aa r X i v : . [ m a t h . GN ] M a r Ascoli and sequentially Ascoli spaces
S. Gabriyelyan
Department of Mathematics, Ben-Gurion University of the Negev, Beer-Sheva, P.O. 653, Israel
Abstract
A Tychonoff space X is called ( sequentially ) Ascoli if every compact subset (resp. convergentsequence) of C k ( X ) is evenly continuous, where C k ( X ) denotes the space of all real-valued con-tinuous functions on X endowed with the compact-open topology. Various properties of (sequen-tially) Ascoli spaces are studied, and we give several characterizations of sequentially Ascoli spaces.Strengthening a result of Arhangel’skii we show that a hereditary Ascoli space is Fr´echet–Urysohn.A locally compact abelian group G with the Bohr topology is sequentially Ascoli iff G is compact.If X is totally countably compact or near sequentially compact then it is a sequentially Ascolispace. The product of a locally compact space and an Ascoli space is Ascoli. If additionally X is a µ -space, then X is locally compact iff the product of X with any Ascoli space is an Ascoli space.Extending one of the main results of [18] and [16] we show that C p ( X ) is sequentially Ascoli iff X has the property ( κ ). We give a necessary condition on X for which the space C k ( X ) is sequentiallyAscoli. For every metrizable abelian group Y , Y -Tychonoff space X , and nonzero countable ordinal α , the space B α ( X, Y ) of Baire- α functions from X to Y is κ -Fr´echet–Urysohn and hence Ascoli. Keywords: C p ( X ), C k ( X ), Baire- α function, Ascoli, sequentially Ascoli, κ -Fr´echet–Urysohn, P -space
1. Introduction
Various topological properties generalizing metrizability have been intensively studied both bytopologists and analysts for a long time. Especially important properties are those ones whichgeneralize metrizability: the Fr´echet–Urysohn property, sequentiality, the k -space property andcountable tightness. These properties are intensively studied for function spaces, free locally convexspaces, ( LM )-spaces, strict ( LF )-spaces and their strong duals, and Banach and Fr´echet spaces inthe weak topology etc., see for example [1, 10, 23, 24, 26, 31] and references therein.A much weaker notion than Fr´echet–Urysohness was introduced by Arhangel’skii and studiedin [25]: a topological space X is said to be κ -Fr´echet–Urysohn if for every open subset U of X andevery x ∈ U , there exists a sequence { x n } n ∈ ω ⊆ U converging to x .For Tychonoff topological spaces X and Y , we denote by C k ( X, Y ) and C p ( X, Y ) the space C ( X, Y ) of all continuous functions from X to Y endowed with the compact-open topology or thepointwise topology, respectively. If Y = R , we set C k ( X ) := C k ( X, R ) and C p ( X ) := C p ( X, R ).Being motivated by the classical Ascoli theorem we introduced in [5] a new class of topologicalspaces, namely, the class of Ascoli spaces. A Tychonoff space X is Ascoli if every compact subset K of C k ( X ) is evenly continuous, that is the map X × K ∋ ( x, f ) f ( x ) ∈ R is continuous. By Email address: [email protected] (S. Gabriyelyan)
Preprint submitted to Elsevier April 2, 2020 scoli’s theorem [9, Theorem 3.4.20], each k -space is Ascoli. The Ascoli property in topologicalspaces, topological groups, (free) locally convex spaces, function spaces etc. was intensively studiedin [3, 5, 11, 13, 14, 16, 17, 18, 19]. Being motivated by the classical notion of c -barrelled locallyconvex spaces, in [15] we defined a Tychonoff space X to be sequentially Ascoli if every convergentsequence in C k ( X ) is equicontinuous. Clearly, every Ascoli space is sequentially Ascoli, but theconverse is not true in general, see [15]. Below we formulate some of the most interesting results(although the clauses (i)–(iv) were proved for the property of being an Ascoli space, their proofsand Proposition 2.10 show that one can replace “Ascoli” by “sequentially Ascoli”). Theorem 1.1. (i) ([13])
A Fr´echet space E with the weak topology is a sequentially Ascoli spaceif and only if E = F n or E = F ω , where F = R or C is the field of E . (ii) ([3, 19]) The closed unit ball of a Banach space E endowed with the weak topology is a se-quentially Ascoli space if and only if E does not contain an isomorphic copy of ℓ . (iii) ([14]) A strict ( LF ) -space E is a sequentially Ascoli space if and only if E is a Fr´echet spaceor E = φ . (iv) ([14]) The space of distributions D ′ (Ω) is not a sequentially Ascoli space. (v) ([16, 18]) C p ( X ) is an Ascoli space if and only if X has the property ( κ ) . The following diagram describes the relationships between the aforementioned properties: κ -Fr´echet–Urysohn + Ascoli + sequentiallyAscolimetric + Fr´echet–Urysohn + K S sequential + k -space K S None of these implications is reversible, see [5, 9, 16, 15]. For further generalizations of Ascolispaces see [4].Now we describe the content of the paper whose main purpose is the further study of Ascoliand sequentially Ascoli spaces. In Section 2 we establish some categorical properties of sequentiallyAscoli spaces analogous to the corresponding properties of Ascoli spaces considered in [5]. Inparticular, various characterizations of sequentially Ascoli spaces are given in Theorem 2.7. We showthat the square of a Fr´echet–Urysohn space can be not Ascoli (Proposition 2.14). Strengthening aresult of Arhangel’skii we prove in Theorem 2.21 that a hereditary Ascoli space is Fr´echet–Urysohn,however every non-discrete P -space is hereditary sequentially Ascoli but Ascoli (Proposition 2.18).In Section 3 we show that a locally compact abelian group G endowed with the Bohr topologyis sequentially Ascoli if and only if G is compact (Theorem 3.1). In Theorem 3.2 we prove that if X is totally countably compact or near sequentially compact, then it is a sequentially Ascoli space.However there are countably compact spaces which are not sequentially Ascoli (Proposition 3.3).In Section 4 we prove that the product of a locally compact space and an Ascoli space is Ascoli(Theorem 4.2). In Theorem 4.4 we partially reverse this result by showing that a µ -space X islocally compact if and only if the product of X with an arbitrary Ascoli (even Fr´echet–Urysohn)space is an Ascoli space.In the last Section 5 we study various function spaces which are sequentially Ascoli. In Theorem5.13 we extend (v) of Theorem 1.1 by showing that C p ( X ) is sequentially Ascoli if and only if X has the property ( κ ). Let Y be a metrizable abelian group, X be a Y -Tychonoff space, andlet α be a nonzero countable ordinal. In Theorem 5.14 we show that the space B α ( X, Y ) of allBaire- α functions from X to Y is κ -Fr´echet–Urysohn and hence Ascoli. To obtain these results we2ctively use the idea successfully applied in [16] and which is that instead of the whole functionspace C p ( X, Y ) we consider only some of its sufficiently rich and saturated subspaces in the senseof Definition 2.1. Finally in Proposition 5.18 we obtain a necessary condition on X for which thespace C k ( X ) is sequentially Ascoli.
2. Sequentially Y -Ascoli spaces In this section we characterize sequentially Ascoli spaces and prove some standard categoricalassertions. Also we provide several examples which clarify relationships between the considerednotions and pose open questions.Recall that a family I of compact subsets of a topological space X is called an ideal of compactsets if S I = X and for any sets A, B ∈ I and any compact subset K ⊆ X we get A ∪ B ∈ I and A ∩ K ∈ I . Let F ( X ) and K ( X ) be the family of all finite subsets or all compact subsets of X ,respectively. Denote by S ( X ) the family of all finite unions of convergent sequences in X . Clearly, F ( X ), S ( X ) and K ( X ) are ideals of compact sets in X and F ( X ) ⊆ S ( X ) ⊆ K ( X ).We shall consider the following separation axioms. Definition 2.1.
Let Y be a topological space. A topological space X is called • ([6]) Y -Tychonoff if for every closed subset A of X , point y ∈ Y and function f : F → Y defined on a finite subset F of X \ A , there exists a continuous function ¯ f : X → Y such that¯ f ↾ F = f and ¯ f ( A ) ⊆ { y } ; • ([6]) Y -normal if X is a T -space and for any closed set F ⊆ X and each continuous function f : F → Y with finite image f ( F ) there exists a continuous function ¯ f : X → Y such that¯ f | F = f ; • ([20]) Y I -Tychonoff for an ideal I of compact subsets of X if for every closed subset A of X ,point y ∈ Y and function f : K → Y with finite image f ( K ) defined on a subset K ∈ I of X \ A , there exists a continuous function ¯ f : X → Y such that ¯ f ↾ K = f and ¯ f ( A ) ⊆ { y } .For I = F ( X ), S ( X ) or K ( X ) we shall say that X is Y p -Tychonoff, Y s -Tychonoff or Y k -Tychonoff,respectively. Clearly, X is Y p -Tychonoff if and only if it is Y -Tychonoff, and Y k -Tychonoff + Y I -Tychonoff + Y -Tychonoff . If Y is path-connected and admits a non-constant continuous function χ : Y → R , then, byProposition 2.9 of [20], X is Tychonoff if and only if it is Y k -Tychonoff.Let X and Y be topological spaces. For every y ∈ Y , we denote by y ∈ Y X the constantfunction defined by y ( x ) := y for every x ∈ X . Each ideal I of compact subsets of X determinesthe I -open topology τ I on the power space Y X . A subbase of this topology consists of the sets[ K ; U ] = { f ∈ Y X : f ( K ) ⊆ U } , where K ∈ I and U is an open subset of Y . In particular, for every g ∈ Y X , finite subfamily F = { F , . . . , F n } ⊆ I and each family U = { U , . . . , U n } of open subsets of Y such that g ( F i ) ⊆ U i for every i = 1 , . . . , n , the sets of the form W [ g ; F , U ] := n \ i =1 W [ g ; F i , U i ] , where W [ g ; F i , U i ] := (cid:8) f ∈ Y X : f ( F i ) ⊆ U i (cid:9) . I -open topology τ I at the function g . The space C ( X, Y ) of all continuousfunctions from X to Y endowed with the I -open topology induced from ( Y X , τ I ) will be de-noted by C I ( X, Y ). If I = F ( X ) , S ( X ) or K ( X ), we write τ I = τ p , τ s or τ k and C I ( X, Y ) = C p ( X, Y ) , C s ( X, Y ) or C k ( X, Y ), respectively. In the case when Y = R we write simply C p ( X ), C s ( X ) or C k ( X ) and observe that the sets of the form[ K ; ε ] := [ K ; ( − ε, ε )] , where K ⊆ X belongs to I and ε >
0, form a base at the zero function of the topology τ I . Inthe case when Y is a discrete abelian group and also for simplicity of notations, we set[ K ; 0] := [ K ; { } ] . Proposition 2.2.
Let Y be a T -space containing at least two points. Then: (i) τ p ≤ τ s ≤ τ k . (ii) If X is a Y -Tychonoff space, then τ s = τ p if and only if every convergent sequence in X isessentially constant. (iii) If X is a Y s -Tychonoff space, then τ s = τ k if and only if each compact subset of X is containedin a finite union of convergent sequences. Proof. (i) is clear. Below we prove only (ii) because (iii) has a similar proof.(ii) Assume that τ s = τ p and suppose for a contradiction that X contains an infinite convergentsequence S = { x n } n ∈ ω with x n → x . Since | Y | >
1, fix two distinct points z, t ∈ Y . As Y is T , choose an open neighborhood V of z such that t V . Then [ S ; V ] is a τ s -neighborhood ofthe constant function z . To get a contradiction (i.e. that τ p = τ s ), we show that W [ z ; F, W ] * W [ z ; S, V ] for every finite subset F of X and each open neighborhood W ⊆ Y of z . We assume that W ⊆ V . Indeed, since S is infinite, there exists an m ∈ ω such that x m F . As X is Y -Tychonoff,there is f ∈ C ( X, Y ) such that f ( F ) ⊆ W and f ( x m ) V . Then f ∈ W [ z ; F, W ] \ W [ z ; S, V ].Conversely, if every convergent sequence in X is essentially constant then the equality τ s = τ p holdstrivially. (cid:3) For a cardinal number κ , let V ( κ ) be the Fr´echet-Urysohn fan which is, by definition, thetopological space obtained from κ many copies of pairwise disjoint one-to-one convergent sequencesby identifying their limit points, endowed with the quotient topology. Specifically, V ( κ ) = S i ∈ κ S i ,where each S i is homeomorphic to a convergent sequence with its limit x ∞ and S i ∩ S j = { x ∞ } whenever i = j . All points of V ( κ ) except for x ∞ are isolated, while a neighborhood of x ∞ is any U ⊆ V ( κ ) such that x ∞ ∈ U and S i \ U is finite for every i ∈ κ . It is well-known (and easy tocheck) that V ( κ ) is a Fr´echet–Urysohn Tychonoff space. Note that every compact subset of V ( κ )has empty intersection with all but finitely many sets of the form S − i = S i \ { x ∞ } . Example 2.3.
Let Y = R . Now, if X = β N , then τ p = τ s < τ k . If X = V ( κ ), then τ p < τ s = τ k .For X = [0 , τ p < τ s < τ k . (cid:3) Recall that a subset A of C k ( X, Y ) is called evenly continuous if the evaluation map ψ : A × X → Y, ψ ( f, x ) := f ( x ), is continuous, i.e., for every f ∈ A , x ∈ X and neighborhood U f ( x ) of f ( x ) thereare neighborhoods V f ⊆ A of f in A ⊆ C k ( X, Y ) and O x of x such that ψ ( g, z ) = g ( z ) ∈ U f ( x ) forevery g ∈ V f and each z ∈ O x . Definition 2.4.
Let X and Y be Tychonoff spaces. The space X is called a sequentially Y -Ascolispace if each convergent sequence S ⊆ C k ( X, Y ) is evenly continuous.4n Theorem 2.7 below we show that X is a sequentially Ascoli space if and only if X is sequentially R -Ascoli. Also we shall repeatedly use without additional mentioning the following remark: Sincea convergent sequence S ⊆ C k ( X, Y ) has only one non-isolated point, say f , to check that S isevenly continuous it is sufficient to prove that S is evenly continuous at any point ( f , x ) with x ∈ X .To characterize sequentially Y -Ascoli spaces we need additional notations. For any topologicalspaces X and Y the canonical valuation map δ : X → C ( C k ( X, Y ) , Y ) , δ ( x )( f ) := f ( x ) , assigns to each point x ∈ X the Y -valued Dirac measure δ ( x ) : C k ( X, Y ) → Y concentrated at x .The map δ is well-defined since the function δ ( x ) : C k ( X, Y ) → Y is continuous at each function f ∈ C k ( X, Y ) (indeed, for any open neighborhood V ⊆ Y of δ ( x )( f ) = f ( x ), the set [ { x } ; V ] is anopen neighborhood of f in C k ( X, Y ) with δ ( x ) (cid:0) [ { x } ; V ] (cid:1) ⊆ V ).All topological groups are assumed to be T and hence Tychonoff. Let Y be an abelian topologicalgroup, and let X be a topological space. A subset F of C ( X, Y ) is called equicontinuous if for every x ∈ X and each neighborhood U of zero 0 ∈ Y there is a neighborhood O x of x such that f ( t ) − f ( x ) ∈ U for all t ∈ O x and f ∈ F . It is clear that every equicontinuous set is evenly continuous, but the converse is not true in general.We proved in [5] that X is Y -Ascoli if and only if the canonical map δ is continuous as amap from X to C k ( C k ( X, Y ) , Y ). Below we obtain an analogous characterization of sequentially Y -Ascoli spaces. Theorem 2.5.
Let Y be a Tychonoff space containing at least two points, X be a Y -Tychonoffspace, and let δ : X → C s (cid:0) C k ( X, Y ) , Y (cid:1) be the canonical map. Then the following assertions areequivalent: (i) X is a sequentially Y -Ascoli space; (ii) δ is continuous; (iii) δ is an embedding; (iv) each point x ∈ X is contained in a dense sequentially Y -Ascoli subspace of X .If in addition Y is an abelain group, then (i)-(iv) are equivalent to (v) every convergent sequence in C k ( X, Y ) is equicontinuous. Proof. (i) ⇒ (ii) Assume that X is sequentially Y -Ascoli. We have to show that δ is continuousat each point x ∈ X . Fix a sub-basic neighborhood[ S ; V ] ⊆ C s ( C k ( X, Y ) , Y )of δ ( x ), where S = { f n : n ∈ ω } ⊆ C k ( X, Y ) with f n → f and V ⊆ Y open. It follows from δ ( x ) ∈ [ S ; V ] that f n ( x ) = δ ( x )( f n ) ∈ V for every n ∈ ω . Since X is sequentially Y -Ascoli, thereare N ∈ ω and an open neighborhood O of x such that f n ( O ) ⊆ V for every n > N . Takinga smaller neighborhood O ⊆ X of x we can assume that f n ( O ) ⊆ V for every n ∈ ω . But thismeans that δ ( x ) ∈ [ S ; V ] for every x ∈ O . Thus the canonical map δ is continuous at x .(ii) ⇒ (i) Assume that δ is continuous. To show that X is sequentially Y -Ascoli, we have to checkthat every convergent sequence S = { f n : n ∈ ω } ⊆ C k ( X, Y ) with f n → f is evenly continuous5t ( f , x ) with x ∈ X . So fix an x ∈ X and an open neighborhood O f ( x ) ⊆ Y of f ( x ). Using theregularity of Y , choose an open neighborhood e O f ( x ) of f ( x ) such that cl Y (cid:0) e O f ( x ) (cid:1) ⊆ O f ( x ) . Set U := (cid:2) { x } ; e O f ( x ) (cid:3) ∩ S and S := cl S ( U ). Then S is a convergent sequence such that S \ S isfinite and δ ( x )( g ) = g ( x ) ∈ cl Y (cid:0) e O f ( x ) (cid:1) ⊆ O f ( x ) , for every g ∈ S , and therefore δ ( x ) ∈ [ S ; O f ( x ) ]. Since δ is continuous, there is a neighborhood O x of x such that δ ( O x ) ⊆ (cid:2) S ; O f ( x ) (cid:3) . So, for every t ∈ O x and each g ∈ U ⊆ S , we have g ( t ) = δ ( t )( g ) ∈ O f ( x ) ,which means that S is evenly continuous at ( f , x ).(ii) ⇒ (iii) Fix two distinct points y, z ∈ Y , and choose an open neighborhood V of y such that z V . Since X is Y -Tychonoff, the map δ is injective. So it remains to show that the map δ − : δ ( X ) → X is continuous. Fix an x ∈ X and an open neighborhood U of x . Since X is Y -Tychonoff, there is an f ∈ C ( X, Y ) such that f ( x ) = y and f ( X \ U ) ⊆ { z } . (2.1)Set S := { f n } n ∈ ω , where f n = f for every n ∈ ω , and consider the following open basic neighborhoodof δ ( x ): W := W [ δ ( x ); S, V ] ⊆ C s (cid:0) C k ( X, Y ) , Y (cid:1) . Then x ∈ δ − (cid:0) W ∩ δ ( X ) (cid:1) if and only if δ ( x ) ∈ W [ δ ( x ); S, V ] if and only if δ ( x )( f ) = f ( x ) ∈ V .Now, since z V , (2.1) implies that x ∈ U . Thus δ − | δ ( X ) is continuous.The implications (iii) ⇒ (ii) and (i) ⇒ (iv) are trivial.(iv) ⇒ (i) Assume that each point x ∈ X is contained in a dense sequentially Y -Ascoli subspaceof X . We have to show that X is sequentially Y -Ascoli. Take a convergent sequence S = { f n : n ∈ ω } ⊆ C k ( X, Y ) with f n → f . To show that S is evenly continuous, fix a point x ∈ X and anopen neighborhood O = O f ( x ) of f ( x ). By our assumption, the point x is contained in a densesequentially Y -Ascoli subspace Z ⊆ X . The density of Z in X implies that the restriction operator ζ : C k ( X, Y ) → C k ( Z, Y ) , ζ : g g | Z , is injective. Since the space Z is sequentially Y -Ascoli, there is a neighborhood W ⊆ Z of x and N ∈ ω such that ζ ( f n )( x ) ∈ O for every x ∈ W and each n ≥ N. (2.2)The closure W of W in X is a (closed) neighborhood of z in X , and (2.2) implies f n ( x ) ∈ O for every x ∈ W and each n ≥ N. Taking into account that x and O were arbitrary and Y is Tychonoff, we obtain that S is evenlycontinuous.(v) ⇒ (i) follows from the easy fact that every equicontinuous subset of C k ( X, Y ) is evenly con-tinuous.(i) ⇒ (v) Let K be a convergent sequence in C k ( X, Y ). So K is evenly continuous. Fix a point x ∈ X and an open neighborhood U of zero in Y . Choose an open neighborhood V of 0 ∈ Y such that V + V ⊆ U . For every f ∈ K , choose an open neighborhood O f ⊆ X of x and an openneighborhood U f ⊆ K of f such that g ( t ) ∈ f ( x ) + V for every t ∈ O f and g ∈ U f . (2.3)6ince K is compact, there are f , . . . , f n ∈ K such that K = S ni =1 U f i . Set O := T ni =1 O f i . Now if t ∈ O and g ∈ K , choose f i such that g ∈ U f i and then g ( t ) − g ( x ) = (cid:0) g ( t ) − f i ( x ) (cid:1) + (cid:0) f i ( x ) − g ( x ) (cid:1) ( . ) ∈ V + V ⊆ U. Thus K is equicontinuous. (cid:3) Let X be a topological space. A family { A i } i ∈ I of subsets of X is called compact-finite if forevery compact K ⊆ X , the set { i ∈ I : K ∩ A i } is finite; and { A i } i ∈ I is called locally finite if forevery point x ∈ X there is a neighborhood X of x such that the set { i ∈ I : U ∩ A i } is finite. Afamily { B i } i ∈ I of subsets of X is called strongly ( functionally ) compact-finite if for every i ∈ I , thereis an (functionally) open neighborhood U i of B i such that the family { U i } i ∈ I is compact-finite. Proposition 2.6.
Let Y be an abelain topological group containing at least two points, and let X be a Y -Tychonoff space. Consider the following assertions: (i) every strongly compact-finite sequence of closed subsets of X is locally finite; (ii) every strongly functionally compact-finite sequence of functionally closed subsets of X is locallyfinite; (iii) X is a sequentially Y -Ascoli space.Then (i) ⇒ (ii) ⇒ (iii) . If in addition X is Y -normal, then (iii) ⇒ (i) . Proof.
The implication (i) ⇒ (ii) is trivial.(ii) ⇒ (iii) Suppose for a contradiction that X is not sequentially Y -Ascoli. Then, by the equiv-alence (i) ⇔ (v) of Theorem 2.5, there exists a null-sequence { f n : n ∈ ω } in C k ( X, Y ) which is notequicontinuous at some point x ∈ X . So there is a functionally open neighborhood W of zero in Y such that the closed sets A n := { x ∈ X : f n ( x ) − f n ( x ) ∈ Y \ W } ( n ∈ ω )satisfy the condition x ∈ S n ∈ ω A n \ S n ∈ ω A n , i.e. the family { A n : n ∈ ω } is not locally finite.Observe also that the sets A n are functionally closed in X . Choose a functionally open neighborhood W ⊆ Y of zero such that W + W ⊆ W . For every n ∈ ω , set U n := { x ∈ X : f n ( x ) − f n ( x ) ∈ Y \ W } , so U n is a functionally open neighborhood of A n . Hence to get a contradiction it remains to showthat the family U = { U n : n ∈ ω } is compact-finite. Let K be a compact subset of X . Since f n → in the compact-open topology, there exists an N ∈ ω such that f n ∈ [ K ∪ { x } ; W ] for every n ≥ N , where the open symmetric neighborhood W ⊆ Y of 0 is chosen so that W + W ⊆ W .Therefore, for every n ≥ N and each x ∈ K , we have f n ( x ) − f n ( x ) ∈ W + W ⊆ W and hence U n ∩ K = ∅ . Thus U is compact-finite.(iii) ⇒ (i) Assume that X is a Y -normal space. Fix two distinct points y, z ∈ Y , and choosean open neighborhood V of y such that z V . Suppose for a contradiction that there exists astrongly compact-finite sequence { A n : n ∈ ω } of closed subsets in X which is not locally finite.For every n ∈ ω , choose an open neighborhood U n of A n such that the sequence U = { U n : n ∈ ω } is compact-finite. Since X is Y -normal, for every n ∈ ω there is a continuous function f n : X → Y such that f n ( A n ) = { z } and f n ( X \ U n ) ⊆ { y } . Since U is compact-finite, we obtain that f n → y in C k ( X, Y ). As { A n : n ∈ ω } is not locally-finite, there exists a point x ∈ X such that, for every7eighborhood U of x , the set { n ∈ ω : A n ∩ U = ∅} is infinite. Hence, for every neighborhood U of x , there is an n ∈ ω such that f n ( x ) = z for some x ∈ A n . Therefore the convergent sequence { f n : n ∈ ω } ∪ { y } is not evenly continuous at ( x , y ). Thus X is not sequentially Y -Ascoli, acontradiction. (cid:3) If Y = R , we have the following characterization of sequentially Ascoli spaces. Theorem 2.7.
Let X be a Tychonoff space, and let δ : X → C s (cid:0) C k ( X ) (cid:1) be the canonical map.Then the following assertions are equivalent: (i) X is a sequentially R -Ascoli space; (ii) δ is continuous; (iii) δ is an embedding; (iv) each point x ∈ X is contained in a dense sequentially Ascoli subspace of X ; (v) every convergent sequence in C k ( X ) is equicontinuous, i.e. X is sequentially Ascoli; (vi) every strongly functionally compact-finite sequence of functionally closed subsets of X is locallyfinite. Proof.
Since a topological space is Tychonoff if and only if it is R -Tychonoff, the equivalences(i)-(v) follow from Theorem 2.5. The implication (vi) ⇒ (i) follows from Proposition 2.6.(v) ⇒ (vi) Suppose for a contradiction that there exists a strongly functionally compact-finitesequence of functionally closed subsets { A n : n ∈ ω } in X which is not locally finite. For every n ∈ ω , choose a functionally open neighborhood U n of A n such that the sequence U = { U n : n ∈ ω } is compact-finite. Since X is Tychonoff, Theorem 1.5.13 of [9] implies that for every n ∈ ω thereis a continuous function f n : X → [0 ,
1] such that f n ( A n ) = { } and f n ( X \ U n ) = { } . Since U iscompact-finite, we obtain that f n → in C k ( X ). As { A n : n ∈ ω } is not locally-finite, there existsa point x ∈ X such that, for every neighborhood U of x , the set { n ∈ ω : A n ∩ U = ∅} is infinite.Hence, for every neighborhood U of x , there is an n ∈ ω such that f n ( x ) = 1 for some x ∈ A n .Therefore the convergent sequence { f n : n ∈ ω } ∪ { } is not equicontinuous at the point x . (cid:3) Now we establish some elementary properties of sequentially Y -Ascoli spaces. Recall that asubspace Z of a topological space X is a retract of X if there is a continuous map r : X → Z suchthat r ( z ) = z for all z ∈ Z . The proof of the next proposition is actually a partial case (when acompact set is a convergent sequence) of Proposition 5.2 of [5] and hence is omitted. Proposition 2.8.
Let Y be a Tychonoff space. (i) If X is a sequentially Y -Ascoli space, then each retract Z of X is sequentially Y -Ascoli. (ii) For any family { X i } i ∈ I of sequentially Y -Ascoli spaces the topological sum X := L i ∈ I X i issequentially Y -Ascoli. (iii) Each sequentially Y -Ascoli space X is sequentially Y κ -Ascoli for every cardinal κ . (iv) Each sequentially Y -Ascoli space X is sequentially Z -Ascoli for every subspace Z ⊆ Y . Since any (zero-dimensional) Tychonoff space embeds into some power R κ (respectively, 2 κ ),Theorem 2.7 and Proposition 2.8 imply Corollary 2.9. (i) If X is a sequentially Ascoli space, then X is sequentially Y -Ascoli for everyTychonoff space Y . (ii) If X is a sequentially -Ascoli space, then X is sequentially Y -Ascoli for every zero-dimensional T -space Y .
8n what follows we shall use repeatedly the following proposition.
Proposition 2.10 ([15, 19]).
Assume that for a set I (respectively, I = ω ), the Tychonoff space X admits a family U = { U i : i ∈ I } of open sets, a subset A = { a i : i ∈ I } ⊆ X and a point z ∈ X such that (i) a i ∈ U i for every i ∈ I ; (ii) (cid:12)(cid:12) { i ∈ I : C ∩ U i = ∅} (cid:12)(cid:12) < ∞ for each compact subset C of X ; (iii) z is a cluster point of A .Then X is not a (sequentially) Ascoli space. Analogously we have the following result. The characteristic function of a subset A of a set Ωis denoted by A . Proposition 2.11.
Assume that for a set I (respectively, I = ω ), the Tychonoff space X admitsa family U = { U i : i ∈ I } of clopen sets and a point z ∈ X such that (i) (cid:12)(cid:12) { i ∈ I : C ∩ U i = ∅} (cid:12)(cid:12) < ∞ for each compact subset C of X ; (ii) z is a cluster point of S i ∈ I U i .Then X is not a (sequentially) -Ascoli space. Proof.
We have to find a compact subset (or a convergent sequence) K in C k ( X, ) which is notequicontinuous. Since all U i are clopen, for every i ∈ I , the characteristic function U i is continuous.Now we define K := { U i : i ∈ I } ∪ { } . By (ii), the family K is not equicontinuous at the point z . So, to show that X is not a (sequentially) 2-Ascoli space, it suffices to prove that K ⊆ C k ( X, )is compact (or converges to , respectively). Fix an arbitrary standard neighborhood [ C ; 0] of ∈ C k ( X, ). Then, by (i), the set J := { i ∈ I : C ∩ U i = ∅} is finite. Therefore, for every i J we have U i ∈ [ C ; 0]. Thus K is compact (or converges to ). (cid:3) Example 2.12.
A closed subspace of a κ -Fr´echet–Urysohn space can be not a sequentially Ascolispace. Indeed, let A be a countable subset on the unit sphere of ℓ constructed in Proposition 4.1of [19]. It immediately follows from [19, Proposition 4.1] and Proposition 2.10 that the space A is not sequentially Ascoli. On the other hand, the space A being Lindel¨of is homeomorphic to aclosed subspace of some product R λ . It remains to note that R λ is a κ -Fr´echet–Urysohn space byCorollary 2.3 of [16]. (cid:3) Example 2.13.
A product of a Polish space and a sequential k ω -space can be not sequentiallyAscoli. Indeed, let φ be the inductive limit of finite-dimensional vector spaces. It is well knownthat the space φ is a sequential non-Fr´echet–Urysohn k ω -space (see for example Proposition 2.1 in[14]). Taking into account Proposition 2.10, in Theorem 1.2 of [14] it is proved that the product ℓ × φ is not a sequentially Ascoli space. In particular, the three space property does not hold inthe class of (sequentially) Ascoli spaces. (cid:3) Gruenhage and Tanaka proved that the square V ( ω ) × V ( ω ) of the Fr´echet–Urysohn fan hasuncountable tightness and hence is not a k -space, see [2, Lemma 7.6.22]. Below we show that V ( ω ) × V ( ω ) is not a 2-Ascoli space, and hence the square of a Fr´echet–Urysohn space may evennot be 2-Ascoli. Proposition 2.14.
The product V ( ω ) × V ( ω ) is not a -Ascoli space. roof. We shall use the idea of Gruenhage and Tanaka and consider two families { A α : α ∈ ω } and { B α : α ∈ ω } of infinite subsets of ω satisfying the following two conditions:(a) A α ∩ B β is finite for all α, β ∈ ω ,(b) for no D ⊆ ω , all sets A α \ D and B α ∩ D , α ∈ ω , are finite.Set z := ( x ∞ , x ∞ ). For all α, β ∈ ω and n ∈ A α ∩ B β , we put U n,α,β := { ( x n,α , x n,β ) } (so U n,α,β isa clopen subset of V ( ω ) × V ( ω )). To apply Proposition 2.11 to the family U = { U n,α,β } and thepoint z we have to check conditions (i)-(ii). The condition (ii) is proved in Lemma 7.6.22 of [2].To check (i), let C be a compact subset of V ( ω ) × V ( ω ). Then there is a finite subset Γ of ω such that C is contained in the product (cid:16) S γ ∈ Γ S γ (cid:17) × (cid:16) S γ ∈ Γ S γ (cid:17) . Therefore, if ( x n,α , x n,β ) ∈ C ,then α, β ∈ Γ and, by construction, n ∈ A α ∩ B β . Now the finiteness of the sets Γ and A α ∩ B β (see (a)) imply that the set { ( n, α, β ) : C ∩ U n,α,β = ∅} = { ( n, α, β ) : ( x n,α , x n,β ) ∈ C } is finite, and hence (i) of Proposition 2.11 holds true. Thus V ( ω ) × V ( ω ) is not 2-Ascoli. (cid:3) We do not know whether the product V ( ω ) × V ( ω ) is not sequentially Ascoli. Problem 2.15.
Is there a (sequentially) Ascoli group whose square is not (sequentially) Ascoli?
The next proposition complements Proposition 2.8.
Proposition 2.16.
Let Y be a Tychonoff space containing at least two points, X and Z be Y -Tychonoff spaces, and let p : X → Z be an open surjective map. If X is a (sequentially) Y -Ascolispace, then so is Z . Proof.
Denote by p ∗ : C k ( Z, Y ) → C k ( X, Y ), p ∗ ( f ) := f ◦ p , the adjoint map of p . Let K be acompact subset (or a convergent sequence) of C k ( Z, Y ). Fix f ∈ K , s ∈ Z and a neighborhood O f ( s ) ⊆ Y of f ( s ). Choose t ∈ X such that p ( t ) = s . Since p ∗ ( K ) is a compact subset (or aconvergent sequence) of C k ( X, Y ), there are open neighborhoods U ⊆ X of t and U p ∗ ( f ) of p ∗ ( f ) in C k ( X, Y ) such that g ( x ) ∈ O f ( s ) for every x ∈ U and g ∈ U p ∗ ( f ) ∩ p ∗ ( K ) . (2.4)As p is open, p ( U ) is an open neighborhood of s . Set U f := ( p ∗ ) − ( U p ∗ ( f ) ) ∩ K , so U f is aneighborhood of f in K . Now, for every z = p ( x ) ∈ p ( U ) with x ∈ U and each F ∈ U f , (2.4)implies F ( z ) = F ◦ p ( x ) = p ∗ ( F )( x ) ∈ O f ( s ) . Thus K is evenly continuous. (cid:3) Since quotient maps of topological groups are open (see [22, Theorem 5.26]) Proposition 2.16implies
Corollary 2.17.
Let Y be a Tychonoff space containing at least two points, and let H be a normalclosed subgroup of a Y -Tychonoff group G such that the quotient group G/H is Y -Tychonoff. If G is (sequentially) Y -Ascoli, then so is G/H . In particular, a Hausdorff quotient group of a(sequentially) Ascoli group is a (sequentially) Ascoli group. X is called a P -space if each G δ -set in X is open. InProposition 2.9 of [15] we proved that any non-discrete Tychonoff P -space is sequentially Ascolibut not Ascoli. Below we extend this result. Proposition 2.18.
Let X be a non-discrete Tychonoff P -space. Then: (i) X is a hereditary sequentially Ascoli space; (ii) X is not -Ascoli; (iii) if additionally X is a one-point Lindel¨ofication of an uncountable discrete set, then X is ahereditary Baire space. Proof. (i) Let Z be a subspace of X . Then Z is also a Tychonoff P -space. Hence, by Proposition2.9 of [15], Z is a sequentially Ascoli space. Thus X is a hereditary sequentially Ascoli space.(ii) Every Tychonoff P -space is zero-dimensional and every compact subset of X is finite (seeLemma 2.8 in [15]). Thus X is not 2-Ascoli by Proposition 5.12 of [5].(iii) Let Z be a subspace of X . If Z does not contain the unique non-isolated point x ∞ of X oris countable, then Z with the induced topology is discrete and hence Baire. If Z is uncountable andcontains x ∞ , then Z is topologically isomorphic to a one-point Lindel¨ofication of the uncountablediscrete space Z \{ x ∞ } , and hence Z is homeomorphic to a one point Lindel¨ofication of someuncountable discrete set. So it is sufficient to show that X is Baire. It is easy to see that any opendense subset of X is either X or X \{ x ∞ } . Therefore any intersection of dense open subsets of X contains the dense subset X \{ x ∞ } . Thus X is Baire. (cid:3) Proposition 2.18 motivates the following problem.
Problem 2.19.
Is every -Ascoli zero-dimensional space X sequentially Ascoli? If X is 2-normal, Problem 2.19 has a positive answer even in a stronger form. Proposition 2.20.
Let X be a zero-dimensional space. If X is -normal, then X is a sequentially -Ascoli space if and only if it is sequentially Ascoli. Proof.
Every sequentially Ascoli space is sequentially 2-Ascoli. Conversely, assume that X isa sequentially 2-Ascoli space. To prove that X is a sequentially Ascoli space, let { f n } n ∈ ω be anull-sequence in C k ( X ) and suppose for a contradiction that it is not equicontinuous at some point x ∈ X . Without loss of generality we can assume that f n ( x ) = 0 for all n ∈ ω replacing f n by f n − f n ( x ) if needed. Then there is δ > A n := { x ∈ X : | f n ( x ) | ≥ δ } ( n ∈ ω )satisfy the following condition x ∈ [ n ∈ ω A n . (2.5)For every n ∈ ω , define B n := { x ∈ X : | f n ( x ) | ≤ δ } . Then A n and B n are closed and disjoint.Since X is 2-normal, there is a function g n ∈ C ( X, ) such that g n ( B n ) = { } and g n ( A n ) ⊆ { } ( n ∈ ω ) . (2.6)Now (2.5) and (2.6) imply that the sequence { g n } n ∈ ω is not equicontinuous at x . Therefore, byTheorem 2.5(v), to get a desired contradiction it suffices to show that g n → in C k ( X, ). To thisend, fix an arbitrary compact subset K of X . Since f n → in C k ( X ), choose an m ∈ ω such that f n ∈ W [ ; K, ( − δ, δ )] for every n ≥ m. Then, for every n ≥ m , we have K ⊆ { x ∈ X : | f n ( x ) | < δ } ⊆ B n and hence g n ( K ) = { } . Thus g n → in C k ( X, ). (cid:3) X is a hereditary k -space (i.e., every subspace of X is a k -space), then X is Fr´echet–Urysohn. Below we essentially strengthen this remarkable resultby proving the following theorem. Theorem 2.21.
A Tychonoff space is hereditary -Ascoli if and only if it is Fr´echet–Urysohn.Consequently, a Tychonoff space is hereditary Ascoli if and only if it is Fr´echet–Urysohn. Proof.
If we shall show that every hereditary 2-Ascoli space is a k -space, we shall prove thatit is a hereditary k -space, and applying Arhangel’skii’s theorem [9, 3.12.15] we obtain that it isFr´echet–Urysohn. So, let X be a hereditary 2-Ascoli space and suppose for a contradiction that X is not a k -space. Then there exists a non-closed subspace A of X such that A ∩ K is closed in K for every compact subspace K of X . Choose an arbitrary point z ∈ A \ A and consider the subspace Y := A ∪ { z } of X .We claim that if K is a compact subspace of Y then either z K or z is an isolated point of K . Indeed, if z is a non-isolated point of K , then z ∈ A ∩ K . Since z A we obtain that A ∩ K isnot closed in K that contradicts the choice of A .Consider the family U of all collections { U i } i ∈ I of pairwise disjoint open sets in Y such that z S i ∈ I U i . By Zorn’s lemma, the family U has a maximal (under inclusion) collection { V i } i ∈ I .The maximality of { V i } i ∈ I and the fact that z is not isolated in Y imply that z ∈ S i ∈ I V i .Consider the subspace Z := { z } ∪ S i ∈ I V i of Y and hence of X . Since X is hereditary 2-Ascoli, the space Z is a 2-Ascoli space. Observe that z is a non-isolated point of Z , and the fact z S i ∈ I V i implies that all subspaces V i are clopen in Z . Therefore, for every i ∈ I , the function V i is continuous on Z . Set K := { } ∪ { V i } i ∈ I . Let us show that K is a compact subset of C k ( Z, ). Indeed, let [ K ; 0] be a standard open neighborhood of the zero function ∈ C k ( Z, ),where K is a compact subset of Z . By the claim we can assume that K = { z } ∪ K , where K is acompact subset of S i ∈ I V i and z K . Then the set J := { i ∈ I : V i ∩ K = ∅} is finite. Clearly, if i J , then V i ∈ [ K ; 0]. Thus K is compact. Since z ∈ S i ∈ I V i , it follows that the compact set K is not equicontinuous at the point z . Therefore the space Z is not 2-Ascoli, a contradiction. Thusthe space X must be a k -space. (cid:3) We end this section with the question to find natural classes of sequentially Ascoli spaces X which are Ascoli. Problem 2.22.
Let X be a Tychonoff space such that every compact subset of C k ( X ) is metrizable(for example, X is a cosmic space or more generally X contains a dense σ -compact subspace). Isit true that X is sequentially Ascoli if and only if it is an Ascoli space? The condition for the compact subsets of C k ( X ) of being metrizable in Problem 2.22 is essential(also one cannot replace the condition to have a dense σ -compact subspace by Lindel¨ofness of X as Proposition 2.18 shows). Example 2.23.
There is a sequentially Ascoli non-Ascoli space X such that every compact subsetof C k ( X ) is Fr´echet–Urysohn. Proof.
Let X be a non-discrete Lindel¨of P -space. Then, by Proposition 2.18, X is a sequentiallyAscoli non-Ascoli space. On the other hand, since every compact subset of a P -space is finite wehave C k ( X ) = C p ( X ). Therefore, by Theorem II.7.16 of [1], the space C k ( X ) and hence all itscompact subspaces are Fr´echet–Urysohn. (cid:3) . Sequentially Ascoli spaces in some compact-type classes of Tychonoff spaces Let G be a locally compact abelian (lca for short) group. The group G endowed with theBorh topology τ b induced from the Bohr compactification bG of G is denoted by G + . Numeroustopological and algebraic-topological properties of the precompact group G + are well-studied, werefer the reader to Chapter 9 of [2] and references therein. It is an easy consequence of the Glicksbergtheorem (which states that the groups G and G + have the same compact sets) that the group G + is a k -space if and only if G is compact. Below we essentially extend this result. Theorem 3.1. If G is an lca group, then G + is sequentially Ascoli if and only if G is compact. Proof. If G is compact, then G + = G is an Ascoli space. Conversely, assume that G + is sequen-tially Ascoli. We have to show that G is compact. By Theorem 24.30 of [22], G is topologicallyisomorphic to R n × G , where n ∈ ω and G is an lca group containing an open compact subgroup G . Claim 1. n = 0 and hence G = G . Indeed, suppose for a contradiction that n >
0. Then R + is a direct summand of G + (see for example Problem 9.9.N of [2]). Hence, by Corollary 2.17, R + is sequentially Ascoli. So to get a contradiction we shall show below that the space R + is not sequentially Ascoli. Since R + is Tychonoff, for every natural number n >
0, there is a continuousfunction f n : R + → [0 ,
2] such that f n (cid:0) [ − n, n ] (cid:1) = { } and f n (cid:0) [ − n, − n − ∪ [ n + 1 , n ] (cid:1) = { } . By the Glicksberg theorem, for every compact subset K of R + there is an n ∈ ω such that K ⊆ [ − n, n ]. This fact and the definition of the functions f n imply that f n → f := in C k ( R + ).We show that the null-sequence S = { f n } n ∈ ω is not equicontinuous at zero 0 ∈ R + (which meansthat R + is not sequentially Ascoli, see Theorem 2.7). Indeed, let ε = 1 and fix an arbitrary openneighborhood U of 0 ∈ R + . Since the group R + is not locally compact, for every n ∈ ω theintersection U ∩ (cid:0) ( −∞ , − n − ∪ [ n + 1 , ∞ ) (cid:1) is not empty. So there are n > x n ∈ U ∩ (cid:0) [ − n, − n − ∪ [ n + 1 , n ] (cid:1) . Then f n ( x n ) − f n (0) =2 > ε . Thus S is not equicontinuous. Claim 2. G = G is compact. Indeed, suppose for a contradiction that G is not compact.Then the quotient group G/G is infinite and discrete. Applying Problem 9.9.N of [2] (or a moregeneral Lemma 2.3 of [12]), we obtain ( G/G ) + = G + /G and hence, by Corollary 2.17, the group( G/G ) + is sequentially Ascoli. By 16.13(c) of [22], the group G/G has a subgroup H such thatthe discrete group T := ( G/G ) /H is countably infinite. Once again applying Problem 9.9.N of [2]and Corollary 2.17, we obtain that the countably infinite group T + is sequentially Ascoli. By theGlicksberg theorem, every compact subset of T + is finite. Therefore, by Proposition 2.10 of [15], T + is not sequentially Ascoli. This contradiction shows that the group G must be compact. (cid:3) Every Hausdorff compact space is Ascoli. So it is natural to consider other classes of compact-type spaces. Let us recall that a topological space X is called • sequentially compact if every sequence in X has a convergent subsequence; • ω -bounded if every sequence in X has compact closure; • totally countably compact if every sequence in X has a subsequence with compact closure; • near sequentially compact if for any sequence ( U n ) n ∈ ω of open sets in X there exists a sequence( x n ) n ∈ ω ∈ Q n ∈ ω U n containing a convergent subsequence ( x n k ) k ∈ ω ;13 countably compact if every sequence in X has a cluster point.Near sequentially compact spaces were introduced and studied by Dorantes-Aldama and Shakhma-tov [8] as selectively sequentially pseudocompact spaces. However we shall use the term “nearsequentially compact spaces” since it is more self-suggesting and is a partial case of a more generalnotion of “near sequentially compact subset” introduced in [7]. In the next diagram we summarizethe relationships between the above-defined notions and note that none of these implications isreversible (see [8] and [34, Section 2]):compact + ω -bounded + totally countablycompact + countablycompact (cid:11) (cid:19) sequentiallycompact + ; ♦♦♦♦♦♦♦♦♦♦♦ ♦♦♦♦♦♦♦♦♦♦♦ near sequentiallycompact + pseudocompact Theorem 3.2.
Let X be a Tychonoff space satisfying at least one of the next conditions: (i) X is totally countably compact; (ii) X is near sequentially compact.Then X is a sequentially Asoli space. Proof.
Suppose for a contradiction that X is not sequentially Ascoli. Then, by the equivalence(i) ⇔ (v) of Theorem 2.5, there exists a null-sequence { f n : n ∈ ω } in C k ( X ) which is not equicontin-uous at some point x ∈ X . Replacing f n by f n − f n ( x ) if needed we can assume that f n ( x ) = 0for every n ∈ ω . Then there is an ε > A n := { x ∈ X : f n ( x ) ∈ ( −∞ , − ε ] ∪ [ ε, ∞ ) } ( n ∈ ω )and the open sets B n := { x ∈ X : f n ( x ) ∈ ( −∞ , − ε ) ∪ ( ε, ∞ ) } ( n ∈ ω )satisfy the condition x ∈ S n ∈ ω B n \ S n ∈ ω A n . Passing to a subsequence if needed we can assumealso that all sets B n s are nonempty. Let us show that the family A = { A n : n ∈ ω } is compact-finite. Indeed, let K be a compact subset of X . Since f n → in the compact-open topology, thereexists an N ∈ ω such that f n ∈ [ K ; ε ] for every n ≥ N . Therefore, A n ∩ K = ∅ for every n ≥ N .Thus A is compact-finite. Now we consider the cases (i) and (ii).(i) Assume that X is totally countably compact. For every n ∈ ω choose a point a n ∈ A n . Since X is totally countably compact, there is a subsequence { a n k } k ∈ ω of { a n } n ∈ ω which has compactclosure C . It is clear that the compact set C intersects infinitely many A n s, and hence A is notcompact-finite. This is a desired contradiction.(ii) Assume that X is near sequentially compact. Since all B n s are open in X and X is nearsequentially compact, there is a sequence ( x n ) n ∈ ω ∈ Q n ∈ ω B n containing a subsequence ( x n k ) k ∈ ω that converges to some point x ∗ ∈ X . So the convergent sequence ( x n k ) k ∈ ω ∪ { x ∗ } intersectsinfinitely many A n s, and hence A is not compact-finite. This contradiction finishes the proof. (cid:3) However, there are countably compact Tychonoff spaces X which are not sequentially Ascoli asthe next proposition shows (for a concrete example of a separable space X satisfying the conditionsof this proposition see [34, 2.13]). 14 roposition 3.3. Let X be an infinite countably compact zero-dimensional T -space. If everycompact subset of X is finite, then X is not sequentially -Ascoli. Proof.
Since X is infinite and Tychonoff, there is a sequence { U n } n ∈ ω of clopen nonempty subsetsof X such that U n ∩ U m = ∅ for all distinct n, m ∈ ω . For every n ∈ ω take a point x n ∈ U n .Since X is countably compact, there is a cluster point z of the sequence { x n } n ∈ ω . Observe that z S n ∈ ω U n because all U n s are open and pairwise disjoint.For every n ∈ ω , set f n := U n . Since all compact subsets of X are finite, the pairwise disjointnessof U n s implies that f n → in C k ( X, ). On the other hand, since any neighborhood V of z containssome x k we have f k ( x k ) − f k ( z ) = 1 . Therefore the null-sequence { f n } n ∈ ω is not equicontinuous.Thus X is not sequentially 2-Ascoli. (cid:3) Proposition 2.18 and Theorem 3.2 motivate the following problem.
Problem 3.4.
Does there exist a totally countably compact (( near ) sequentially compact or ω -bounded ) Tychonoff space X which is not Ascoli?
4. A characterization of locally compact µ -spaces Essentially using a result of Whitehead [35, Lemma 4], Cohen proved that, if Z is a locallycompact Hausdorff space, then Z × X is a k -space for every k -space X . In [27] Michael proved theconverse assertion. So we obtain a characterization of locally compact spaces, see [9, 3.12.14(c)]: ATychonoff space Z is locally compact if and only if the product Z × X is a k -space for every k -space X . This result motivates the following problem:
Problem 4.1.
Is it true that a Tychonoff space Z is locally compact if and only if the product Z × X is a (sequentially) Ascoli space for every (sequentially) Ascoli space X ? In Theorem 4.4 below, for the case of Ascoli spaces, we obtain a positive answer to this problemunder additional assumption that X is a µ -space. Recall that a topological space is called a µ -space if every functionally bounded subset of X is relatively compact (a subset A of X is called functionally bounded if f ( A ) is a bounded subset of R for every f ∈ C ( X )). First we prove thefollowing analogue of the Whitehead–Cohen result mentioned above. Theorem 4.2.
Let Z be a locally compact space. If X is an Ascoli space, then so is Z × X . Proof.
Let K be a compact subset of C k ( Z × X ). We have to show that K is equicontinuous.Fix δ > z , x ) ∈ Z × X .Choose an open neighborhood U of z with compact closure. Define a map T : U × K → C k ( X )by T ( z, f )( x ) := f ( z, x ) ∈ C k ( X ). We claim that T is continuous. Indeed, fix an arbitrary point( y, g ) ∈ U × K and an open standard neighborhood T ( y, g ) + [ C ; ε ] of the function T ( y, g ) ∈ C k ( X ),where C ⊆ X is compact and ε >
0. For every x ∈ C , choose an open neighborhood V x ⊆ U of y and an open neighborhood O x ⊆ X of x such that | g ( z, t ) − g ( y, x ) | < ε for all z ∈ V x and t ∈ O x . (4.1)Since C is compact, there are x , . . . , x s ∈ C such that C ⊆ S si =1 O x i . For every x ∈ C , choose i ( x ) ∈ { , . . . , s } such that x ∈ O x i ( x ) and set V := T si =1 V x i ⊆ U . Then V is an open neighborhood15f y in U . Now, for every z ∈ V , x ∈ C and each function g + h ∈ (cid:0) g + (cid:2) U × C ; ε (cid:3)(cid:1) ∩ K , (4.1)implies | T ( z, g + h )( x ) − T ( y, g )( x ) | = | g ( z, x ) + h ( z, x ) − g ( y, x ) |≤ | g ( z, x ) − g ( y, x i ( x ) ) | + | g ( y, x i ( x ) ) − g ( y, x ) | + | h ( z, x ) | < ε + ε + ε = ε. Thus T is continuous, and hence the set T ( U × K ) ⊆ C k ( X ) is compact.Since X is an Ascoli space, it follows that the compact set T ( U × K ) is equicontinuous. Inparticular, there is an open neighborhood O of x such that | f ( z, x ) − f ( z, x ) | = | T ( z, f )( x ) − T ( z, f )( x ) | < δ for all x ∈ O , z ∈ U and f ∈ K. (4.2)Consider the restriction map R : C k ( Z × X ) → C k ( Z ) defined by R ( f ) := f ( z, x ). Then R ( K )is a compact subset of C k ( Z ). The space Z being locally compact is Ascoli. Therefore there is anopen neighborhood W ⊆ U of z such that | f ( z, x ) − f ( z , x ) | = | R ( f )( z ) − R ( f )( z ) | < δ for all z ∈ W and f ∈ K. (4.3)Now, for every ( z, x ) ∈ W × O and each f ∈ K , (4.2) and (4.3) imply | f ( z, x ) − f ( z , x ) | ≤ | f ( z, x ) − f ( z, x ) | + | f ( z, x ) − f ( z , x ) | < δ. Therefore K is equicontinuous. Thus Z × X is an Ascoli space. (cid:3) We say that a topological space X is locally functionally bounded if every point of X has afunctionally bounded neighborhood. To prove Theorem 4.4 we need the following assertion. Proposition 4.3.
If a Tychonoff space Z is not locally functionally bounded, then there is aFr´echet–Urysohn fan X such that the product Z × X is not an Ascoli space. Proof.
Let z be a point of Z which does not have a functionally bounded neighborhood. Fixan open base { V α } α ∈A of z . For every α ∈ A , choose a function f α ∈ C ( X ) such that f α ↾ V α is unbounded. Then we can choose a discrete sequence { V n,α } n ∈ ω in X such that all V n,α areopen subsets of V α \{ z } . For every n ∈ ω , choose an arbitrary point z n,α ∈ V n,α . Consider theFr´echet–Urysohn fan X := V ( A ) with the unique non-isolated point x ∞ (so S α = { x n,α } ∪ { x ∞ } and x n,α → x ∞ as n → ∞ ). For each α ∈ A and every n ∈ ω , set a n,α := ( z n,α , x n,α ) ∈ Z × V ( A ) , U n,α := V n,α × { x n,α } . It is clear that U n,α is an open neighborhood of a n,α and ( z , x ∞ ) is a cluster point of the family { a n,α : α ∈ A , n ∈ ω } . Therefore, to apply Proposition 2.10 we have to check that for every compactsubset C of Z × V ( A ), the set A := { ( n, α ) ∈ ω × A : U n,α ∩ C = ∅} is finite. To this end, consider the projection K := π V ( A ) ( C ) of C into V ( A ). Then K is a compactsubset of V ( A ), and hence there is a finite subset I ⊆ A such that { x n,α } ∩ K = ∅ only if α ∈ I .Now, for every α ∈ I the family { π Z ( U n,α ) = V n,α : n ∈ ω } of projections onto Z is discrete in Z .Therefore, for every α ∈ I , the family { n ∈ ω : V n,α ∩ π Z ( C ) = ∅} is finite. Thus A is finite. Finally, Proposition 2.10 implies that the space Z × V ( A ) is not Ascoli. (cid:3) heorem 4.4. For a µ -space Z the following assertions are equivalent: (i) Z is locally compact; (ii) Z × X is an Ascoli space for each Ascoli space X ; (iii) Z × X is an Ascoli space for each Fr´echet–Urysohn space X . Proof.
The implication (i) ⇒ (ii) follows from Theorem 4.2, and (ii) ⇒ (iii) is trivial.(iii) ⇒ (i) By Proposition 4.3, the space Z is locally functionally bounded. Fix a point z ∈ Z andits functionally bounded neighborhood V z . Since Z is a µ -space, it follows that the closure V z of V z is a compact subset of Z . So z has a compact neighborhood V z . Thus Z is locally compact. (cid:3)
5. Sequentially Ascoli function spaces
Let X be a Tychonoff space and let ( Y, ρ ) be a metric space. We denote by Y X the family ofall maps from X to Y , and let E be a subspace of Y X . In particular, if E is endowed with thepointwise topology, then for a function f ∈ E , a finite subset A of X and ε >
0, we set W [ f ; A, ε ] := (cid:8) g ∈ E : ρ (cid:0) f ( x ) , g ( x ) (cid:1) < ε for every x ∈ A (cid:9) , so W [ f ; A, ε ] is a standard open neighborhood of f in E .If additionally Y is a (metrizable) locally convex space, there is a long tradition to investigatelocally convex properties of the space C ( X, Y ) endowed with the pointwise topology or the compact-open topology by means of topological properties of X and locally convex properties of Y . Of coursethe most important case is the case when Y is a Banach space. For numerous results obtained inthe eighties of the last century and historical remarks we refer the reader to the well known lecturenotes of Schmets [32].The study of topological properties of spaces C p ( X, Y ) and C k ( X, Y ), when X = R , [0 ,
1] or thedoubleton = { , } , is one of the main direction in General Topology. We refer the reader to thebooks [1, 26, 33] and the papers [5, 16, 11, 18, 19, 20, 28, 29, 30] and references therein.Although the spaces C p ( X, Y ) and C k ( X, Y ) are the most important, there are other importantclasses of (discontinuous) functions from X to Y widely studied in General Topology and Analysisas, for example, the classes of bounded/precompact continuous functions and the classes of Bairefunctions (we recall their definitions below). So the question of the study of topological propertiesof these classes of functions arises naturally. It turns out that the study of the aforementionedclasses of functions can be realized simultaneously using the following idea successfully applied in[16]: instead the whole space C ( X, Y ) we consider only some of its sufficiently rich subspaces inthe sense of Definitions 2.1 and 5.3.The main purpose of this section is to characterize those spaces X for which the functionspace C p ( X, Y ) and classes of Baire functions are (sequentially) Ascoli, also we obtain a necessarycondition on X for which the space C k ( X ) is sequentially Ascoli. To this end, we shall essentiallyuse the ideas from [16, 18] and the papers of Sakai [29, 30]; in fact (the proofs of) Lemmas 5.1and 5.2, Proposition 5.4 and Proposition 5.11 given below are corresponding modifications andextensions (of the proofs) of Lemmas 2.1 and 2.2 of [18], Theorem 1.1 of [18] and Theorem 2.1of [29], and of Theorem 2.1 of [29], respectively (nevertheless we give their detailed proofs for thereader convenience and to make the paper self-contained). Lemma 5.1.
Let ( Y, ρ ) be a metric space, X be a set, E be a dense subspace of the topologicalproduct Y X , and let g ∈ E . For every n ∈ ω , let U n be an open subset of E such that g ∈ U n ,and let W n be an open cover of U n . Then for every n ∈ ω , there exists W n ∈ W n such that g ∈ S { W n : n ∈ ω } . roof. We proceed by induction on n ∈ ω . For n = 0, choose arbitrarily W ∈ W and h ∈ W .Since W is open, we can choose a finite subset A of X and ε > W [ h ; A , ε ] ⊆ W .For n = 1, choose W ∈ W such that W ∩ W [ g ; A ,
1] is not empty (this is possible since g ∈ U and W covers U ). Take arbitrarily an h ∈ W ∩ W [ g ; A , A of X containing A and ε , 0 < ε < ε , such that W [ h ; A , ε ] ⊆ W ∩ W [ g ; A , . Continuing this process we can construct an increasing sequence { A n : n ∈ ω } of finite subsets of X , a decreasing null-sequence { ε n : n ∈ ω } of positive reals, a sequence { W n ∈ W n : n ∈ ω } ofopen subsets of E and a sequence { h n : n ∈ ω } in E such that W (cid:2) h n +1 ; A n +1 , ε n +1 (cid:3) ⊆ W n +1 ∩ W (cid:2) g ; A n , n +1 (cid:3) for all n ∈ ω. (5.1)We claim that { W n : n ∈ ω } is as required. Indeed, fix a finite F ⊆ X and ε >
0. Choose a naturalnumber n > F ∩ ( S n ∈ ω A n ) ⊆ A n and n + ε n +1 < ε . Since E is dense in Y X , thereis an h ∈ E such that ρ (cid:0) h ( x ) , h n +1 ( x ) (cid:1) < ε n +1 for every x ∈ A n +1 , (so, by (5.1), ρ (cid:0) h ( x ) , g ( x ) (cid:1) < n +1 + ε n +1 < ε for every x ∈ F ∩ A n = F ∩ A n +1 ) and ρ (cid:0) h ( x ) , g ( x ) (cid:1) < ε for every x ∈ F \ A n +1 . Then h ∈ W (cid:2) h n +1 ; A n +1 , ε n +1 (cid:3) ⊆ W n +1 and h ∈ W [ g ; F, ε ]. Thus h ∈ W n +1 ∩ W [ g ; F, ε ]. (cid:3) Lemma 5.2.
Let ( Y, ρ ) be a metric space containing at least two points, X be a set, E be a densesubspace of the topological product Y X , and let g ∈ E . Assume that E is a sequentially Ascoli spaceand { U n : n ∈ ω } is a sequence of open subsets of E such that g ∈ S { U n : n ∈ ω } but g U n forall n ∈ ω . Then there exists a compact subspace K of E such that the set { n ∈ ω : K ∩ U n = ∅} isinfinite. Proof.
Suppose for a contradiction that for every compact subset K of E , K ∩ U n = ∅ only forfinitely many n ∈ ω . Since E is Tychonoff, for every n ∈ ω and each h ∈ U n , we can choose afunction φ h,n ∈ C (cid:0) E (cid:1) such that φ h,n ( h ) > φ h,n ↾ E \ U n = 0 , and set W h,n := { f ∈ E : φ h,n ( f ) > } , it is clear that W h,n is an open subset of U n . For every n ∈ ω , set e U n := S l ≥ n U l and W n := { W h,l : l ≥ n and h ∈ U l } . Then g ∈ e U n and W n is an open cover of e U n . Applying Lemma 5.1 for the sequence { e U n : n ∈ ω } we obtain the following: for every n ∈ ω there exists W n ∈ W n such that g ∈ S { W n : n ∈ ω } . Let l n ≥ n and φ n be witnesses for W n ∈ W n (observe also that g W n for all n ∈ ω ).We claim that φ n converges to the zero function ∈ C k ( E ). Indeed, fix a compact K ⊆ E and ε >
0. By our assumption the set F := { n ∈ ω : K ∩ U l n = ∅} is finite. Therefore, for every n ∈ ω \ F , we have φ n ↾ K = 0 and hence φ n ∈ W [ ; K, ε ]. Thus φ n → in C k ( E ).On the other hand, given any open V ⊆ E containing g and m ∈ ω , the facts g ∈ S { W n : n ∈ ω } and g W n imply that there exist n ≥ m and f ∈ V ∩ W n such that φ n ( f ) >
1. This proves thatthe convergent sequence { φ n : n ∈ ω } ∪ { } ⊆ C k ( E )is not equicontinuous at g . Thus E is not sequentially Ascoli, a contradiction. (cid:3) efinition 5.3 ([20]). Let X and Y be topological spaces. A subspace E of C p ( X, Y ) is called relatively Y p -Tychonoff if for every closed subset A of X , finite subset F ⊆ X contained in X \ A ,point y ∈ Y and each f ∈ C p ( X, Y ) there is a function ¯ f ∈ E such that ¯ f ↾ F = f ↾ F and ¯ f ( A ) ⊆{ y } . (cid:3) In other words, E is a relatively Y p -Tychonoff subspace of C p ( X, Y ) if for every closed subset A of X , every function f ∈ C p ( X, Y ) and each finite F ⊆ X such that F ⊆ X \ A , the restriction f ↾ F canbe extended to a function ¯ f ∈ E with an additional condition ¯ f ( A ) ⊆ { y } for some point y ∈ Y .For a subset A of a topological space X , let [ A ] := A and for each α ∈ ω , let [ A ] α be the setof limits of convergent sequences in S i ∈ α [ A ] i . The set [ A ] s := S α ∈ ω [ A ] α is called the sequentialclosure of A .A family { A i } i ∈ I of subsets of a set X is called point-finite if for every x ∈ X , the set { i ∈ I : x ∈ A i } is finite. A family { A i } i ∈ I of subsets of a topological space X is said to be strongly point-finite if for every i ∈ I there is an open set U i of X such that A i ⊆ U i and the family { U i : i ∈ I } ispoint-finite.Following Sakai [29], a topological space X is said to have the property ( κ ) if every pairwisedisjoint sequence of finite subsets of X has a strongly point-finite subsequence. Proposition 5.4.
Let Y be a metrizable non-compact space, X be a Y -Tychonoff space, and let E be a relatively Y p -Tychonoff subspace of C p ( X, Y ) containing all constant functions. Consider thefollowing statements: (i) E is a sequentially Ascoli space; (ii) the sequential closure of any open set of E is closed; (iii) X has the property ( κ ) .Then (i) ⇒ (iii) and (ii) ⇒ (iii). Proof.
Since Y is not compact, one can find a sequence { g n } n ∈ ω ⊆ Y and a sequence { V n } n ∈ ω ofopen subsets in Y such that(a) g n ∈ V n for all n ∈ ω , and(b) the family V = { V n } n ∈ ω is discrete in Y .(i) ⇒ (iii) Assume that E is a sequentially Ascoli space. Consider a sequence A = { A n : n ∈ ω } of finite subsets of X such that A n ∩ A m = ∅ for all n = m . We have to find an infinite J ⊆ ω andopen sets U j ⊇ A j for all j ∈ J , such that the family { U j : j ∈ J } is point-finite.For every k ∈ ω , set O k := E ∩ T a ∈ A k W [ g k ; { a } , V k ]. It follows from (a) and (b) that g O k forall k >
0. Since X is Y -Tychonoff and E is dense in C p ( X, Y ) (so E is dense in Y X ), the disjointnessof the sequence A implies g ∈ S { O k : k > } E . By Lemma 5.2, there exists a compact set K ⊆ E intersecting infinitely many of the O k ’s, so the set J := { n > K ∩ O n = ∅} is infinite. For every j ∈ J , choose a function h j ∈ K ∩ O j and set U j := { x ∈ X : h j ( x ) ∈ V j } , and note that U j is an open subset of X containing A j . We show that { U j } j ∈ J is point-finite. Tothis end, fix an arbitrary x ∈ X and set J ′ x := { j ∈ J : x ∈ U j } . We have to prove that J ′ x is finite.Suppose for a contradiction that J ′ x is infinite. Since h j ( x ) ∈ V j , (b) implies that the sequence { h j ( x ) : j ∈ J ′ x } is closed and discrete in Y . On the other hand, since { h j : j ∈ J ′ x } is a subset of the19ompact set K , the sequence { h j ( x ) : j ∈ J ′ x } ⊆ Y must be relatively compact. This contradictionshows that J ′ x must be finite as desired.(ii) ⇒ (iii) Assume that the sequential closure of any open set of E is closed. Consider a disjointsequence { A n : n ∈ ω } of finite subsets of X . We have to find an infinite J ⊆ ω and open sets U j ⊇ A j for all j ∈ J , such that { U j : j ∈ J } is point-finite. Suppose for a contradiction that nosubsequence of { A n : n ∈ ω } is strongly point-finite.For each n ∈ ω , set U n := { f ∈ E : f ( A n ) ⊆ V n } = E ∩ \ a ∈ A n [ { a } ; V n ] , and let U := S n> U n . Since each U n is open in E , the set U is also open in E . Observe that U n ⊆ { f ∈ E : f ( A n ) ⊆ V n } . Claim 1. [ U ] s = S n> [ U n ] s ⊆ S n> U n . Indeed, let f j ∈ U n j for 0 < n < n < · · · . Then A n j ⊆ f − j ( V n j ), and hence, by our supposition that no subsequence of { A n : n ∈ ω } is stronglypoint-finite, the family (cid:8) f − j ( V n j ) (cid:9) j ∈ ω is not point-finite. Therefore there exists a point z ∈ X whichis contained in f − j ( V n j ) for infinitely many j ’s. This means that f j ( z ) ∈ V n j for infinitely many j ’s. But then, taking into account that the family V ′ := { V n j } is discrete and hence S V ′ = S V n j ,we obtain that f j ( z ) does not converge in Y . Thus, if a sequence S = { f j } j ∈ ω ⊆ U converges in E , there is an m ∈ ω such that S ∩ U m is infinite and hence S converges to a point of [ U m ] s . Theclaim is proved.By (a) and (b) we have g V n for every n >
0. It follows that g S n> U n . Hence, by Claim1, g [ U ] s . Therefore, to get a desired contradiction it suffices to prove that g ∈ U (which meansthat the sequential closure of U is not closed).Fix a finite subset F of X and δ >
0. Since A n ∩ A m = ∅ for all n = m , there is k ∈ ω such that F ∩ A k = ∅ . Since X is Y -Tychonoff and E is a relatively Y p -Tychonoff subspace of C p ( X, Y ), onecan find a function f ∈ E such that f ↾ F = g ↾ F and f ( A k ) = { g k } ⊆ V k . Thus f ∈ W [ g ; F, δ ] ∩ U k and hence g ∈ U . (cid:3) A class of spaces with the property ( κ ) is given below. Proposition 5.5.
Every Tychonoff P -space X has the property ( κ ) . Proof.
First we recall that every Tychonoff P -space is zero-dimensional. Observe that everycountable subset of a Tychonoff P -space being an F σ -set is closed and discrete. Now let { A n : n ∈ ω } be a disjoint sequence of finite subsets of X . Set A := S n ∈ ω A n . For every a ∈ A choose a clopenneighborhood V a of a such that the sequence { V a : a ∈ A } is disjoint. For every n ∈ ω , set U n := S a ∈ A n V a . It is clear that the sequence { U n : n ∈ ω } is point-finite. Thus X has the property( κ ). (cid:3) Now we show that in the case when Y is a discrete metrizable group the property ( κ ) is alsosufficient in Proposition 5.4. Proposition 5.6.
Let Y be a discrete abelian group containing more than one point, X be aninfinite Y -Tychonoff space, and let E be a relatively Y p -Tychonoff subgroup of C p ( X, Y ) containingall constant functions. If X has the property ( κ ) , then E is a κ -Fr´echet–Urysohn space. roof. Let U be an open subset of E and let g ∈ U \ U . Since the group E is homogeneous,without loss of generality we can assume that g is the zero-function . So, let ∈ U \ U . We haveto find a sequence in U converging to . Since Y is discrete, for every f ∈ U , there is a finite subset F ( f ) of X such that T x ∈ F ( f ) W [ f ; { x } , ⊆ U . Set A ( f ) := { x ∈ F ( f ) : f ( x ) = 0 } and B ( f ) := { x ∈ F ( f ) : f ( x ) = 0 } , and note that B ( f ) = ∅ for every f ∈ U because U .Now we construct a sequence { f n } n ∈ ω ⊆ U inductively, as follows. Take an arbitrary f ∈ U .For n = 1, the inclusion ∈ U implies that there is f ∈ U ∩ T x ∈ F ( f ) W [ f ; { x } , n > f n ∈ U ∩ T (cid:8) W (cid:2) ; { x } , (cid:3) : x ∈ S n − i =0 F ( f i ) (cid:9) .By the choice of f n , for every n > B ( f n ) ∩ (cid:0) B ( f ) ∪ · · · ∪ B ( f n − ) (cid:1) ⊆ B ( f n ) ∩ n − [ i =0 F ( f i ) = ∅ . So the sequence { B ( f n ) } n> is pairwise disjoint and hence, by the property ( κ ), we can take astrongly point-finite subsequence { B ( f n j ) } j ∈ ω . Let { O j } j ∈ ω be a point-finite open family of X such that B ( f n j ) ⊆ O j and O j ∩ A ( f n j ) = ∅ ( j ∈ ω ) . (5.2)Since X is Y -Tychonoff and E is a relatively Y p -Tychonoff subspace of C p ( X, Y ), for every j ∈ ω there is a function t j ∈ E such that t j (cid:0) X \ O j (cid:1) = { } and t j ( z ) = f n j ( z ) for all z ∈ B ( f n j ) . (5.3)Observe that if x ∈ A ( f n j ), (5.2) implies that t j ( x ) = 0 = f n j ( x ). Therefore t j ↾ F ( f nj ) = f n j ↾ F ( f nj ) for every j ∈ ω . Thus t j ∈ T x ∈ F ( f nj ) W [ f n j ; { x } , ⊆ U for all j ∈ ω . Finally, since { O j } j ∈ ω ispoint-finite, (5.3) implies that t j → . (cid:3) Corollary 5.7.
Let Y be a discrete space containing more than one point, and let X be a zero-dimensional T -space. If X has the property ( κ ) , then C p ( X, Y ) is a κ -Fr´echet–Urysohn space. Proof.
By Proposition 2.7 of [6], the space X is Y -Tychonoff. Since every discrete space is home-omorphic to a discrete abelian group of the same cardinality, the assertion follows from Proposition5.6 applied to E = C p ( X, Y ). (cid:3) The proof of Proposition 5.6 shows that only the following two moments are essential: (1) thesequence { B ( f n ) } n> is disjoint, and (2) one can find a function t j such that t j ↾ F ( f nj ) = f n j ↾ F ( f nj ) .To get useful variations of these two properties in the case of non-discrete group Y we need thefollowing notion which is stronger than the property of being a Y I -Tychonoff space. Definition 5.8.
Let X and Y be topological spaces and let I be an ideal of compact sets of X . Asubspace H of Y X is said to have an I -approximation property at a point y if the point y has anopen base V of neighborhoods such that for every V ∈ V , closed subset A of X , F ∈ I containedin X \ A , and a function f : F → V with finite image there is a function ¯ f ∈ H such that¯ f ( X ) ⊆ V, ¯ f ↾ F = f and ¯ f ( A ) ⊆ { y } . The space X has the I -approximation property if it has the I -approximation property at each point y ∈ Y . (cid:3)
21s usual if I = F ( X ) , S ( X ) or K ( X ), we shall say that the space X has the p -, s - or k -approximationproperty , respectively. If Y = R (or any locally arc-connected space), it follows from Proposition 2.9of [20] that a topological space X has the I -approximation property if and only if X is Tychonoff. Example 5.9.
Let X be a zero-dimensional T -space, and let I be an ideal of compact sets of X . Then X has the I -approximation property for every nonempty Tychonoff space Z since, byProposition 2.10 of [20], X is Y I -Tychonoff for every T -space Y . (cid:3) Let X be a Tychonoff space and let E be a locally convex space. We denote by C b ( X, E ) and C rc ( X, E ) the spaces of all functions f ∈ C ( X, E ) such that f ( X ) is a bounded or a relativelycompact subset of E , respectively. Proposition 5.10.
Let E be a non-trivial locally convex space, X be a Tychonoff space, and let I be an ideal of compact sets of X . Then: (i) X is E I -Tychonoff; (ii) C b I ( X, E ) and C rc I ( X, E ) are relatively E I -Tychonoff subspaces of C I ( X, E ) ; (iii) C I ( X, E ) , C b I ( X, E ) and C rc I ( X, E ) have the I -approximation property. Proof. (i) and (iii) follow from (ii) of Proposition 2.9 in [20] (recall that any locally convexspace has a base at zero consisting of absolutely convex open neighborhoods), and (ii) is proved inProposition 4.4 of [20]. (cid:3)
Proposition 5.11.
Let Y be a non-discrete metric abelian group, X be an infinite Y -Tychonoffspace, and let E be a subgroup of C p ( X, Y ) satisfying the following properties: (i) E is a relatively Y p -Tychonoff subspace of C p ( X, Y ) containing constant functions; (ii) E has the p -approximation property.If X has the property ( κ ) , then E is a κ -Fr´echet–Urysohn space. Proof.
Let U be an open subset of E and let g ∈ U \ U . Since the group E is homogeneous,without loss of generality we can assume that g is the zero-function . So, let ∈ U \ U . We haveto find a sequence in U converging to . Fix a decreasing basis { V n } n ∈ ω of open neighborhoods of0 ∈ Y witnessing the p -approximation property.Now we construct a sequence { f n } n ∈ ω ⊆ U inductively as follows. Take an arbitrary f ∈ U .Choose a finite subset F ( f ) of X and open sets { U x : x ∈ F ( f ) } in Y such that E ∩ \ x ∈ F ( f ) W [ f ; { x } , U x ] ⊆ U (in what follows we shall omit “ E ∩ ” to simplify the notations). Define A ( f ) := { x ∈ F ( f ) : f ( x ) = 0 } and B ( f ) := { x ∈ F ( f ) : f ( x ) = 0 } , and note that B ( f ) = ∅ because U .For n = 1, the inclusion ∈ U implies that there is an h ∈ U ∩ T x ∈ F ( f ) W [ ; { x } , V ]. Choosea finite subset F of X and open sets { U x : x ∈ F } in Y such that F ( f ) $ F (this is possible since X is infinite) and \ x ∈ F W [ h ; { x } , U x ] ⊆ U ∩ \ x ∈ F ( f ) W [ ; { x } , V ] .
22n particular, h ( x ) ∈ V for every x ∈ B ( f ) . (5.4)Set A ( h ) := { x ∈ F : h ( x ) = 0 } and B ( h ) := { x ∈ F : h ( x ) = 0 } , and note that B ( h ) = ∅ because U . If the set B ( h ) \ B ( f ) is not empty, we set f := h and C := B ( f ) \ B ( f ). Assume now that B ( h ) \ B ( f ) = ∅ . Then h ( x ) = 0 for every x ∈ F \ B ( f ),and hence the set V ∩ T x ∈ F \ B ( f ) U x is an open neighborhood of 0 ∈ Y . Since Y is non-discrete,take an arbitrary nonzero point y ∈ V ∩ T x ∈ F \ B ( f ) U x . As X is Y -Tychonoff and E is a relatively Y p -Tychonoff subspace of C p ( X, Y ), there is an f ∈ E such that (recall that F ( f ) $ F ) f ( x ) = y if x ∈ F \ B ( f ) , and f ( x ) = h ( x ) for x ∈ F ∩ B ( f ) = B ( f ) . By construction we have f ∈ T x ∈ F W [ h ; { x } , U x ] ⊆ U . So there is a finite subset F ( f ) of X andopen sets { U x, : x ∈ F ( f ) } in Y such that F ⊆ F ( f ) and \ x ∈ F ( f ) W [ f ; { x } , U x, ] ⊆ U ∩ \ x ∈ F ( f ) W [ ; { x } , V ] . Set A ( f ) := { x ∈ F ( f ) : f ( x ) = 0 } , B ( f ) := { x ∈ F ( f ) : f ( x ) = 0 } , and C := B ( f ) \ B ( f ). Note that C = ∅ because, by construction, ∅ 6 = F \ B ( f ) ⊆ C . If z ∈ B ( f ) ∩ B ( f ), then the choice of f implies that f ( z ) = h ( z ) and hence, by (5.4), f ( z ) ∈ V .Continuing this process, for every n > f n ∈ U ∩ \ ( W (cid:2) ; { x } , V n (cid:3) : x ∈ n − [ i =0 F ( f i ) ) , a finite set F ( f n ) ⊆ X , and open sets { U x,n : x ∈ F ( f n ) } such that(a) S n − i =0 F ( f i ) $ F ( f n ) and T x ∈ F ( f n ) W [ f n ; { x } , U x,n ] ⊆ U ;(b) the set C n := B ( f n ) \ (cid:0) B ( f ) ∪ · · · ∪ B ( f n − ) (cid:1) is not empty;(c) if z ∈ B ( f n ) ∩ (cid:0) B ( f ) ∪ · · · ∪ B ( f n − ) (cid:1) , then f n ( z ) ∈ V n .For every n >
0, set D n := B ( f n ) \ C n = B ( f n ) ∩ (cid:0) B ( f ) ∪ · · · ∪ B ( f n − ) (cid:1) . For the pairwise disjoint sequence { C n } n> , we can take a strongly point-finite subsequence { C n j } j ∈ ω . Let { O j } j ∈ ω be a point-finite open family of X such that C n j ⊆ O j and O j ∩ (cid:0) D n j ∪ A ( f n j ) (cid:1) = ∅ ( j ∈ ω ) . For every j ∈ ω , choose an open subset W j of X such that D n j ⊆ W j and W j ∩ (cid:0) O j ∪ A ( f n j ) (cid:1) = ∅ . Since X is Y -Tychonoff and E is a relatively Y p -Tychonoff subspace of C p ( X, Y ), for every j ∈ ω there is a function t j ∈ E such that t j (cid:0) X \ O j (cid:1) = { } and t j ( z ) = f n j ( z ) for all z ∈ C n j . (5.5)Since { O n } n ∈ ω is point-finite, it follows that t j → in C p ( X, Y ).23ow the choice of the sequence { V n } n ∈ ω , (ii) and (c) imply that for every j ∈ ω , there is an h j ∈ E such that h j ( X ) ⊆ V n j , h j (cid:0) X \ W j (cid:1) = { } and h j ( z ) = f n j ( z ) for all z ∈ D n j . (5.6)In particular, the sequence { h j } j ∈ ω uniformly converges to . Since E is a group, we obtain that { t j + h j } j ∈ ω ⊆ E and t j + h j → in E . The choice of O and W j and (5.5) and (5.6) imply( t j + h j ) ↾ F ( f nj ) = f n j ↾ F ( f nj ) , and hence t j + h j ∈ \ x ∈ F ( f nj ) W [ f n j ; { x } , U x,n j ] ( a ) ⊆ U for all j ∈ ω. Thus E is κ -Fr´echet–Urysohn. (cid:3) Now we prove the main result of this section.
Theorem 5.12.
Let Y be a metrizable non-compact abelian group, X be a Y -Tychonoff space, andlet E be a subgroup of C p ( X, Y ) satisfying the following properties: (a) E is a relatively Y p -Tychonoff subspace of C p ( X, Y ) containing all constant functions; (b) if Y is non-discrete, then E has the p -approximation property.. Then the following assertions are equivalent: (i) E is a κ -Fr´echet–Urysohn space; (ii) E is an Ascoli space; (iii) E is a sequentially Ascoli space; (iv) the sequential closure of any open set of E is closed; (v) X has the property ( κ ) . Proof.
The implication (i) ⇒ (ii) follows from Theorem 2.5 of [16] which states that every κ -Fr´echet–Urysohn space is Ascoli. The implications (i) ⇒ (iv) and (ii) ⇒ (iii) are trivial. The impli-cations (iii) ⇒ (v) and (iv) ⇒ (v) follow from Proposition 5.4. Finally, (v) implies (i) by Proposition5.11. (cid:3) Theorem 5.13.
Let Y be a non-trivial metrizable locally convex space, X be a Tychonoff space,and let E = C p ( X, Y ) , C bp ( X, Y ) or E = C rcp ( X, Y ) . Then the following assertions are equivalent: (i) E is a κ -Fr´echet–Urysohn space; (ii) E λ is κ -Fr´echet–Urysohn for every cardinal λ > ; (iii) E is an Ascoli space; (iv) E λ is Ascoli for every cardinal λ > ; (v) E is a sequentially Ascoli space; (vi) E λ is sequentially Ascoli for every cardinal λ > ; (vii) X is either finite or has the property ( κ ) ; (viii) E λ is κ -Fr´echet–Urysohn (Ascoli or sequentially Ascoli) for some cardinal λ > . roof. If X is finite, then E = Y | X | is metrizable. Therefore E λ is κ -Fr´echet–Urysohn for everycardinal λ >
0, see [25] or [16]. Now we assume that X is infinite.The equivalences (i) ⇔ (iii) ⇔ (v) ⇔ (vii) follow from Theorem 5.12 and Proposition 5.10. Theimplication (ii) ⇒ (iv) follows from Theorem 2.5 of [16], and (iv) ⇒ (vi) is clear. The implications(ii) ⇒ (i), (iv) ⇒ (iii) and (vi) ⇒ (v) are also clear.Let us prove that (vii) implies (ii). Assume that X has the property ( κ ). Then, by Proposition7.3 of [29], the topological sum Z = L λ X of λ copies of the space X also has the property ( κ ).Therefore, by the equivalence (i) ⇔ (vii), the space C rcp ( Z, Y ) is κ -Fr´echet–Urysohn. It is clear thatthe space C rcp ( Z, Y ) is naturally homeomorphic to a dense subset of the locally convex space E λ .Thus E λ is κ -Fr´echet–Urysohn by Corollary 2.2 of [16].The implications (ii) ⇒ (viii), (iv) ⇒ (viii) and (vi) ⇒ (viii) are clear. Finally, the implication(viii) ⇒ (i) ((viii) ⇒ (iii) or (viii) ⇒ (v)) follows from the fact that E is a direct summand of E λ and Proposition 3.3 of [25] (or Corollary 2.17, respectively). (cid:3) Now we give an application of Theorem 5.12 to spaces of Baire functions. Let X and Y betopological spaces. We say that a sequence { f n } n ∈ ω ⊆ Y X stably converges to a function f ∈ Y X if for every x ∈ X the set { n ∈ ω : f n ( x ) = f ( x ) } is finite. Set B ( X, Y ) = B st ( X, Y ) := C p ( X, Y ).For every countable nonzero ordinal α , denote by B α ( X, Y ) ⊆ Y X ( B stα ( X, Y ) ⊆ Y X ) the familyof all functions f : X → Y which are pointwise limits of function sequences from S β<α B β ( X, Y )( S β<α B stβ ( X, Y ), respectively). The family B ( X, Y ) := S α ∈ ω B α ( X, Y ) is the class of all Bairefunctions from X to Y . It is clear that C p ( X, Y ) ֒ → B st ( X, Y ) ֒ → B ( X, Y ) ֒ → B ( X, Y ) . For the first countable spaces X the following theorem was proved in (a) of Theorem 3.5 in [16]. Theorem 5.14.
Let Y be a metrizable abelian group, X be a Y -Tychonoff space, and let H be asubgroup of Y X containing B st ( X, Y ) . Then H is a κ -Fr´echet–Urysohn space. Proof.
Observe that B st ( X, Y ) is a dense subgroup of H . Hence, by Corollary 2.2 of [16], if B st ( X, Y ) is a κ -Fr´echet–Urysohn space then so is H . Therefore it is sufficient to show that H = B st ( X, Y ) is a κ -Fr´echet–Urysohn space.It is well known (see for example Proposition 5.3 in [20]) that the space B ( X, Y ) and hence itssubgroup H can be considered as a subspace of C p ( X ℵ , Y ), where X ℵ is the P -modification of X (recall that X ℵ is zero-dimensional). Further, Proposition 5.15 of [20] states that for every finitesubset D of Y , the intersection H ∩ C p ( X ℵ , D ) is a relatively D p -Tychonoff subspace of C p ( X ℵ , D ).In particular, the group H is a relatively Y p -Tychonoff subspace of C p ( X ℵ , Y ). Note also that thespace X ℵ is Y -Tychonoff (because it is zero-dimensional) and has the property ( κ ) by Proposition5.5. Now we consider two cases. Case 1. The group Y is discrete. Then H is κ -Fr´echet–Urysohn by Proposition 5.6. Case 2. The group Y is not discrete. By Example 5.9 and the aforementioned Proposition 5.15of [20], the group H has also the p -approximation property. Now Proposition 5.11 implies that H is a κ -Fr´echet–Urysohn space. (cid:3) Let X be a Tychonoff space and E be a locally convex space. A map f : X → E is called bounded ( relatively compact ) if the image f ( X ) is a bounded (respectively, relatively compact) subset of E .For every countable ordinal α , we denote by B bα ( X, E ), B rcα ( X, E ), B st,bα ( X, E ) or B st,rcα ( X, E )25he family of all functions from B α ( X, E ) or B stα ( X, E ) which are bounded or relatively compact,respectively. It is clear that B st,rcα ( X, E ) ⊆ B st,bα ( X, E ) ⊆ B stα ( X, E ) and B rcα ( X, E ) ⊆ B bα ( X, E ) ⊆ B α ( X, E ) . The next assertion generalizes (b) of Theorem 3.5 in [16].
Theorem 5.15.
Let E be a metrizable locally convex space, X be a Tychonoff space, and let H bea subgroup of E X containing B st,rc ( X, E ) . Then H is a κ -Fr´echet–Urysohn space. Proof.
As in the proof of Theorem 5.14, we can assume that H = B st,rc ( X, E ) and that H canbe considered as a subgroup of C p ( X ℵ , E ). By the aforementioned Proposition 5.15 of [20], thegroup H is a relatively E p -Tychonoff subspace of C p ( X ℵ , E ) and has the p -approximation property.Since X ℵ is a Tychonoff P -space, Proposition 5.5 implies that X ℵ has the property ( κ ). Finally,Proposition 5.11 implies that H is a κ -Fr´echet–Urysohn space. (cid:3) Now we consider the following problem:
Problem 5.16.
Let Y be a metrizable space (for example, Y = R , or [0 , ). Characterize Y -Tychonoff spaces X for which the function space C k ( X, Y ) is (sequentially) Ascoli. In two partial and important cases when Y = and X is a zero-dimensional metric space or Y = R and X is a metrizable space, Problem 5.16 is solved completely in [19] and [11], respectively(the clauses (a) in (i) and (ii) follow from Proposition 2.10 and the proofs of Lemma 3.2 in [18] andLemma 2.5 in [11], respectively). Theorem 5.17. (i) ([18]) If X is a paracompact space of point-countable type, then the followingassertions are equivalent: (a) C k ( X ) is sequentially Ascoli, (b) C k ( X ) is Ascoli, (c) C k ( X, I ) is sequentially Ascoli, (d) C k ( X, I ) is Ascoli, (e) X is locally compact. (ii) ([11]) For a metrizable zero-dimensional space X the following assertions are equivalent: (a) C k ( X, ) is sequentially -Ascoli, (b) C k ( X, ) is Ascoli, (c) X is locally compact or X is notlocally compact but the set X ′ of non-isolated points of X is compact. For the general case of an arbitrary Tychonoff space X we are able to prove the next necessarycondition. Proposition 5.18.
If a Tychonoff space X is such that C k ( X ) is sequentially Ascoli, then everycompact-finite sequence of compact subsets of X has a strongly compact-finite subsequence. Proof.
Let { K n } n ∈ ω be a compact-finite sequence of nonempty compact sets in X . We have toshow that there is a subsequence of { K n } n ∈ ω which is strongly compact-finite. For every n ∈ ω ,define a function F n on C k ( X ) by F n ( f ) := k f − ( n + 1) · k K n = sup (cid:8) | f ( x ) − ( n + 1) | : x ∈ K n (cid:9) . Since F n is the composition of the restriction map C k ( X ) → C k ( K n ) and the norm of the Banachspace C ( K n ), we obtain that F n is continuous. For every n ∈ ω , define A n := (cid:8) f ∈ C k ( X ) : F n ( f ) ≤ (cid:9) and O n := (cid:8) f ∈ C k ( X ) : F n ( f ) < (cid:9) . so A n is a functionally closed subset of C k ( X ) and O n is a functionally open neighborhood of A n .26 laim 1. ∈ S k ∈ ω A n k \ S n ∈ ω A n k , for every sequence n < n < · · · in ω . So each subsequenceof { A n } n ∈ ω is not locally finite. Indeed, let [ K ; ε ] be a standard neighborhood of the zero-function in C k ( X ), where K ⊆ X is compact and ε >
0. As { K n } n ∈ ω is compact-finite, there is an m ∈ ω such that K ∩ K n m = ∅ . Then any function h ∈ C k ( X ) such that h ↾ K = 0 and h ↾ K nm = n m + 1belongs to [ K ; ε ] ∩ A n m . The claim is proved.Now, since C k ( X ) is sequentially Ascoli, Claim 1 and (vi) of Theorem 2.7 imply that the sequence { O n } n ∈ ω is not compact-finite. Therefore there is a compact subset K of C k ( X ) such that the set J := { n ∈ ω : K ∩ O n = ∅} is infinite. For every j ∈ J , choose a function h j ∈ K ∩ O j and set U j := { x ∈ X : | h j ( x ) − ( j + 1) | < } . Note that U j is an open neighborhood of K j . Therefore to show that the sequence { K j } j ∈ J isstrongly compact-finite it is sufficient to prove that the sequence U = { U j } j ∈ J is compact-finite.To this end, fix a compact subset K of X and let π K : C k ( X ) → C k ( K ) be the restriction map. Set J := { j ∈ J : U j ∩ K = ∅} and suppose for a contradiction that J is infinite. For every j ∈ J ,set g j := π K ( h j ). Then g j ∈ π K ( K ), where π K ( K ) is a compact subset of the Banach space C ( K ).Therefore there is d > k g j k K ≤ d for every j ∈ J . However, since K ∩ U j = ∅ by oursupposition, we obtain that k g j k K ≥ j + → ∞ . This contradiction shows that J is finite. Thus U is compact-finite. (cid:3) We end the paper with the following problem.
Problem 5.19.
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