Generalization of sequences and convergence in metric spaces

Topology and its Applications 171 (2014) 63–70

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Topology and its Applications www.elsevier.com/locate/topol

Generalization of sequences and convergence in metric spaces Alexey Ostrovsky

a r t i c l e

i n f o

Article history: Received 26 July 2013 Received in revised form 19 April 2014 Accepted 22 April 2014 Available online 6 May 2014

a b s t r a c t Adding to the previous results by the author and using some generalization of sequences, we study a special case of countable decomposability of functions: representation of functions as open, closed and continuous ones with the possible exception of countably many points. © 2014 Elsevier B.V. All rights reserved.

MSC: primary 26A21, 54C10 secondary 54C08, 28A20 Keywords: Sequence Open Closed Continuous Decomposition of functions

1. Introduction As we have shown in [1–3], the functions that take open or closed sets into a combination of n open or closed sets for n = 2 are decomposable into countably many closed, open and continuous functions. In this paper we will be focusing on an important special case of countable decomposability of functions: representation of functions as open, closed or continuous ones with the possible exception of countably many points. We will use countable compact sets Sn (y) and their subsets as a generalization of sequences in metrizable spaces. 1.1. Countable compact sets Sn (y) We will denote by Sn (y) a standard countable compact subset of a space Y such that its n-th derived set (Sn (y))n is a singleton of y. We denote by In (y) the set of isolated points of Sn (y). E-mail address: [email protected]. URL: https://www.researchgate.net/profile/Alexey_Ostrovsky/. http://dx.doi.org/10.1016/j.topol.2014.04.012 0166-8641/© 2014 Elsevier B.V. All rights reserved.

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In particular, I1 (y) = {yk }∞ k=1 such that yk → y as k → ∞; and S1 (y) = {y} ∪ {yk } is a standard sequence with its limit point, and I2 (y) = {yk,n }∞ n=1 such that yk,n → yk as k → ∞, and S2 (y) = {y} ∪ {yk } ∪ {yk,n } 1.2. Binary classes An and Mn Starting from classes A1 and M1 of open sets and closed sets lying in a metric space, we define the following additive and multiplicative classes: An = {A ∪ B : A ∈ Mn−1 , B ∈ An−1 } and Mn = {A ∩ B : A ∈ Mn−1 , B ∈ An−1 } By definition, the elements of classes An and Mn , called An -sets and Mn -sets, form a class Bn . Obviously, ∅ ∈ Bn and Bn ⊂ Bn+1 . We say that f : X → Y is open-Bn (resp., closed-Bn ) iff the image of every open (resp., closed) subset is a Bn -set. We say that a function f : X → Y is Bn -measurable iff the preimage of every open subset is a Bn -set. In this case f −1 is a (multivalued) open-Bn function. Below we will consider only single-valued functions. Continuous (resp., open, closed) functions are A1 -measurable (resp., open-A1 , closed-M1 ). Their countable characteristics in terms of S1 (y) are well known. For example, a function f : X → Y is continuous iff one of the following equivalent conditions holds: • for each open set U ⊂ Y , its preimage f −1 (U ) is an A1 -set in X; • for each closed set F ⊂ Y , its preimage f −1 (F ) is an M1 -set in X; • for each S1 (x) ⊂ X, f (x) ∈ cl Y f (I1 (x)). 2. s2 -Functions We will consider a standard base B of a zero-dimensional metric space X: this is any base B consisting of clopen subsets such that, for any set F and n ∈ N+ , the family {V ∈ B : V ∩ F = ∅, diam(V ) > 1/n} is finite. Obviously, each subspace of the Cantor set C has a standard base. In this section, we generalize the notions of closed, open and continuous functions with the use of S2 (x)-sets. 2.1. s2 -Continuous functions A function f : X → Y is said to be sn -continuous if, for any Sn (x) ⊂ X, f (x) is a limit point of f (In (x)). It is obvious that a function is continuous iff it is s1 -continuous. Theorem 1. Let f : X → Y be an s2 -continuous function between X, Y ⊂ C. Then there exists a countable set Y0 ⊂ Y such that the restriction f |(X \ f −1 (Y0 )) is continuous.

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Proof. For any V ∈ B, each point x of the set T = cl X f −1 (V ) \ f −1 (V ) is isolated in T , and, hence, T is countable. Indeed, if we suppose the opposite, i.e., that x is not isolated, then there exists an S2 (x) = {x} ∪ {xk } ∪ {xk,n } ⊂ cl X f −1 (V ) such that x, xk ∈ T and xk,n ∈ f −1 (V ). By the definition of an s2 -continuous function, f (x) is a limit point for {f (xk,n )}. Since V is closed and f (xk,n ) ∈ V , we obtain f (x) ∈ V , which contradicts x ∈ / f −1 (V ). Since T is countable, using the countability of the base B, we find that the sets X0 =



cl X f −1 (V ) \ f −1 (V ) : V ∈ B



and Y0 = f (X0 ) are countable. Obviously, we can assume that B is closed under taking complements; then, for any V ∈ B, f −1 (V ) \ cl X f −1 (Y \ V ) is open and its trace on X \ f −1 (Y0 ) coincides with the trace of f −1 (V ) on X \ f −1 (Y0 ). 2 2.2. s2 -Open functions A function f : X → Y is said to be sn -open if, for any Sn (y) ⊂ Y , x ∈ f −1 (y), and, for an open ball U (x) centered at x, f (O(x)) intersects at least one point of In (y). It is obvious that a function is s1 -open iff it is open in the standard sense: the image of each open set is open. Theorem 2. Let f : X → Y be an s2 -open function between X, Y ⊂ C. Then there exists a countable set Y0 ⊂ Y such that the restriction f |(X \ f −1 (Y0 )) is open. Proof. For each V ∈ B, each point of the set   D = f (V ) ∩ cl Y Y \ f (V ) is isolated in D. Indeed, suppose, contrary to our claim, that the set D ⊂ f (V ) has a non-isolated (in D) point y; hence, there exist yk ∈ D such that yk → y. Since yk ∈ cl Y (Y \ f (V )), there exist yk,n ∈ Y \ f (V ) such that yk,n −→ yk as n → ∞. In other words, for S2 (y) = {y} ∪ {yk } ∪ {yk,n }, we have y, yk ∈ f (V ) and yk,n ∈ Y \ f (V ).

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Take a point x ∈ V ∩ f −1 (y). By the definition of an s2 -open function, the neighborhood V intersects some f −1 (yk,n ), which contradicts yk,n ∈ Y \ f (V ). Remark 1. f (V ) = (O1 ∩ F1 ) ∪ O2 , where each of O1 or O2 is open and F1 is closed in Y . Indeed, since the countable set D consists of isolated points pi in D, there exist disjoint neighborhoods  O(pi ) of points pi such that D = i O(pi ) ∩ cl Y (Y \ f (V )). Hence, D is the intersection of an open and a closed set in Y . Obviously, f (V ) = D ∪ Int(f (V )). 2 Returning to the proof of Theorem 2, notice that, by the definition of D, f (V ) is a union of a countable set D and an open in f (V ) subset B = f (V ) \ cl Y (Y \ f (V )). Denote by Y0 the union of all countable sets D for V ∈ B. By our construction, every set V is clopen in X; hence, the set V \ f −1 (Y0 ) is clopen in X \ f −1 (Y0 ), and f (V ) \ Y0 = B \ Y0 is open in Y \ Y0 . If U is an open subset of X \ f −1 (Y0 ), then, obviously, U \ f −1 (Y0 ) is a countable union of some sets V \ f −1 (Y0 ), and f (U \ f −1 (Y0 )) is open in Y \ Y0 . This completes the proof of our theorem. 2 The following corollary is a quite obvious consequence of the method of Theorem 2, so we present it without proof. Corollary 1. Let f : X → Y be an sn -open function. Then there exists a countable set Y0 ⊂ Y such that the restriction f |(X \ f −1 (Y0 )) is an open function onto Y \ Y0 . 2 2.3. s2 -Closed functions A function f : X → Y is said to be sn -closed iff, for every Sn (y) ⊂ Y , every set T ⊂ X for which f (T ) = In (y) has a limit point in f −1 (y). It is obvious that a function is s1 -closed iff it is closed in the standard sense: the image of each closed set is closed. Theorem 3. Let f : X → Y be an s2 -closed function with compact fibers between X, Y ⊂ C. Then there is a countable set Y0 ⊂ Y such that the restriction f |(X \ f −1 (Y0 )) is a closed function onto Y \ Y0 . Proof. First of all, we prove that, for any V ∈ B, each point of the set T = cl Y f (V ) \ f (V ) is isolated in T , and, hence, T is countable. Indeed, suppose, contrary to our claim, that the set T has a non-isolated (in T ) point y. Then there exist yk ∈ T such that yk → y ∈ T . Since yk ∈ T = cl Y f (V ) \ f (V ), there exist yk,n ∈ f (V ) \ T such that yk,n → yk , and, hence, there is an S2 (y) ⊂ cl Y f (V ) such that y, yk ∈ S2 (y) ∩ T

and yk,n ∈ S2 (y) ∩ f (V ) \ T.

Take the points xk,n ∈ V ∩ f −1 (yk,n ). Since f is s2 -closed, there is a limit point x ∈ f −1 (y) for {xk,n }. Since V is closed and xk,n ∈ V , we have y = f (x) ∈ f (V ), which contradicts y ∈ T = cl Y f (V ) \ f (V ). The union Y0 of all the sets T obtained for V ∈ B is countable. Obviously, the set V \ f −1 (Y0 ) is clopen in X \ f −1 (Y0 ).

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Since the set of limit points of f (V ) in Y is a subset of T , we find that f (V \ f −1 (Y0 )) is closed in Y \ Y0 . Let F be a closed subset of X \ f −1 (Y0 ). We will prove that its image is closed in Y \ Y0 . Indeed, we can suppose additionally that B is a ternary clopen base formed by finite clopen covers γn consisting of pairwise disjoint sets V such that diam V = 31n and each element of γn+1 lies in an element of γn .  In this case, the set Fn = {V : V ∈ γn , V ∩ F = ∅} \ f −1 (Y0 ) is clopen in X \ f −1 (Y0 ); hence, the intersection ∞ 

f (Fn ) ⊃ f (F )

n=1

is closed in Y \ Y0 . Since for every V ∈ γn , diam V =

1 3n ,

we can easily establish that ∞ 

Fn = F.

n=1

Since every f −1 (y) is compact and every Fn is closed in X \ f −1 (Y0 ), we obtain

f (F ) =

∞ 

f (Fn ).

n=1

Consequently, f (F ) is closed in Y \ Y0 , and Theorem 3 is proved. 2 3. Open-B2 and closed-B2 functions We say that f is open-An (resp., closed-An ) iff the image of every open (resp., closed) subset is an An -set. Open-Mn and closed-Mn functions are defined analogously. Obviously, a function is open (resp., closed) iff it is open-A1 (resp., closed-M1 ). The definition of open-M1 and closed-A1 functions makes it clear that if a continuous function f : X → Y is open-M1 or closed-A1 , then each point in Y is isolated. Hence, all continuous open-B1 functions are open and all continuous closed-B1 functions are closed. Now, we consider closed-B2 and open-B2 functions: Theorem 4. Every continuous closed-B2 (resp., open-B2 ) function f : X → Y between separable metrizable spaces is closed (resp., open) with the possible exception of countably many points of Y . Theorem 4 follows from Propositions 3.1–3.4 since each subspace Z of a separable metrizable space Y is dense-in-itself with the possible exception of countably many points. Proposition 3.1. Every continuous closed-A2 function X → Y onto a metrizable space Y without isolated points is closed. Proof. Indeed, suppose the contrary: f is not closed. Then there is an S1 (y) ⊂ Y and a set T of points xk ∈ f −1 (yk ) without limit point in f −1 (y); hence, y ∈ cl Y f (T ) \ f (T ). Since Y is without isolated points, there exists a compact set S2 (y) ⊃ S1 (y) with yk,n → yk . Since T is closed in X, its image f (T ) = {yk } is a union of a closed set F and an open (in Y ) set O. If O = ∅, then

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O contains a point yk,n ∈ / f (T ), which is impossible; hence, f (T ) = F is closed in Y , which contradicts y∈ / f (T ). 2 Proposition 3.2. Every continuous open-M2 function X → Y onto a metrizable space Y without isolated points is open. Proof. Suppose f is not open. Then there is an S1 (y) ⊂ Y such that, for some x ∈ f −1 (y) and some open ball O(x) centered at x, the intersection f −1 (yk ) ∩ O(x) is empty for every yk . Since Y is without isolated points, there exists an S2 (y) ⊃ S1 (y) with yk,n → yk . It is clear that the restriction f |f −1 (S2 (y)) is an open-M2 function onto S2 (y). Since I2 (y) is open in S2 (y), the set (O(x) ∪ f −1 (T )) ∩ f −1 (S2 (y)) is open in f −1 (S2 (y)), and, by assumption, its image {y} ∪ I2 (y) is an intersection of a closed (in S2 (y)) set F and an open (in S2 (y)) set O. Since F ⊃ I2 (y), F = S2 (y), and, hence, O = {y}∪I2 (y), which is impossible since each neighborhood of y contains some point yk . 2 Proposition 3.3. A continuous function f : X → Y between metrizable spaces is open-A2 iff X is a union of two disjoint sets X1 and X2 such that f |X1 : X1 → f (X1 ) is open and f (X2 ) is a discrete subspace that is closed in Y . Proof. The implication ⇐ is obvious; below, we will prove the inverse implication ⇒. Denote X1 = X \ f −1



 y : f is not open at some point of f −1 (y)

and X2 = X \ X1 To prove that f (X2 ) is a discrete subspace that is closed in Y , suppose the contrary: f (X2 ) is not discrete or not closed. Then there is an S1 (y) ⊂ Y with yk → y ∈ Y and yk ∈ f (X2 ). Take xk ∈ f −1 (yk ) ∩ X2 . Since xk ∈ / X1 , we can find an S2 (y) ⊃ S1 (y) and open pairwise disjoint neighborhoods O(xk ) that do not intersect f −1 (y) and f −1 (yk,n ). ∞ It is easy to see that the set S2 (y) ∩ f ( k=1 O(xk )) = {yk }∞ k=1 is not an A2 -set in S2 (y). This contradicts the assumption. 2 Proposition 3.4. A continuous function f : X → Y between metrizable spaces is closed-M2 iff X is a union of two disjoint sets X1 and X2 such that f |X1 : X1 → f (X1 ) is closed and f (X2 ) is a discrete subspace that is closed in Y . Proof. The implication ⇐ is obvious. Let us prove the inverse implication ⇒. 1. Suppose there is a point y ∈ Y with open base Uk (y) at y such that f is not s1 -closed on Uk (y) \ {y}. Then there is a sequence {yk }∞ k=1 such that yk → y as k → ∞, and, for each k ∈ N, two conditions must be fulfilled at the same time: (1) there exists a sequence {yk,n }∞ n=1 such that yk,n → yk as n → ∞; (2) there exist xk,n ∈ f −1 (yk,n ) without limit point in f −1 (yk ). Denote S2 (y) = {y} ∪ {yk } ∪ {yk,n }.

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Taking some tails of the sequences {yk,n }, we can suppose that S2 (y) is compact (see Step 1 in [3]). Now, we define a set X1 = f −1 (y) ∪ {xk,n }. It can be easily verified that this set is already closed in X. The set f (X1 ) = {y} ∪ {yk,n } is not the intersection of a closed set F and an open set O. Indeed, suppose the contrary. Since y ∈ O, some yk ∈ O, and, since yn,k ∈ F , also yk ∈ O ∩ F . However, yk ∈ / f (X1 ), which gives a contradiction. 2. Hence, for each point y ∈ Y , there is an open ball O(y) centered at y such that f is closed at each point of O(y) \ {y}. Using the paracompactness of metric spaces, for the cover {O(y) : y ∈ Y }, we consider its open refinement. We can find a closed, discrete (in Y ) subspace Q consisting of some centers of the balls, an open (in X) subspace X1 = f −1 (Y \ Q) and a discrete subspace X2 = f −1 (Q) satisfying the assertion of our proposition. 2 4. Countable characteristics of Bn -sets and Bn -measurable functions We will consider below the metrizable spaces X, Y . At the end of Section 1.2 we mentioned a countable characteristic of continuous functions on the basis of the well-known countable (namely, S1 (x)) characteristic of closed subsets X ⊂ Y : • X is closed in Y iff, for each S1 (x) ⊂ Y , S1 (x) ∩ X = {xk }, or • X is not closed in Y iff there is an S1 (x) ⊂ Y , S1 (x) ∩ X = {xk }. A natural question arises whether it would be possible to give a countable (namely, Sn (x)) characteristic for constructible, resolvable sets, in particular, Bn -sets and for corresponding Bn -measurable, open-Bn and closed-Bn functions? The simplest examples of this are given by Proposition 4.1 and Corollary 2: Proposition 4.1. For each X ⊂ Y , the following conditions are equivalent: (a) X is not an M2 -set in Y ; (b) there is an S2 (x) ⊂ Y such that S2 (x) ∩ X = {x} ∪ {xk,n }. Proof. It is clear that (b) ⇒ (a). To prove (a) ⇒ (b), note that if T = cl Y X \ X is closed in cl Y X, then obviously X is an M2 -set in Y as the difference of two closed sets; but this contradicts our assumption. Hence, T is not closed in cl Y X, and there are xk ∈ T , xk −→ x, xk,n −→ xk such that x, xk,n ∈ T \ X; hence, we have (a). 2 The following corollary gives a countable characteristic of M2 -measurable functions: Corollary 2. For a function f : X → Y , the following are equivalent: (1) for each open U ⊂ Y , the set f −1 (U ) is an M2 -subset of X; (2) for each S2 (x) ⊂ X such that f |S2 (x) is one-to-one, there is no open U ⊂ Y such that f −1 (U )∩S2 (x) = {x} ∪ {xk,n }.

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Proof. To prove the implication (2) ⇒ (1), suppose, contrary to (1), that there is an open U ⊂ Y such that the set f −1 (U ) is not an M2 -subset of X; then, by the above proposition, there is an S2 (x) ⊂ X such that S2 (x) ∩ f −1 (U ) = {x} ∪ {xk,n }. If, in the proof of this proposition, we consider T = cl Y f −1 (U ) \ f −1 (U ), then, obviously, the points f (xk ) and f (xk.n ) can be assumed to be pairwise distinct; hence, f |S2 (x) is one-to-one, but (2) does not hold. To prove (1) ⇒ (2), suppose that there is an S2 (x) ⊂ X and an open U ⊂ Y such that S2 (x) ∩ f −1 (U ) = {x} ∪ {xk,n }. According to (b) of the above proposition, this contradicts (1). 2 The proofs presented above contain results that may be interesting in and of themselves for further study. Here the following countable characteristic may be helpful. Proposition 4.2. If a continuous function f : X → Y takes each open set into the union of an M2 -set and an A1 -set, then there is no S3 (y) ⊂ Y and open O ⊂ X intersecting f −1 (y), such that (a) f (O) ∩ ((S3 (y))0 \ (S3 (y))1 ) = f (O) ∩ I3 (y) = ∅; (b) f (O) ∩ ((S3 (y))1 \ (S3 (y))2 ) is an infinite set; (c) f (O) ∩ ((S3 (y))2 \ (S3 (y))3 ) = ∅. Proof. Suppose the contrary. According to (a), f (O) does not contain isolated points from S3 (y) and hence does not contain open subsets of S3 (y); then f (O) ∩ (S3 (y))1 is a union of an M2 -set and a closed set. According to (c), we can suppose that the union of y ∈ f (O) with I2 (y) is an M2 -subset of S2 (y) = (S3 (y))1 , which is impossible. 2 References [1] A. Ostrovsky, Preservation of the Borel class under open-LC functions, Fundam. Math. 213 (2011) 191–195. [2] A. Ostrovsky, An alternative approach to the decomposition of functions, Topol. Appl. 159 (2012) 2004–2008. [3] A. Ostrovsky, Closed-constructible functions are piecewise closed, Topol. Appl. 160 (2013) 1675–1680.