A Note on Completely Metrizable Spaces of Continuous Injections

Volume 36, 2010 Pages 141–143

http://topology.auburn.edu/tp/

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 퐼(푋, 푌 ), of continuous injections from a Tychonoﬀ space 푋 to a complete metric space 푌 equipped with the compact-open topology.

The space 퐶푘(푋, 푌 ) of all continuous functions from a Tychonoﬀ space 푋 to a complete metric space (푌, 푑) equipped with the compact- open topology is completely metrizable if 푋 is a hemicompact 푘ℝ- space. For each 푓 ∈ 퐶(푋, 푌 ), compact set 퐴 in 푋, and 휖 > 0, deﬁne ⟨푓, 퐴, 휖⟩ = {푔 ∈ 퐶(푋, 푌 ): 푑(푓(푥), 푔(푥)) < 휖 for all 푥 ∈ 퐴}.

Then for each 푓 ∈ 퐶푘(푋, 푌 ), the collection {⟨푓, 퐴, 휖⟩ : 퐴 is a compact set in 푋, 휖 > 0} forms a neighborhood base at 푓 in 퐶푘(푋, 푌 ). Since complete metrizability is not a hereditary property, some nat- ural subspaces of 퐶푘(푋, 푌 ) may not be completely metrizable even if 퐶푘(푋, 푌 ) is completely metrizable. In the theorem given in [2], a suﬃcient condition has been given on 푋 for the subspace 퐼(푋, 푌 ) of all injections in 퐶푘(푋, 푌 ) to be completely metrizable. This theorem says that if 푋 is a hemicompact metric space and 푌 is a complete metric space, then 퐼(푋, 푌 ) is completely metrizable. In

2010 Mathematics Subject Classiﬁcation. Primary 54C35; Secondary 54E50.

Key words and phrases. completely metrizable space, function space, 푘ℝ- space, submetrizable. ⃝c 2010 Topology Proceedings. 141 142 P. GARG this paper, we show that the condition of metrizability on 푋 in R. A. McCoy’s theorem can be removed. The idea of the proof lies in the fact that, if 퐼(푋, 푌 ) is non-empty, then 푋 is submetrizable. A space 푋 is said to be submetrizable if it has a weaker metrizable topology. If 퐼(푋, 푌 ) is nonempty, then there exists a continuous injection from 푋 to the complete metric space 푌 . This existence shows that 푋 is submetrizable. Also it can be easily shown that every submetrizable space is a 푘ℝ-space. Theorem. Suppose 푋 is a hemicompact space and 푌 is completely metrizable, then 퐼(푋, 푌 ) is completely metrizable.

Proof: Without loss of generality, we may assume that 퐼(푋, 푌 ) ∕= ∅. So there exists a one-to-one continuous function 푓 : 푋 −→ 푌 . So 푋 is submetrizable. Consequently, 푋 is a hemicompact 푘ℝ-space. Hence, 퐶푘(푋, 푌 ) is completely metrizable. Since 푋 is hemicompact, then there exists a sequence {퐴푛 : 푛 ∈ ℕ} of compact subsets of 푋 such that 푋 = ∪퐴푛 and whenever 퐴 is compact in 푋, 퐴 ⊆ 퐴푛 for some 푛. Since each 퐴푛 is compact submetrizable, each 퐴푛 is separable. Hence, 푋 is a countable union of separable spaces. Consequently, 푋 is also separable. Let 푓 ∈ 퐼(푋, 푌 ). Then 푓 : 푋 −→ 푓(푋) is a continuous bijec- tion. Since 푋 is separable, 푓(푋) is separable. So 푋 has a separable metrizable compression. Hence, 퐶푘(푋) is separable. (Here 퐶푘(푋) = 퐶푘(푋, ℝ), where ℝ is the real line with the usual topology.) Let 퐷 = {휙푛 : 푛 ∈ ℕ} be a dense subset of 퐶푘(푋). Then 퐷 separates the points of 푋. For each compact set 퐴 in 푋, 휙 ∈ 퐶푘(푋), and 휖 > 0, deﬁne [휙, 퐴, 휖] = {푓 ∈ 퐶(푋, 푌 ): ∣휙(푥) − 휓푓(푥)∣ < 휖 ∀ 푥 ∈ 퐴, for some continuous function 휓 : 푌 −→ ℝ}

Claim 1. Each [휙, 퐴, 휖] is open in 퐶푘(푋, 푌 ). Let 푓 ∈ [휙, 퐴, 휖]. Then there exists 휓 ∈ 퐶(푌 ) such that ∣휙(푥) − 휓푓(푥)∣ < 휖 for all 푥 ∈ 퐴. Since 퐴 is compact and 휙 − 휓푓 is continuous, there exists 휖1 > 0 such that ∣휙(푥) − 휓푓(푥)∣ < 휖1 < 휖 ∀ 푥 ∈ 퐴. Choose 휖2 > 0 such that 휖1 + 휖2 < 휖. Since 푓(퐴) is compact, 휓 is uniformly continuous on 푓(퐴). So for 휖2 > 0, there exists 훿 > 0 such that for every 푦1, 푦2 ∈ 푓(퐴), ∣휓(푦1) − 휓(푦2)∣ < 휖2 whenever 푑(푦1, 푦2) < 훿. A NOTE 143

Now we show that ⟨푓, 퐴, 훿⟩ ⊆ [휙, 퐴, 휖]. Let 푔 ∈ ⟨푓, 퐴, 훿⟩, then 푑(푓(푥), 푔(푥)) < 훿 for all 푥 ∈ 퐴. So ∣휓푓(푥) − 휓푔(푥)∣ < 휖2 for all 푥 ∈ 퐴. Let 푥 ∈ 퐴 be arbitrary. Then ∣휙(푥) − 휓푔(푥)∣ = ∣휙(푥) − 휓푓(푥) + 휓푓(푥) − 휓푔(푥)∣ ≤ ∣휙(푥) − 휓푓(푥)∣ + ∣휓푓(푥) − 휓푔(푥)∣ < 휖1 + 휖2 < 휖. So 푔 ∈ [휙, 퐴, 휖]. 1 Claim 2. 퐼(푋, 푌 ) = ∩{[휙푛, 퐴푚, 푝 ]: 푚, 푛, 푝 ∈ ℕ}. 1 Suppose 푓 ∈ ∩{[휙푛, 퐴푚, 푝 ]: 푚, 푛, 푝 ∈ ℕ}. Let 푥, 푦 ∈ 푋 such that 푥 ∕= 푦. Since 퐷 separates the points of 푋, there exists 푛 ∈ ℕ such 2 that 휙푛(푥) ∕= 휙푛(푦). Choose 푝 ∈ ℕ such that 푝 < ∣휙푛(푥) − 휙푛(푦)∣. 1 Choose 푚 ∈ ℕ such that {푥, 푦} ⊆ 퐴푚. Since 푓 ∈ [휙푛, 퐴푚, 푝 ], there exists a continuous function 휓 : 푌 −→ ℝ such that ∣휙푛(푥) − 1 2 휓푓(푥)∣ < 푝 ∀ 푥 ∈ 퐴푚. If possible, let 푓(푥) = 푓(푦). Then 푝 < ∣휙푛(푥) − 휙푛(푦)∣ = ∣휙푛(푥) − 휓푓(푥) + 휓푓(푦) − 휙푛(푦)∣ ≤ ∣휙푛(푥) − 1 1 2 휓푓(푥)∣ + ∣휓푓(푦) − 휙푛(푦)∣ < 푝 + 푝 = 푝 . Consequently, 푓(푥) ∕= 푓(푦). Hence, 푓 ∈ 퐼(푋, 푌 ). Conversely, let 푓 ∈ 퐼(푋, 푌 ). Then 푓 : 퐴푚 −→ 푓(퐴푚) is a −1 homeomorphism. Thus, 휙푛푓 : 푓(퐴푚) −→ ℝ is a continuous real- valued function on 푓(퐴푚). So there exists a continuous extension −1 1 휓 : 푌 −→ ℝ of 휙푛푓 . Then ∣휙푛(푥) − 휓푓(푥)∣ = 0 < 푝 for all 1 푥 ∈ 퐴푚. So 푓 ∈ [휙푛, 퐴푚, 푝 ] for all 푚, 푛, 푝 ∈ ℕ. Consequently, 1 퐼(푋, 푌 ) = ∩{[휙푛, 퐴푚, 푝 ]: 푚, 푛, 푝 ∈ ℕ}. Since 퐼(푋, 푌 ) is a 퐺훿-subset of the completely metrizable space 퐶푘(푋, 푌 ), 퐼(푋, 푌 ) 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.

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