arXiv:1701.00075v1 [math.GN] 31 Dec 2016 Extending Baire-one functions on compact spaces Olena Karlova, Volodymyr Mykhaylyuk Department of Mathematical Analysis, Faculty of Mathematics and Informatics, Yurii Fedkovych Chernivtsi National University, Kotsyubyns'koho str., 2, Chernivtsi, Ukraine Abstract We answer a question of O. Kalenda and J. Spurny´ from [8] and give an example of a completely regular hereditarily Baire space X and a Baire-one function f : X [0, 1] which can not be extended to a Baire-one function on X. Keywords: extension; Baire-one function; fragmented function; countably fragmented function 2000 MSC: Primary 54C20, 26A21; Secondary 54C30, 54C50 1. Introduction The classical Kuratowski's extension theorem [14, 35.VI] states that any map f : E Y of the first Borel class to a Polish space Y can be extended to a map g : X Y of the first Borel class if E is a G-subspace of a metrizable space X. Non-separable version of Kuratowski's theorem was proved by Hansell [4, Theorem 9], while abstract topological versions of Kuratowski's theorem were developed in [5, 8, 9]. Recall that a map f : X Y between topological spaces X and Y is said to be - Baire-one, f B1(X, Y ), if it is a pointwise limit of a sequence of continuous maps fn : X Y ; - functionally F-measurable or of the first functional Borel class, f K1(X, Y ), if the preimage f -1(V ) of any open set V Y is a union of a sequence of zero sets in X. Notice that every functionally F-measurable map belongs to the first Borel class for any X and Y ; the converse inclusion is true for perfectly normal X; moreover, for a topological space X and a metrizable separable connected and locally path-connected space Y we have the equality B1(X, Y ) = K1(X, Y ) (see [10]). Kalenda and Spurny´ proved the following result [8, Theorem 13]. Theorem A. Let E be a Lindel¨of hereditarily Baire subset of a completely regular space X and f : E R be a Baire-one function. Then there exists a Baire-one function g : X R such that g = f on E. The simple example shows that the assumption that E is hereditarily Baire cannot be omitted: if A and B are disjoint dense subsets of E = Q [0, 1] such that E = A B and X = [0, 1] or X = E, then the characteristic function f = A : E R can not be extended to a Baire-one function on X. In connection with this the following question was formulated in [8, Question 1]. Question 1. Let X be a hereditarily Baire completely regular space and f a Baire-one function on X. Can f be extended to a Baire-one function on X? We answer the question of Kalenda and Spurny´ in negative. We introduce a notion of functionally countably fragmented map (see definitions in Section 2) and prove that for a Baire-one function f : X R on a completely regular space X the following conditions are equivalent: (i) f is functionally countably fragmented; (ii) f can be extended to a Baire-one function on X. In Section 3 we give an example of a completely regular hereditarily Baire (even scattered) space X and a Baire-one function f : X [0, 1] which is not functionally countably fragmented and consequently can not be extended to a Baire-one function on X. 2. Extension of countably fragmented functions Let X be a topological space and (Y, d) be a metric space. A map f : X Y is called -fragmented for some > 0 if for every closed nonempty set F X there exists a nonempty relatively open set U F such that diamf (U ) < . If f is -fragmented for every > 0, then it is called fragmented. Let U = (U : [0, ]) be a transfinite sequence of subsets of a topological space X. Following [6], we define U to be regular in X, if Email addresses: maslenizza.ua@gmail.com (Olena Karlova), vmykhalyuk@ukr.net (Volodymyr Mykhaylyuk) (a) each U is open in X; (b) = U0 U1 U2 · · · U = X; (c) U = < U for every limit ordinal [0, ). Proposition 1. Let X be a topological space, (Y, d) be a metric space and > 0. For a map f : X Y the following conditions are equivalent: (1) f is -fragmented; (2) there exists a regular sequence U = (U : [0, ]) in X such that diamf (U+1 \ U) < for all [0, ). Proof. (1)(2) is proved in [12, Proposition 3.1]. (2)(1). We fix a nonempty closed set F X. Denote = min{ [0, ] : F U = }. Property (c) implies that = + 1 for some < . Then the set U = U F is open in F and diamf (U ) diamf (U+1 \ U) < . If a sequence U satisfies condition (2) of Proposition 1, then it is called -associated with f and is denoted by U(f ). We say that an -fragmented map f : X Y is functionally -fragmented if U(f ) can be chosen such that every set U is functionally open in X. Further, f is functionally -countably fragmented if U(f ) can be chosen to be countable and f is functionally countably fragmented if f is functionally -countably fragmented for all > 0. Evident connections between kinds of fragmentability and its analogs are gathered in the following diagram. functional countable fragmentability functional fragmentability fragmentability countable fragmentability Baire-one continuity functional F-measurability Notice that none of the inverse implications is true. Remark 1. (a) If X is hereditarily Baire, then every Baire-one map f : X (Y, d) is barely continuous (i.e., for every nonempty closed set F X the restriction f |F has a point of continuity) and, hence, is fragmented (see [14, 31.X]). (b) If X is a paracompact space in which every closed set is G, then every fragmented map f : X (Y, d) is Baire-one in the case either dimX = 0, or Y is a metric contractible locally path-connected space [11, 12]. (c) Let X = R be endowed with the topology generated by the discrete metric d(x, y) = 1 if x = y, and d(x, y) = 0 if x = y. Then the identical map f : X X is continuous, but is not countably fragmented. For a deeper discussion of properties and applications of fragmented maps and their analogs we refer the reader to [1, 2, 7, 13, 15]. Proposition 2. Let X be a topological space, (Y, d) be a metric space, > 0 and f : X Y be a map. If one of the following conditions hold (1) Y is separable and f is continuous, (2) X is metrizable separable and f is fragmented, (3) X is compact and f B1(X, Y ), then f is functionally countably fragmented. Proof. Fix > 0. (1) Choose a covering (Bn : n N) of Y by open balls of diameters < . Let U0 = , Un = f -1( kn Bk) for every n N and U0 = n=0 Un. Then the sequence (U : [0, 0]) is -associated with f. (2) Notice that any strictly increasing well-ordered chain of open sets in X is at most countable and every open set in X is functionally open. (3) By [12, Proposition 7.1] there exist a metrizable compact space Z, a continuous function : X Z and a function g B1(Z, Y ) such that f = g . Then g is functionally -countably fragmented by condition (2) of the theorem. It is easy to see that f is functionally -countably fragmented too. Lemma 3. Let X be a topological space, E X and f B1(E, R). If there exists a sequence of functions fn B1(X, R) such that (fn) n=1 converges uniformly to f on E, then f can be extended to a function g B1(X, R). Proof. Without loss of generality we may assume that f0(x) = 0 for all x E and |fn(x) - fn-1(x)| 1 2n-1 for all n N and x E. Now we put gn(x) = max{min{(fn(x) - fn-1(x)), 2-n+1}, -2-n+1} and notice that gn B1(X, R). Moreover, the series n=1 gn (x) is uniformly convergent on X for a function g B1(X, R). Then g is the required extension of f . Recall that a subspace E of a topological space X is z-embedded in X if for any zero set F in E there exists a zero set H in X such that H E = F ; C-embedded in X if any bounded continuous function f on E can be extended to a continuous function on X. Proposition 4. Let E be a z-embedded subspace of a completely regular space X and f : E R be a functionally countably fragmented function. Then f can be extended to a functionally countably fragmented function g B1(X, R). Proof. Let us observe that we may assume the space X to be compact. Indeed, E is z-embedded in X, since X is C-embedded in X [3, Theorem 3.6.1], and if we can extend f to a functionally countably fragmented function h B1(X, R), then the restriction g = h|E is a functionally countably fragmented extension of f on X and g B1(X, R). Fix n N and consider 1 n -associated with f sequence U = (U : ). Without loss of the generality we can assume that all sets U+1 \ U are nonempty. Since E is z-embedded in X, one can choose a countable family V = (V : ) of functionally open sets in X such that V V for all , V E = U for every and V = < V for every limit ordinal . For every [0, ) we take an arbitrary point y f (U+1 \ U). Now for every x X we put fn(x) = y, x V+1 \ V, y0, x X \ V. Observe that fn : X R is functionally F-measurable, since the preimage f -1(W ) of any open set W R is an at most countable union of functionally F-sets from the system {V+1 \ V : [0, )} {X \ V}. Therefore, fn B1(X, R). It is easy to see that the sequence (fn) n=1 is uniformly convergent to f on E. Now it follows from Lemma 3 that f can be extended to a function g B1(X, R). According to Proposition 2 (3), g is functionally countably fragmented. Corollary 5. Every functionally countably fragmented function f : X R defined on a topological space X belongs to the first Baire class. Proof. For every n N we choose a 1 n -associated with f family Un = (Un, : n) of functionally open sets Un, and corresponding family (n, : n) of continuous functions n, : X [0, 1] such that Un, = -n,1((0, 1]). We consider the at most countable set = n=1{n, : 0 n} and the continuous mapping : X [0, 1], (x) = ((x)). Show that f (x) = f (y) for every x, y X with (x) = (y). Let x, y X with (x) = (y). For every n N we choose n n such that x Un,n+1 \ Un,n . Then y Un,n+1 \ Un,n and |f (x) - f (y)| diam(Un,n+1 \ Un,n ) 1 n for every n N. Thus, f (x) = f (y). Now we consider the function g : (X) R, g((x)) = f (x). Clearly, that every set (Un,) is open in the metrizable space (X ). Therefore, for every nN the family ((Un,) : n) is 1 n -associated with g. Thus, g is functionally countably fragmented. According to Proposition 4, g B1((X), R). Therefore, f B1(X, R). Combining Propositions 2 and 4 we obtain the following result. Theorem 6. Let X be a completely regular space. For a Baire-one function f : X R the following conditions are equivalent: (1) f is functionally countably fragmented; (2) f can be extended to a Baire-one function on X. 3. A Baire-one bounded function which is not countably fragmented Theorem 7. There exists a completely regular scattered (and hence hereditarily Baire) space X and a Baire-one function f : X [0, 1] which can not be extended to a Baire-one function on X. Proof. Claim 1. Construction of X. Let Q = {rn : n N}, rn = rm for all distinct n, m N and {r2n-1 : n N} = {r2n : n N} = Q, A = {r2n-1} × [0, 1], B = {r2n} × [0, 1]. n=1 n=1 We consider partitions A = (At : t [0, 1]) and B = (Bt : t [0, 1]) of the sets A and B into everywhere dense sets At and Bt, respectively, such that |At| = |Bt| = c. Moreover, let [0, 1] = <1 T with |T| = c for every < 1. For every [0, 1) we put Q = tT At, is even, tT Bt, is odd, Q= Q, X = Q × {} and X = X. <1 <1 Claim 2. Indexing of X. For every [0, 1) we consider the set I = {(i)[,1) : |{ : i = 0}| 0} [0, 1][,1) and notice that |I| = c. Let : I T be a bijection and X = X,j , jI+1 where X,j = A+1(j) × {}, B+1(j) × {}, For all , [0, 1) with > and i I we put is even, is odd, Ji, = {j I : j|[,1) = i}. In particular, if = , = + 1 and i I+1, then we denote the set Ji +1, simply by Ji . Notice that |Ji | = c and we may assume that X,i = {xj : j Ji }. Then X = {xi : i I}, since I = iI+1 Ji . Claim 3. Topologization of X. For all [1, 1), i I and x = xi X we put L