A characterization of the uniform convergence points set of some convergent sequence of functions
aa r X i v : . [ m a t h . GN ] A ug A CHARACTERIZATION OF THE UNIFORM CONVERGENCE POINTS SETOF SOME CONVERGENT SEQUENCE OF FUNCTIONS
OLENA KARLOVA , Abstract.
We characterize the uniform convergence points set of a pointwisely convergent sequence of real-valuedfunctions defined on a perfectly normal space. We prove that if X is a perfectly normal space which can be coveredby a disjoint sequence of dense subsets and A ⊆ X , then A is the set of points of the uniform convergence for someconvergent sequence ( f n ) n ∈ ω of functions f n : X → R if and only if A is G δ -set which contains all isolated points of X . This result generalizes a theorem of J´an Bors´ık published in 2019. Dedicated to the memory of J´an Bors´ık Introduction
Let X be a topological space, ( Y, | · − · | ) be a metric space; B ( a, r ) and B [ a, r ] be an open and aclosed ball in Y with a center a ∈ Y and a radius r >
0, respectively. By ∂A we denote a boundaryof a set A .Let F = ( f n ) n ∈ ω be a sequence of functions f n : X → Y . We denote P C ( F ) the set of allpoints x ∈ X such that the sequence ( f n ( x )) n ∈ ω is convergent in Y . Therefore, we define the limitfunction f ( x ) by the rule f ( x ) = lim n →∞ f n ( x ) for all x ∈ P C ( F ). Let us observe that the set P C ( F ) can be represented in the form P C ( F ) = \ k ∈ ω [ n ∈ ω \ m ∈ ω f − n + m ( B [ f ( x ) , k +1 ]) . (1.1)If every function f n is continuous, then the set P C ( F ) is F σδ in X . Hans Hahn [6] and Wac lawSierpi´nski [11] proved independently that the converse proposition is true for metrizable X and Y = R , that is, for every F σδ -subset A of a metrizable space X there exists a sequence F ofreal-valued continuous functions f n : X → R such that A = P C ( F ).After appearance of this theorem many results were obtained in similar directions: other typesof convergence and other classes of functions were considered (see, for instance, [1, 4, 7, 8, 9, 10, 12,13, 14]). J´an Bors´ık studied in [1], in particular, the uniform convergence points set of a (convergentpointwisely) sequence of functions. Definition 1.
A sequence ( f n ) n ∈ ω of functions f n : X → Y between a topological space X and ametric space ( Y, | · − · | ) is uniformly Cauchy at a point x ∈ X , if for every ε > U of x and a number n ∈ ω such that | f n ( x ) − f m ( x ) | < ε for all n, m ≥ n and x ∈ U .Let U C ( F ) be a set of all points with the uniform Cauchy property for a sequence F = ( f n ) n ∈ ω .It is easy to see that if ( f n ) n ∈ ω is convergent pointwisely on X to a function f : X → Y , then U C ( F ) = { x ∈ X : ∀ ε > ∃ U ∋ x ∃ n ∀ n ≥ n | f n ( y ) − f ( y ) | < ε ∀ y ∈ U } . (1.2)Moreover, in this case U C ( F ) is the set of all points of the uniform convergence of F .Bors´ık proved the following result. Mathematics Subject Classification.
Primary 54C30, 26A21; Secondary 54C50.
Key words and phrases. set of points of uniform convergence; uniformly Cauchy sequence. , Theorem A (Bors´ık, [1]) . Let X be a metric space and A ⊆ X . Then A = U C ( F ) for someconvergent sequence F = ( f n ) n ∈ ω of functions f n : X → R if and only if A is G δ and contains allisolated points of X . This short note is inspirited by the above mentioned paper of J´an Bors´ık. We generalize histheorem on a wider class of topological spaces.
Definition 2.
A topological space X is ω -resolvable if there exists a partition { X n : n ∈ ω } of X by dense subsets.For crowded spaces (i.e., spaces without isolated points) the class of all ω -resolvable spaces in-cludes all metrizable spaces, Hausdorff countably compact spaces, arcwise connected spaces, etc. [2]The main result of our note is the following theorem. Theorem 1.
Let X be a perfectly normal ω -resolvable space and A ⊆ X . The following conditionsare equivalent:(i) A is a set of all points of uniform convergence for some convergent sequence ( f n ) n ∈ ω offunctions f n : X → R ;(ii) A is a G δ -set which contains all isolated points of X . Proof of Theorem 1
The implication (i) ⇒ (ii) follows immediately from the equality (1.2). (ii) ⇒ (i) . Let ( G n ) n ∈ ω be a sequence of open sets in X such that A = T n ∈ ω G n , G n +1 ⊆ G n for every n ∈ ω and let G = X .Since X is perfectly normal, for every n ∈ N there exist continuous functions ϕ n , ψ n : X → [0 , ϕ − n (0) = G n and ψ − n (0) = X \ G n . Then every function α n : X → [ − ,
1] defined by theformula α n = ϕ n − ψ n has the following properties: α n ( x ) < ∀ x ∈ G n ,α n ( x ) = 0 ∀ x ∈ ∂G n ,α n ( x ) > ∀ x ∈ X \ G n . We consider functions β n : X → [ − ,
1] defined by the rule β n ( x ) = max i ≤ n α i ( x )for all x ∈ X . Then we claim that ∂G n = β − n (0)for every n ∈ N . We need to prove β n ( x ) = 0 ⇔ α n ( x ) = 0 for every x ∈ X . Assume that β n ( x ) = 0. Then α i ( x ) ≤ i ≤ n and α k ( x ) = 0 for some k ≤ n . Then x ∈ G i for all i ≤ n . If x ∂G n , then x ∈ G n ⊆ G i for all i ≤ n , consequently, α k ( x ) <
0, a contradiction. Hence, x ∈ ∂G n and α n ( x ) = 0. Conversely, if α n ( x ) = 0, then x ∈ ∂G n ⊆ G n ⊆ G i for all i ≤ n . In consequence, α i ( x ) ≤ i < n and β n ( x ) = 0.Now we put γ n ( x ) = min i ≤ n | β i ( x ) | and notice that γ − n (0) = [ i ≤ n ∂G i . CHARACTERIZATION OF THE UNIFORM CONVERGENCE POINTS SET 3
Finally, let δ n ( x ) = (cid:26) , x ∈ G n ,γ n ( x ) , x ∈ X \ G n . Obviously, the functions β n , γ n and δ n are continuous and δ n ≤ γ n .We put U k, = F k, = ∅ , k ∈ N . For all k, n ∈ N we define U k,n = γ − n ([0 , k )) and F k,n = δ − n ([ k , . The sets U k,n and F k,n satisfy the following conditions:(A) U k,n is open and F k,n is closed in X ,(B) U k +1 ,n ⊆ U k,n ⊆ U k,n +1 ,(C) F k,n ⊆ F k +1 ,n ∩ F k,n +1 ,(D) δ − n ((0 , S k ∈ N F k,n = X \ (cid:0)S ni =1 ∂G i ∪ G n (cid:1) ,(E) γ − n (0) = S ni =1 ∂G i = T k ∈ N U k,n ,(F) U k,n ∩ F k,n = ∅ for all k, n ∈ N .Moreover, the sets G n satisfy the property(G) ( ∂G n \ ∂G n − ) ∩ ∂G i = ∅ , i < n , n ∈ N .Since the most of properties are evident, we prove only ( C ) and ( G ).( C ). It is enough to prove that F k,n ⊆ F k,n +1 . Fix x ∈ F k,n for some k, n ∈ N . Then δ n ( x ) ≥ k and, in consequence, x G n . Therefore, α n ( x ) > α n +1 ( x ) >
0. Hence, β n ( x ) > β n +1 ( x ) >
0. Since γ n ( x ) = δ n ( x ) ≥ k , the inequality | β i ( x ) | ≥ k holds for all i ≤ n . In particular, | β n ( x ) | = β n ( x ) ≥ k . Then | β n +1 ( x ) | = β n +1 ( x ) ≥ β n ( x ) ≥ k . Thus, δ n +1 ( x ) = γ n +1 ( x ) =min i ≤ n +1 | β i ( x ) | ≥ k and x ∈ F k,n +1 .( G ). Fix x ∈ ∂G n \ ∂G n − . Since x ∈ ∂G n , x ∈ G n \ G n . Then x ∈ G n − , because thesequence ( G n ) n ∈ ω is decreasing. Moreover, x ∂G n − and therefore x ∈ G n − . Again, since( G n ) n ∈ ω decreases, x ∈ G i for all i < n . Hence, x ∂G i , i < n .Since X \ A is an open subset of the ω -resolvable space X , it is ω -resolvable also. Hence, thereexists a sequence ( B k ) k ∈ ω of mutually disjoint subsets of X \ A such that X \ A = [ k ∈ ω B k and each set B k in dense in X \ A .For every x ∈ X we put f ( x ) = (cid:26) n , x ∈ ∂G n \ ∂G n − for some n ∈ N , , otherwise . (2.1)Notice that f is correctly defined because of property (G).Now let C k,n = ( U k,n \ U k,n − ) ∪ ( B k ∩ ( F k,n \ F k,n − )) OLENA KARLOVA , and f k ( x ) = (cid:26) n , if x ∈ C k,n for some n ≥ k, , otherwise . In order to show that ( f k ) k ∈ ω converges to f pointwisely on X we fix x ∈ X .If x ∈ S n ∈ ω ∂G n , then we put N = min { n ∈ ω : x ∈ ∂G n } . Therefore, since ∂G = ∅ , property(G) implies that N ∈ N and x ∈ ∂G N \ S i
0. So, there exists m ∈ N such that m < γ n +1 ( x ). But x G n +1 .Therefore, δ n +1 ( x ) = γ n +1 ( x ) > m . Hence, x ∈ int F m ,n +1 .By property (C), there exists a number k > max { k , m , n } such that x ∈ int F k,n +1 . Since x ∈ G n , x F k,n . Then the set G = ( U \ G n +1 ) ∩ (int F k,n +1 \ F k,n )is an open neighborhood of x . Since S n ≤ n +1 ∂G n is nowhere dense in X and B k is dense in X \ A ,there exists a point v ∈ X such that v ∈ ( G \ [ n ≤ n +1 ∂G n ) ∩ B k . Then f k ( v ) = n +1 , since v ∈ C k,n +1 , and f ( v ) ∈ [0 , n +2 ], because v S n ≤ n +1 ∂G n . Hence, | f ( v ) − f k ( v ) | ≥ n +1 − n +2 = ε. Therefore, A = U C ( F ). ✷ Remark 1.
Actually, we use in the proof only the fact that the boundary of every open set ina topological space X is a functionally closed set. It is find out that this is a characterization ofperfectly normal spaces. Moreover, the following conditions are equivalent:(i) X is a perfectly normal space;(ii) every closed nowhere dense subset of X is functionally closed.Evidently, (i) ⇒ (ii). In order to prove (ii) ⇒ (i) we take a closed set F ⊆ X . Since ∂F = F \ int F isclosed and nowhere dense, there exists a continuous function g : X → [0 ,
1] such that ∂F = g − (0).Let us define f : X → [0 ,
1] by f ( x ) = g ( x ) if x ∈ X \ int F and f ( x ) = 0 if x ∈ int F . It is easy tosee that f is continuous and F = f − (0). Therefore, X is perfectly normal by Vedenisoff’s theorem. Remark 2.
By one of reviewers, in Theorem 1 it is sufficient to assume that Y is a non-discretemetric space. Remark 3.
Any topological vector space is ω -resolvable [2]. So, the space of all continuous function C p ([0 , ω -resolvable space which is not metrizable. Remark 4.
Eric K. van Douwen proved in [3, Theorem 5.2] that there exists a crowded countableregular space which cannot be represented as a union of two disjoint dense subsets. It is easy to seethat this space is perfectly normal and not ω -resolvable.3. Acknowledgement
I am very grateful to the reviewers for their careful reading of the manuscript and valuableremarks which allowed to improve the paper.
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