim 1. Each [휙, A; 휖] is open in Ck(X; Y ). Let f 2 [휙, A; 휖]. Then there exists 2 C(Y ) such that j휙(x) − f(x)j < 휖 for all x 2 A. Since A is compact and 휙 − f is continuous, there exists 휖1 > 0 such that j휙(x) − f(x)j < 휖1 < 휖 8 x 2 A. Choose 휖2 > 0 such that 휖1 + 휖2 < 휖. Since f(A) is compact, is uniformly continuous on f(A). So for 휖2 > 0, there exists 훿 > 0 such that for every y1; y2 2 f(A), j (y1) − (y2)j < 휖2 whenever d(y1; y2) < 훿. A NOTE 143 Now we show that hf; A; 훿i ⊆ [휙, A; 휖]. Let g 2 hf; A; 훿i, then d(f(x); g(x)) < 훿 for all x 2 A. So j f(x) − g(x)j < 휖2 for all x 2 A. Let x 2 A be arbitrary. Then j휙(x) − g(x)j = j휙(x) − f(x) + f(x) − g(x)j ≤ j휙(x) − f(x)j + j f(x) − g(x)j < 휖1 + 휖2 < 휖. So g 2 [휙, A; 휖]. 1 Claim 2. I(X; Y ) = \f[휙n;Am; p ]: m; n; p 2 Ng. 1 Suppose f 2 \f[휙n;Am; p ]: m; n; p 2 Ng. Let x; y 2 X such that x 6= y. Since D separates the points of X, there exists n 2 N such 2 that 휙n(x) 6= 휙n(y). Choose p 2 N such that p < j휙n(x) − 휙n(y)j. 1 Choose m 2 N such that fx; yg ⊆ Am. Since f 2 [휙n;Am; p ], there exists a continuous function : Y −! R such that j휙n(x) − 1 2 f(x)j < p 8 x 2 Am. If possible, let f(x) = f(y). Then p < j휙n(x) − 휙n(y)j = j휙n(x) − f(x) + f(y) − 휙n(y)j ≤ j휙n(x) − 1 1 2 f(x)j + j f(y) − 휙n(y)j < p + p = p . Consequently, f(x) 6= f(y). Hence, f 2 I(X; Y ). Conversely, let f 2 I(X; Y ). Then f : Am −! f(Am) is a −1 homeomorphism. Thus, 휙nf : f(Am) −! R is a continuous real- valued function on f(Am). So there exists a continuous extension −1 1 : Y −! R of 휙nf . Then j휙n(x) − f(x)j = 0 < p for all 1 x 2 Am. So f 2 [휙n;Am; p ] for all m; n; p 2 N. Consequently, 1 I(X; Y ) = \f[휙n;Am; p ]: m; n; p 2 Ng. Since I(X; Y ) is a G훿-subset of the completely metrizable space Ck(X; Y ), I(X; Y ) is completely metrizable. □ References [1] Richard F. Arens, A topology for spaces of transformations, Ann. of Math. (2) 47 (1946), 480{495. [2] R. A. McCoy, Completely metrizable spaces of embeddings, Proc. Amer. Math. Soc. 84 (1982), no. 3, 437{442. [3] R. A. McCoy and I. Ntantu, Completeness properties of function spaces, Topology Appl. 22 (1986), no. 2, 191{206. School of Mathematics & Computer Applications (SMCA); Thapar University; P.O. Box 32; Patiala, Pin 147004 India E-mail address: [email protected].
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