closed set. The aim of this paper is to prove the following theorem: Let f : X â Y be a LCâcontinuous function onto a separable metric space Y. Then X can be ...
THE STRUCTURE OF LC–CONTINUOUS FUNCTIONS. ALEXEY OSTROVSKY
Abstract. A function f is LC–continuous if the inverse image of any open set is a locally closed set; i.e., an intersection of an open set and a closed set. The aim of this paper is to prove the following theorem: Let f : X → Y be a LC–continuous function onto a separable metric space Y. Then X can be covered by countably many subsets Tn ⊂ X such that each restriction f |Tn is continuous at all points of Tn . .
1. Introduction All spaces in this paper are supposed to be metrizable. Recall that a subset of a topological space is LC-set or locally closed set if it is the intersection of an open and a closed set. A clopen set is a set which is both open and closed. We will denote by S1 (y) a sequence with its limit point: S1 (y) = {y} ∪ {yi : yi −→ y} Given an arbitrary function f we say that f is -LC-continuous if the inverse image of any open set is an intersection of an open set and a closed set [1]. -countable continuous if X can be partitioned into countably many pairwise disjoint sets Xi such that every restriction f |Xi is continuous [3]. Theorem 1 states that LC-continuous functions are countable continuous. In contrast to all previous works about decomposition of Baire 1 functions, our spaces are not supposed to be Polish or abs. Souslin.
2. The structure of LC–continuous functions in metric spaces Theorem 1. Let f : X → Y be a LC–continuous function onto a separable metric space Y. Then X can be covered by countably many subsets Tn such that each restriction f |Tn is a continuous function. Moreover T0 is Gδ -set and each Tn (n = 1, 2, ...) is a closed subsets of X. Proof. For every n = 1, 2, ... denote 2000 Mathematics Subject Classification. Primary 26A21, 54C50, 54C08; Secondary 26A15. Key words and phrases. LC–continuous, countably continuous, Baire 1 functions, countable homeomorphism. 1
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ALEXEY OSTROVSKY
(*) Xn = {x ∈ X : ∃xi ∈ X such that xi −→ x and dist(f (x), f (xi )) > 1/n}. Lemma 1. Each restriction f |Xn is a continuous function (n = 1, 2, ...). Indeed, suppose, contrary to our claim, that there is x and xj −→ x, where x, xj ∈ Xn , such that 1) f (xj ) 6−→ f (x). According to (*), every xj is a limit point for some sequence {xji } such that 2) dist(f (xj ), f (xji )) > 1/n. Since Y is separable, there is a compact K ⊃ Y. Let us define by: 3) yj∗ ∈ K an accumulation point for f (xji ) (i = 1, 2, ...) : 4) y ∗ ∈ K an accumulation point for yj∗ ; 5) z ∗ an accumulation point for f (xj ). According to 3) and 4) we can suppose: a) f (xji ) −→ yj∗ and yj∗ −→ y ∗ ; It follows from 2) and 5): b) dist(y ∗ , z ∗ ) > 1/2n; It follows from 1) and 5): c) for some d > 0, dist(z ∗ , f (x)) > d. Let us take an open (in K) δ-ball Oδ (y ∗ ) centered at y ∗ , where δ = min{d/2, 1/2n} and take analogously Oδ (f (x)). Throughout the proof, D denotes the open in Y set (Oδ (f (x))∪Oδ (y ∗ ))∩Y. Without loss of generality, we will assume that the union of sets {xji }, {xj } and {x} is a compact S2 (x). It follows immediately from b), c) and definition D that P = f −1 (D) ∩ S2 (x) is a union of sets {x} and {xji }. This contradicts our assumption that f −1 (D) is a locally closed set in X. Indeed, in this case the set P would be a locally closed set in S2 (x). An easy verification shows that it is impossible. The proof, that restriction f |X0 is a continuous function is based S on the the following observation. According to (*), for every x ∈ X \ n Xn , the condition xi −→ x implies that dist(f S (x), f (xi )) = 0, and hence f is continuous at every point of X0 = X \ n>0 Xn . It follows, that f |X0 is continuous. Lemma 2. Every set Xn (n = 1, 2, ....) in Lemma 1 can be considered as closed set. The basic idea of the proof is to consider the set Tn = clX Xn . If x ∈ Tn and xj −→ x, where xj ∈ Xn , we can now proceed analogously to the proof of Lemma 1. It remains to consider the case: x ∈ Tn and xj −→ x, where xj ∈ Tn \ Xn . It is easily seen that by obtaining a contradiction in the proof of previous lemma, we do not use the points xj −→ x in the definition of P. Essential
THE STRUCTURE OF LC–CONTINUOUS FUNCTIONS.
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to the proof is the fact that according to (*), every xj is a limit point for some sequence {xji } such that dist(f (xj ), f (xji )) > 1/n. Corollary 1. Let f : X → Y be a LC-continuous function. Then f is countable continuous. Indeed, let us define in Theorem 1 by H1 = T1 , H2 = T2 \ TS 1 , Hn = S ∞ Tn \ k
