Generalization of sequences and convergence in metric spaces

Generalization of sequences and convergence in metric spaces

Topology and its Applications 171 (2014) 63–70 Contents lists available at ScienceDirect Topology and its Applications www.elsevier.com/locate/topol...

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Topology and its Applications 171 (2014) 63–70

Contents lists available at ScienceDirect

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.