Abstract. We are concerned with C

148 (1995)

Classification of function spaces with the pointwise topology determined

by a countable dense set

by

Tadeusz D o b r o w o l s k i (Norman, Okla.) and Witold M a r c i s z e w s k i (Warszawa)

Abstract. We are concerned with C

(X), the space of continuous real valued func- tions on X considered with the topology of pointwise convergence on A, where A is a countable dense subset of X. We focus on the Borel and the topological classifications of the spaces C

(X). For example, we prove that for countable nondiscrete X, C

(X) is homeomorphic to σ

, the countable product of σ = {(x

) ∈ R

| x

= 0 a.e.}, provided C

(X) ∈ F

.

1. Introduction. All spaces under consideration are completely regu- lar. For a space X, C

(X) denotes the space of all continuous real valued functions on X with the pointwise convergence topology. If A is a dense subset of X, then by C

(X) we denote the space of continuous real valued functions on X with the topology of pointwise convergence on A. Hence, we have C

(X) = {f |A | f is continuous on X} ⊆ R

and C

(X) = C

(X).

Throughout this paper we will assume that A is countable; consequently, C

(X) is a dense linear subspace of R

, a countable product of lines.

Recently, a lot of work has been done on the topological classification of Borel and projective function spaces C

(X) for countable spaces X (for references see [DMM] and [CDM]). It has been shown that C

(X), while Borel, is always of an exact multiplicative class [CDM] and it is conjectured that the topological and the Borel classifications coincide [DMM]. For the countable spaces X the spaces C

(X) seem to be a natural generalization

1991 Mathematics Subject Classification: Primary 46E10, 57N20, 54C35.

Key words and phrases: function space, topology determined by a countable set, bor- elian filters.

Research of the second author partially supported by KBN grant 2 1113 91 01.

The results of this paper were presented by the second author in January 1993 at the 21st Winter School on Abstract Analysis, Podˇebrady, Czech Republic.

[35]

of the spaces C

(X). This paper is to initiate an investigation on the Borel complexity and the topological classification of the spaces C

(X). The Borel structure of the spaces C

(X) seems to be much more complicated than in the case of C

(X). However, we are able to indicate some similarities between the topological classifications of the spaces C

(X) and C

(X).

The spaces C

(X) also allow us to use some methods of descriptive set theory and infinite-dimensional topology (applicable for separable metriz- able spaces) for the investigation of the function spaces on uncountable separable spaces. In this context the spaces C

(X) have appeared in the literature in several natural situations. Before we describe some of them we have to recall a few notions.

A map f : M → N between separable metrizable spaces is of the first Baire class if f

(U ) is an F

-subset of M for every open U ⊂ N (if N is additionally a linear space then this means that f is the pointwise limit of a sequence of continuous maps M → N ). Let P be the space of irrationals.

It turns out that compact spaces that can be embedded in B

(P ), the space of real valued first Baire class functions on P with the topology of pointwise convergence, are of great importance in topology and Banach space theory (see [BFT] and [Ne, Section 1]); they are called Rosenthal compacta. The following result of Godefroy [Go, Theorem 4] shows how C

(X) spaces are involved in dealing with Rosenthal compacta.

1.1. A separable compact space K is a Rosenthal compactum if and only if for every countable dense subset A of K the space C

(K) is analytic (i.e., a continuous image of P ).

The problem of Borel classification of the spaces C

(K) for separable Rosenthal compact spaces K has been disscussed in [Ma1].

Another important fact involving C

(X) spaces is the following factor- ization result [Ma1, Lemma 3.4]:

1.2. Let X and Y be separable spaces and ϕ : C

(X) → C

(Y ) be a homeomorphism. For any countable sets C ⊆ X and D ⊆ Y there exist countable dense sets A ⊆ X and B ⊆ Y such that C ⊆ A, D ⊆ B and the map π

ϕπ

: C

(X) → C

(Y ) is a homeomorphism (π

: C

(X) → C

(X) and π

: C

(Y ) → C

(X) are the standard projections).

This result shows that the problem of the topological classification of the function spaces C

(X) for separable spaces X is related to the problem of the classification of the spaces C

(X).

Let us observe that the topology of C

(X) is precisely the weak topology

on C

(X) induced by the family of evaluation functionals at a, a ∈ A; this

family is countable, consists of continuous linear functionals, and separates

points of C

(X). Any continuous linear functional on C

(X) is a linear com-

bination of finitely many evaluation functionals (at x ∈ X). The following example shows that the weak topology on C

(X) induced by a countable family of continuous linear functionals which separate points of C

(X) is not necessarily that of C

(X) for some countable dense A ⊆ X.

1.3. Example. Let {q

}

be an enumeration of the rationals in the line R. The sequence {ϕ

}

of continuous linear functionals on C

(R) given by ϕ

(f ) = f (q

) + f (q

+ π/n), f ∈ C

(R), n = 1, 2, . . . , separates points of C

(R). However, every evaluation functional on C

(R) is discon- tinuous in the weak topology induced by {ϕ

}

on C

(R).

Recall that if M is a separable metrizable space and α is a countable ordi- nal then A

(M ) (resp., M

(M )) denotes the family of subsets of M that are Borel of additive (resp., multiplicative) class α. By A

(resp., M

) we de- note the class of spaces that are absolute Borel of additive (resp., multiplica- tive) class α. If A ∈ A

\M

(resp., M

\A

or A

∩M

\ S

β<α

(A

∪M

)), then we say that A is of the exact additive (resp., multiplicative or ambi- guous) class α. By P

, n ≥ 0, we denote the nth projective class. A map f : M → N between separable metrizable spaces is Borel of class α if f

(U ) ∈ A

(M ) for every open U ⊂ N (this means that f is Borel of class α = 1 precisely when f is of the first Baire class). The above terminology is that of [Kur].

In Section 2 we address some questions concerning the Borel (projective) structure of spaces C

(X). For instance, fixing X, we wonder how the Borel (projective) class of C

(X) will change when varying A. The following in- teresting question arises. What is the relationship between the exact Borel (projective) classes of C

(X), C

(X) and C

(A) for countable X? For each countable ordinal α, we provide an example of a countable space X

and a dense subset A

so that C

(X

) ∈ M

and C

(X

) 6∈ M

(see 2.6).

In Section 3 we deal with spaces X with exactly one nonisolated point.

Such spaces can be identified with N

= N ∪ {∞} topologized by isolating the points of N = {1, 2, . . .} and using the family {A ∪ {∞} | A ∈ F } as a neighborhood base at ∞, where F is a filter on N. In our considerations, if F is a filter on a countable set T , then we usually identify T and N and most often we let T = ω = {0, 1, . . .}. We will always assume that F contains the Fr´echet filter F

consisting of all cofinite sets in ω. We will write

c

= {f ∈ R

| ∀(ε > 0) (f

((−ε, ε)) ∈ F )}

and

C

= {f ∈ R

| ∃(x ∈ R)∀(ε > 0) (f

((x − ε, x + ε)) ∈ F )}.

It is known [Ma2, Lemma 2.1] that c

is (linearly) homeomorphic to C

(N

).

The space C

can be identified with C

(N

). Moreover, we have

C

= {f ∈ R

| ∀(ε > 0)∃(A ∈ F )∀(a

, a

∈ A) (|f (a

) − f (a

)| < ε)}.

In these definitions the space ω can be replaced by any countable infinite set T . A filter F on T will be treated as a subset of 2

, a copy of the Cantor set. We address the following question. What is the relation between the Borel (projective) exact class of F and that of C

? Our answer is contained in 3.4 and is less satisfactory compared to the case of c

(see [DMM]).

Section 5 is devoted to the topological identification of F

-spaces C

(X).

The main result states that for countable nondiscrete X and dense A ⊆ X, the space C

(X) is homeomorphic to σ

provided C

(X) ∈ F

; here σ = {(x

) ∈ R

| x

= 0 a.e.}. (This fact for A = X is the main result of [DMM].) The same result holds for (not necessarily countable) Fr´echet spaces X. Let us recall that X is a Fr´echet space if given a subset Y ⊂ X and a point x ∈ Y there exists a sequence of elements of Y that converges to x. The main tool to obtain these identification results is the method of absorbing sets (belonging to infinite-dimensional topology, see [BM]).

An application of this technique to function spaces was initiated by van Mill in [vM]. We do not explicitly refer to this method; instead, we use some factorization facts and rely on results of [CDM]. However, we were not able to avoid another notion of infinite-dimensional topology: so-called Z

-spaces. We devote Section 4 to Z

-spaces C

(X). Many of our results concern also spaces C

(X), subsets of C

(X) consisting of all bounded functions.

In Section 6 we settle the case of metrizable X. It turns out that for a metrizable X, C

(X) is analytic only if X is σ-compact (and then, it is homeomorphic to σ

).

In the last section we provide examples of spaces C

with arbitrarily high Borel complexity. This is achieved by employing filters F previously used in [CDM]. Here, as in [CDM], not only do we show that for every α ≥ 2 there exists a filter F such that C

∈ M

\ A

, but actually it follows that C

is an absorbing set for the class of M

. Hence, C

is homeomorphic to the standard, M

-absorbing model Ω

(see [BM]).

2. C

(X) versus C

(X). How much can they differ? For a space X and countable dense subsets A and B of X, we will be interested in how much the Borel (projective) classes of C

(X) and C

(X) can differ. The following estimate was established in [Ma1, Theorem 2.2].

2.1. Proposition. Let X be a Fr´echet space and let A, B ⊆ X be count- able dense sets. For every countable ordinal α and every n ∈ ω, we have:

(a) if C

(X) ∈ M

, then C

(X) ∈ M

,

(b) if C

(X) ∈ A

, then C

(X) ∈ A

,

(c) if C

(X) ∈ P

, then C

(X) ∈ P

.

In particular , for infinite ordinals α and n ∈ ω, the exact Borel (projective) classes of C

(X) and C

(X) coincide.

The following example shows that for non-Fr´echet spaces X the gap between Borel classes can be as big as we wish (for a space Z, βZ denotes the ˇ Cech–Stone compactification of Z).

2.2. Example. We have

(a) C

(βω) ∈ A

\M

(actually, C

(βω) can be identified with the space Σ = {(x

) ∈ R

| (x

) is bounded}),

(b) if p is an ultrafilter on ω (i.e., p ∈ βω \ω), then C

(βω) 6∈ P

∪P

(actually, C

(βω) contains a closed copy of the ultrafilter p).

The statement of 2.2(b) can be reversed in the following way.

2.3. Proposition. Let X be a compact space. Assume there are count- able dense sets A, B ⊆ X such that C

(X) is analytic and C

(X) is non- analytic. Then X contains a copy of βω.

P r o o f. Write S = C

(X) and consider i : X → R

defined by i(x)(f ) = arctan(f (x)), f ∈ S, x ∈ X.

It follows that i is an embedding and i(K) is norm-bounded (here, we con- sider the sup norm on the space of bounded functions on S). Moreover, for a ∈ A, i(a) is continuous on S. Using the Godefroy characterization of Rosenthal compacta and our assumption, we get i(K) * B

(S). Now, our result is a consequence of the following statement from [P, p. 34] (see also [BFT]): “For an arbitrary norm-bounded sequence {f

| j ∈ ω} of continu- ous functions on P (or, more generally, on an analytic space X) one and only one possibility occurs: either all pointwise accumulation points of {f

} are of first Baire class, or there exists a subsequence {f

| j ∈ ω} which behaves on some Cantor set T in P like the sequence of projections p

: ω

→ {0, 1}

(in particular, in the second case, all accumulation points of {f

| j ∈ ω} are non-Borel and the pointwise closure of the set {f

| j ∈ ω} is homeomorphic to the ˇ Cech–Stone compactification of the natural numbers”.

To get our assertion, enumerate A = {a

| j ∈ ω} and apply the above statement to the sequence {i(a

) | j ∈ ω}.

The following result is a direct consequence of 2.3 and 1.1.

2.4. Corollary. Let X be a compact space and let A be a countable dense subset of X. If C

(X) is analytic then either X is a Rosenthal com- pactum or else X contains a copy of βω.

It has been shown [CDM, Theorem 5.1] that for every countable space

X, the space C

(X), when Borel, is of an exact multiplicative class. In

connection with this and 2.2(a) we ask:

2.5. Question. Is there an example of a (countable) space X such that C

(X) ∈ A

\ M

, α > 1, for some countable dense set A ⊆ X?

Next we show that the gap between Borel classes of C

(X) = C

(X) and C

(X) can be large even for countable X.

2.6. Proposition. For every countable ordinal α, there exists a count- able space X

and a dense set A ⊆ X

such that C

(X

) ∈ M

and C

(X

) 6∈ M

.

P r o o f. We will use induction on α. Set X

= A = ω + 1. It is obvious that C

(X

) ∈ M

; by [DGM, Corollary 1.2], C

(X

) is homeomorphic to σ

and hence it belongs to M

\ A

. Let us distinguish x

= ω ∈ X

and B

= ω; note that B

is the set of all isolated points of X

and A = B

∪{x

}.

Suppose the spaces X

have been constructed (for all β < α), sets B

of all isolated points of X

and points x

∈ X

\ B

have been determined so that A

= B

∪ {x

} is a countable dense subset of X

so that C

(X

) 6∈

M

and C

(X

) ∈ M

. For every n ∈ ω, let α

= β provided α = β +1. If α is a limit ordinal, fix a sequence of ordinals {α

}

with α

< α, n ∈ ω, and sup

α

= α. Form a direct sum Y

= ( S

n∈ω

{n} × X

) ⊕ (ω + 1). Consider an equivalence relation R on Y

which identifies (n, x

) with n ∈ ω + 1.

We set X

= Y

/R to be the quotient space. Write q : Y

→ X

for the quotient map. We let

B

= {x ∈ X

| x is isolated in X

}

and x

= q(ω), where ω ∈ ω + 1. By the construction B

= S

n∈ω

q({n} × B

). Finally, we let A

= B

∪ {x

} and observe that A

is countable and dense in X

. (Note that X

and A

are the Arens spaces described in Examples 1.6.19 and 1.6.20 of [En], respectively).

Observe that f is continuous on X

if and only if f |q({n} × X

) and f |q(ω + 1), n ∈ ω, are continuous. It then inductively follows that C

(X

) ∈ M

. We now show that C

(X

) 6∈ M

. Consider the following subspaces E

and F

of C

(X

):

E

= {f ∈ C

(X

) ∩ {0, 1}

| f (x

) = 0}, F

= {f ∈ C

(X

) ∩ {0, 1}

| f (x

) = 1}.

Claim. Write G

= E

∪ F

. The pairs (G

, E

) and (G

, F

) are Wadge (2

, A

∪ M

(2

))-complete (i.e., for every C ∈ A

∪ M

(2

) there exists a map ϕ : 2

→ G

so that ϕ

(E

) = C or ϕ

(F

) = C, respec- tively).

We will provide an inductive proof. If α = 0, then E

and F

are copies of

the rationals and both E

and F

are dense in G

. Let C ∈ M

(2

) (i.e., C

is a closed subset of 2

). Pick a sequence {x

}

⊂ E

such that lim x

=

x

∈ F

. Find a map ϕ : 2

→ {x

| n ∈ ω} such that ϕ

({x

}) = C. If

C ∈ A

(2

) (i.e., C is open in 2

), apply the same argument interchanging the roles of E

and F

. Let C ∈ A

and pick C

∈ M

, C

⊆ C

, n ∈ ω, so that C = S

n∈ω

C

. We have E

= FP(G

, E

) and F

= FP(G

, F

), where for a sequence {(X

, A

)}

of pairs of spaces

FP(X

, A

) = n

(x

) ∈ Y

n∈ω

X

 x

∈ A

a.e.

o

(notation of [CDM, Section 8]). Using the inductive assumption, we can find ϕ

: 2

→ G

⊂ R

such that ϕ

(E

) = C

. Take ϕ = ∆ϕ

: 2

→ R∪

given by ϕ(p) = (ϕ

(p)) and observe that ϕ( S

n∈ω

C

) ⊂ FP(G

, E

) and ϕ(2

\ S

n∈ω

C

) ⊂ Q

n∈ω

F

⊂ F

(cf. [CDM, Proof of 8.1]). This settles the case where C ∈ A

. By symmetry the same argu- ment works for C ∈ M

.

An easy application of the Claim yields E

6∈ M

∪ A

. Since E

can be identified with the set

{f ∈ C

(X

) ∩ {0, 1}

| f (x

) = 0}, a closed subset of C

(X

), our assertion follows.

2.7. R e m a r k. Let us note that even for a compact X the Borel class of C

(X) can be as high as we wish. Such spaces X can be taken as Rosenthal compacta and were provided in [Ma1]. Let us indicate another way of finding such X. By [LvMP, Theorem 4.1], there exists a countable regular space A with exactly one nonisolated point such that the Borel class of C

(A) (and hence, of C

(A)) is as high as we wish. Let X = βA and observe that C

(X) = C

(A).

2.8. R e m a r k. Consider X = N

for a filter F on ω. Applying 3.3, we see that C

(N

) = C

contains a closed copy of F ; and hence can have as complicated Borel structure as we wish. At the same time C

(ω) = R

∈ M

.

The statements 2.6 and 2.8 show that, within the Borel hierarchy, the Borel class of C

(X) cannot be estimated by either the class of C

(X) or C

(A).

2.9. Question. Let X be a countable space and let A be a dense subset of X. Is the Borel (projective) class of C

(X) (resp., C

(A)) determined by that of C

(X)? Is C

(X) Borel (analytic) if C

(X) is? Can the exact Borel (projective) class of C

(X) be greater than that of C

(X)?

2.10. R e m a r k. Let π

: C

(X) → C

(X) be the projection map. Since

C

(X) = π

(C

(X)) and π

is injective, C

(X) is Borel provided C

(X)

is (see [Kur]). We claim that also C

(A) is analytic provided C

(X) is. The

latter can be shown as follows. If C

(X) is analytic, then for every accumu-

lation point a ∈ A the filter F

= {Y ∩A | Y ∈ 2

and a ∈ Int(Y ∪{a})}

is an analytic filter on A \ {a} (cf. the proof of [DMM, Corollary 3.6]). If now by A

we denote the space A topologized by isolating the points of A \ {a} and by using the family {A ∪ {a} | A ∈ F

} as neighborhood base at a, we have C

(A) = T

{C

(A

) | a is an accumulation point of A}.

Since each C

(A

) is analytic, so is C

(A) (this argument was taken from [DMM, Lemma 4.3]).

Let us notice that the last two questions of 2.9 have negative answers for bounded function spaces.

2.11. Example. For any ultrafilter p ∈ βω \ ω we have, C

(ω ∪ {p}) = Σ ∈ A

\ M

and C

(ω ∪ {p}) 6∈ P

∪ P

.

Observe that, for every p from Example 2.11, we also have C

(ω ∪{p}) 6∈

P

∪ P

. Hence, we see that the gap between Borel classes of C

(X) and C

(X) can be as big as we wish.

3. Borel type of spaces C

. For the Fr´echet filter F

on ω we use the classical functional analysis symbols c

and c to denote the spaces c

and C

. The spaces c

and c considered as Banach spaces (with the sup norm) are linearly isomorphic. The following fact shows a dramatic difference if one considers c

and c as subspaces of R

.

3.1. Proposition. The spaces c

, c ⊂ R

are not linearly isomorphic.

P r o o f. Assume T : c

→ c establishes a linear topological isomorphism of c

onto c. Let k · k be the sup-norm on c (and c

). First we note that the graph of T ,

Γ = {(x, T x) | x ∈ c

} ⊂ c

× c,

is closed in the norm topology. This follows from the obvious fact that Γ is closed with respect to the coordinatewise topology on c

×c and the fact that the latter topology is weaker than the norm topology on c

×c. Consequently, by the Closed Graph Theorem, T establishes a linear isomorphism of (c

, k·k) onto (c, k · k).

Consider the continuous linear functional ` : (c, k · k) → R given by

`((x

)) = lim x

. It follows that `◦T : (c

, k·k) → R is continuous and `◦T 6≡

0. This obviously implies that ` ◦ T is continuous with respect to the weak

topology ω on c

. Consequently, for every convex closed neighborhood U of

0 ∈ (c

, k·k), ker(`◦T )∩U is closed in the ω-topology, and ker(`◦T )∩U 6= U .

Let B = {y ∈ c | kyk ≤ 1} be the unit ball in c. We will show that for

U = T

(B), ker(` ◦ T ) ∩ U is dense in U in the ω-topology, contradicting

the above fact. To this end, first notice that given y ∈ B, there exists a

sequence {y

}

⊂ B ∩ c

so that {y

}

converges to y in R

. We see

that {T

y

}

converges to T

y in R

and T

y

∈ ker(` ◦ T ). Since

the ω-topology coincides with the R

-topology on bounded sets, it follows that {T

y

} converges to T

y in the ω-topology.

Before we enter a discussion on Borel types of C

, let us formulate the following general fact which will be employed later on; its proof can be obtained easily by using the argument of the proof of [Ma2, Lemma 2.1].

3.2. Lemma. Let F be a filter on ω which is not the Fr´echet filter. Then C

is homeomorphic to the product C

× R

.

We identify a filter F on a countable set T with a subset of a Cantor set 2

. As shown in [DMM, Lemma 4.2] the exact Borel class of c

is entirely determined by that of F in 2

. Here we try to recover this result for the space C

. The first result shows that Borel (projective) classes of C

and C

are not lower than that of F .

3.3. Proposition. Let F be a filter on a countable set T . For every countable ordinal α and n ∈ ω, we have:

(a) if C

∈ M

, then F ∈ M

, (b) if C

∈ A

, then F ∈ A

, (c) if C

∈ P

, then F ∈ P

.

In (a)–(c), the space C

can be replaced by C

provided F is not an ultra- filter.

P r o o f. We may assume T = ω. Consider

Z = {f ∈ C

| ∀(n ∈ ω) (f (n) = 0 or f (n) = n + 1)},

a closed subset of C

. The map (x

) → ((n + 1)(1 − x

)) embeds 2

in R

and sends F onto Z. Thus, the C

-part of our assertion follows.

Assume that F is not an ultrafilter on ω (see 2.11). There exists a subset M ⊂ ω such that neither M nor N = ω \ M belongs to F . Consider the filters

G = {A ∩ M | A ∈ F } and H = {A ∩ H | A ∈ F } induced by F on M and N , respectively. The sets

Y

= {f ∈ C

∩ {0, 1}

| f |N = 0}, Y

= {f ∈ C

∩ {0, 1}

| f |M = 0}

are closed subsets of C

. It is clear that G and H are homeomorphic to Y

and Y

, respectively. Finally, since F can be identified with the product G × H, our assertion follows.

Here is our counterpart of [DMM, Lemma 4.2] for spaces C

.

3.4. Proposition. Let F be a filter on ω. For every countable ordinal

α ≥ 1 and n ∈ ω, we have:

(a) if F ∈ A

, then C

∈ M

,

(b) if F ∈ M

, then C

∈ M

∩ A

(more exactly, C

is a differ- ence of two sets belonging to M

),

(c) if F ∈ P

, then C

∈ P

.

P r o o f. First we provide an argument which shows (a) and (c). Then we adapt this argument to check that whenever F ∈ M

then C

∩ [0, 1]

∈ M

. Finally, since C

is homeomorphic to

{f ∈ C

| ∀(n ∈ ω) (f (n) ∈ (0, 1) and lim

F

f (n) ∈ (0, 1))}

= (C

∩ [0, 1]

) ∩ (0, 1)

\ (c

∪ (1 + c

)), the space C

is a difference of two sets belonging to M

(by [DMM, Lem- ma 4.2], c

and hence 1 + c

= {(x

+ 1) ∈ R

| (x

) ∈ c

} belong to M

).

Assume F ∈ A

(resp., F ∈ P

). For n ∈ ω and k ∈ Z, write T

=

 2k

2n + 2 , 2k + 1 2n + 2

 . Set T

= S

k∈Z

T

. For m ∈ N and l ∈ Z, define S

= [

k

 T

    dist

 l m , T



≤ 1 m

 .

Let %

: Q

n∈ω

T

→ 2

be given by

%

(f )(n) =

 1 if f (n) ∈ S

, 0 otherwise, for f ∈ Q

n∈ω

T

. One can check that %

is continuous. We claim that

(1) C

∩ Y

n∈ω

T

=

\

[

l∈Z

%

(F ).

(If f = (f (n)) ∈ Q

n∈ω

T

and there exists x ∈ R such that for all ε > 0 we have f

((x − ε, x + ε)) ∈ F , then for every m ≥ 1 one can find l ∈ Z so that |x − l/m| ≤ 1/(2m) and check that {n | f (n) ∈ S

} ∈ F ; this shows %

(f ) ∈ F . Conversely, if for every m ≥ 1 one can find l ∈ Z so that {n | f (n) ∈ S

} ∈ F , then there exists A ∈ F so that whenever i, j ∈ A then |f (i) − f (j)| < 4/m; this implies f ∈ C

.)

Pick a map ψ

: T

→ R that transforms T

affinely onto 

2n+2

,

 , k ∈ Z. Letting ψ = Q

n∈ω

ψ

: Q

n∈ω

T

→ R

we see that ψ is a perfect map and that

(2) C

∩ Y

n∈ω

T

= ψ

(C

).

Applying (1), C

∩ Q

n∈ω

T

∈ M

(resp., C

∩ Q

n∈ω

T

∈ P

). By (2)

and a result of Saint Raymond [SR1], C

∈ M

(resp., C

∈ P

).

Now, let F ∈ M

. We must show that C

∩ [0, 1]

∈ M

. Using the sets T

we define

T

=

n−1

[

k=0

T

.

With these T

, as previously, define for m ∈ N and l ∈ {0, . . . , m} the sets S

and maps %

. One can check

(1)

C

∩ Y

n∈ω

T

=

\

[

%

(F ).

Let ψ

: T

→ [0, 1] be a map such that T

is affinely transformed onto



2n+2

,



. Letting ψ = Q

n∈ω

ψ

: Q

n∈ω

T

→ [0, 1]

, we have

(2)

C

∩ Y

n∈ω

T

= ψ

(C

∩ [0, 1]

).

Using (1)

, C

∩ Q

n∈ω

T

∈ M

. As above, by (2)

and a result of Saint Raymond [SR1], C

∩ [0, 1]

∈ M

.

3.5. Question. Let F ∈ M

be a filter on ω. Is C

∈ M

?

Let F be a filter on ω. Identify the space C

with C

(N

). Using this identification, let ` : C

→ R be given by `(f ) = f (∞). We see that ` is well defined and can be viewed as `(f ) = lim

f (n).

3.6. Proposition. For a filter F ∈ A

∪ M

, and a countable ordinal α, α ≥ 1, ` : C

→ R is Borel of class α.

P r o o f. Let U ⊆ R be an open set. We must show that `

(U ) ∈ A

(C

).

We shall use some of the notation employed in the proof of 3.4. For n ∈ ω and m ∈ N, let

P

= [

{T

| ∀(t ∈ T

) (dist(t, R \ U ) > 1/m)}.

Recall that T

= S

k∈Z

T

and define ζ

: Q

n∈ω

T

→ {0, 1}

by letting for f ∈ Q

n∈ω

T

,

ζ

(f )(n) =

n 1 if f (n) ∈ P

, 0 otherwise.

Clearly, ζ

is continuous, m ≥ 1. Let ψ : Q

n∈ω

T

→ R

be that of the proof of 3.4.

Suppose F ∈ A

. Consider S = S

m=1

ζ

(F ). We see that S ∈ A

(S ∈ A

( Q

n∈ω

T

) for α = 1). We claim that ψ

(ψ(S)) = S. Let f, g ∈ Q

n∈ω

T

be such that g ∈ S

m=1

ζ

(F ) and ψ(f ) = ψ(g). Since ψ(f ) = ψ(g), we have |g(n) − f (n)| < 1/n. Take m

such that g ∈ ζ

(F ). If n > 2m

and g(n) ∈ P

, then f (n) ∈ P

. It follows that ζ

(f ) ⊃ ζ

(g) \ {0, 1, . . . , 2m

}; and hence ζ

(f ) ∈ F . This shows that f ∈ ζ

(F ) ⊂ S.

Suppose F ∈ M

. Let J = {A ⊂ ω | ω \ A ∈ F } be the dual ideal of F ; J is homeomorphic to F . Consider T = ( Q

n∈ω

T

) \ T

m=1

ζ

(J ). We see that T ∈ A

. We claim that ψ

(ψ(T )) = T . Let f, g ∈ Q

n∈ω

T

be such that f 6∈ T

m=1

ζ

(J ) and ψ(f ) = ψ(g). Take m

such that f 6∈ ζ

(J ).

If n > 2m

and f (n) ∈ P

, then g(n) ∈ P

. As previously, it follows that ζ

(g) 6∈ J ; hence g ∈ T .

We now claim that

(1) C

∩ S = C

∩ T =  Y

n∈ω

T



∩ `

(U ).

Namely, we have f ∈ S (resp., f ∈ T ) if and only if there exists m ≥ 1 such that f ∈ ζ

(F ) (resp., f 6∈ ζ

(J )). Suppose f ∈ ( Q

n∈ω

T

) ∩ `

(U ) and f (∞) = x ∈ U . Let ε = dist(x, R \ U ). For m, n > 4/ε we have (x − ε/2, x + ε/2) ∩ T

⊆ P

. It follows that f ∈ ζ

(F ) and hence f 6∈ ζ

(J ). Suppose now that f ∈ C

and f ∈ ζ

(F ). Since f 6∈ ζ

(J ) the set R = {n | f (n) ∈ P

} is not in J . Take x 6∈ U . Then for every n ≥ 1, (x − 1/(2m), x + 1/(2m)) ∩ P

= ∅. Therefore for Q = {n | f (n) ∈ (x − 1/(2m), x + 1/(2m))} we have R ∩ Q = ∅. Since R 6∈ J , we have Q 6∈ F . It follows that f (∞) 6= x, so f (∞) ∈ U .

Since lim

(f (n)−ψ(f )(n)) = 0, one sees that lim

f (n) = lim

ψ(f )(n);

and consequently

`

(U ) = ψ  Y

n∈ω

T



∩ `

(U )

 .

This together with (1) and the fact that ψ

(C

) = C

∩ Q

n∈ω

T

yields

`

(U ) = ψ(C

∩ S) = C

∩ ψ(S),

`

(U ) = ψ(C

∩ T ) = C

∩ ψ(T ).

The facts that ψ

(ψ(S)) = S and ψ

(ψ(T )) = T together with a result of Saint Raymond [SR1] show that, under our assumption, C

∩ ϕ(S) and C

∩ ϕ(T ) belong to A

(C

).

The following is a partial answer to Question 3.5.

3.7. Corollary. For n ∈ ω, let F

be a filter on ω with F

∈ S

β<α

A

, α ≥ 1. Then for the filter F = Q

n∈ω

F

on ω × ω, we have C

∈ M

. P r o o f. We obviously have

F = {A ⊂ ω × ω | ∀(n ∈ ω) ({m | (m, n) ∈ A} ∈ F

)}.

Moreover, for f = (f (m, n)) ∈ R

, f ∈ C

if and only if f (·, n) ∈ C

, n ∈ ω, and lim

f (·, n) = lim

f (·, k) for n, k ∈ ω. Now our assertion follows by application of 3.4 and 3.6.

The assertion of Proposition 3.9 below shows that the result of [DDM, Lemma 4.2] is fully recovered for the spaces C

. The proof of 3.9 requires a particular case of the following general fact.

3.8. Proposition. Let X be a separable metrizable space such that X = S

n=1

X

, X

∈ M

, α ≥ 2 and X

∈ A

(X), where β < α. Then X ∈ M

. P r o o f. Let Y be a metrizable compactification of X and let Y

∈ A

be such that Y

∩ X = X

. Consider Z = S

n=1

Y

∈ A

⊆ M

. We have Y

\X

∈ A

; hence, S

n=1

(Y

\X

) ∈ A

and X = Z\ S

n=1

(Y

\X

) ∈ M

.

3.9. Proposition. Let F be a filter on ω, α be a countable ordinal, α ≥ 1, and let n ∈ ω. We have:

(a) if F ∈ A

\ M

, then C

∈ M

\ A

, (b) if F ∈ M

\ A

, then C

∈ M

\ A

, (c) if F ∈ M

∩ A

\ S

β<α

(A

∪ M

), then C

∈ M

\ A

, (d) if F ∈ P

, then C

∈ P

.

P r o o f. The proof of the second part of 3.4 yields that whenever F ∈ M

, C

∩ [−n, n]

∈ M

for n ≥ 1. Since C

= S

n=1

(C

∩ [−n, n]

) and each C

∩ [−n, n]

is closed in C

, it follows from 3.8 that C

∈ M

(observe that here α ≥ 2 since there are no filters of class M

). If additionally F 6∈ A

then, by 3.3, C

∈ M

\A

; hence (b) follows. Assume F ∈ M

∩A

\ S

β<α

(A

∪M

). Consider the decomposition F = G×H of F into filters G and H described in the proof of 3.3 (obviously, being Borel, F is not an ultrafilter). We may assume that G ∈ M

∩A

\ S

β<α

M

(if not, H has this property). Recall that Σ = {(x

) ∈ R

| (x

) is bounded}. Applying [DMM, Lemma 4.2(3)], we find that c

∈ M

\ A

. Since c

= c

∩ Σ and c

is homeomorphic to c

∩ (−1, 1)

, we conclude that c

∈ M

\ A

. Finally, since C

contains c

as a closed set, it follows that C

∈ M

\ A

; this shows (c).

To get (a), suppose F ∈ A

\ M

. Since C

= C

∩ Σ, 3.4(a) yields C

∈ M

. Considering, as previously, F = G × H, we may assume G ∈ A

\ M

. By [DDM, Lemma 4.2(4)], c

∈ M

\ A

. As before, this implies that c

∈ M

\ A

and consequently C

∈ M

\ A

.

The assertion (d) follows from 3.4(c) and the fact that C

= C

∩ Σ.

We now provide a counterpart of [DMM, Lemma 4.3]. Let X be a count-

able space. For every accumulation point a of X we consider the filter F

on X \ {a} consisting of all neighborhoods of a; hence F

= {Y ∈ 2

| a ∈ Int(Y ∪ {a})}.

3.10. Lemma. Let X be a countable space and A ⊆ X be a dense set.

If there exists a countable ordinal α, α ≥ 1 (resp., an integer n ∈ ω) such that for every accumulation point a ∈ X the filter F

is in A

∪ M

(resp., F

∈ P

), then C

(X) ∈ M

(resp., C

(X) ∈ P

).

P r o o f (cf. [DMM, Lemma 4.3]). Let a be an accumulation point of X.

Define a filter on A by letting

G

= {Y ∩ A | Y ∈ F

}.

The filter G

is homeomorphic to

{Y ∈ F

| X \ (A ∪ {a}) ⊆ Y },

a closed subset of F

. Therefore G

∈ A

∪ M

(resp., G

∈ P

). Let X

be the space A ∪ {a} topologized by isolating the points of A \ {a} and using the family {Y ∪ {a} | Y ∈ G

} as a neighborhood base at a (it may happen that a ∈ A). Then either C

(X

) is isomorphic to c

(in case a ∈ A), or else C

(X

) is isomorphic to C

(in case a 6∈ A). From Proposition 3.4 and [DMM, Lemma 4.2] (see also [CDM, Corollary 5.3(d)]) it follows that in both cases C

(X

) ∈ M

(resp., C

(X

) ∈ P

). Now, 3.10 is a consequence of the following observation:

C

(X) = \

{C

(X

) | a is an accumulation point of X}.

3.11. R e m a r k. Let X

be the space from Proposition 2.6. Then C

(X

)

∈ F

. Moreover, according to Lemma 3.10, there exists a ∈ X

such that F

6∈ S

β<α

A

∪ M

; in fact, one can show that a = x

has this prop- erty. (As shown in [DMM, Corollary 3.6] the filters F

are analytic provided C

(X) is analytic.) Observe that the filter F

has a base which is of the F

-type. Namely, the filter base {Y ⊂ X \ {a} | Y ∪ {a} is a clopen subset of X containing a} is homeomorphic to

{f ∈ C

(X

) ∩ {0, 1}

| f (a) = 1}, a closed subset of C

(X

).

4. Z

-property of C

(X). We recall that a closed subset A of an absolute neighborhood retract M is a Z-set if every map f : K → M of a compactum K into M can be approximated by maps f : K → M \ A.

A space which is a countable union of its own Z-sets is called a Z

-space.

Obviously, every Z

-space is of the first category. It is a well-known fact

that a dense convex subset of a convex Z

-space is itself a Z

-space. Since

C

(X) is dense in R

and since Σ = {(x

) ∈ R

| (x

) is bounded} is a

Z

-space, we have:

4.1. Lemma. Let X be a space and let A be a countable dense subset of X.

(a) The space C

(X) is a Z

-space.

(b) If X is compact, then C

(X) is a Z

-space.

4.2. Proposition. Let X be a countable nondiscrete space and let A be a dense subset of X. If C

(X) is analytic, then C

(X) and C

(X) are Z

-spaces.

Let us recall that a filter F on a countable set T is of the first category if F belongs to the σ-field generated by the open sets and the first category sets of 2

. We need the following fact.

4.3. Lemma. Let F be a first category filter on a countable set T . Then for every 0 < r ≤ ∞ the space C

∩ [−r, r]

is a Z

-space.

P r o o f. By a result of Talagrand [Ta, Theorem 2.1] there exists a se- quence of pairwise disjoint finite sets A

⊂ T such that every A ∈ F meets all but finitely many A

, n ∈ N.

For finite r, let

X

= {f ∈ C

∩ [−r, r]

| ∀(k > n)∃(m ∈ A

) (|f (m)| ≤ 2r/3)}

and

Y

= {f ∈ C

∩ [−r, r]

| ∀(k > n)∃(m ∈ A

) (|f (m)| ≥ r/3)}.

An argument of the proof of [DMM, Proposition 3.3] shows that each X

and Y

, n ≥ 1, is a Z-set. Moreover, C

∩ [−r, r]

= S

n=1

X

∪ Y

. For r = ∞, let

X

= {f ∈ C

| ∀(k > n)∃(m ∈ A

) (|f (m)| ≤ l)}.

As previously, each X

is a Z-set in C

and C

= S

n,l=1

X

.

P r o o f o f 4.2. 1

Assume A has an accumulation point a ∈ A. Consider E = {f ∈ C

(X) ∩ {0, 1}

| f (a) = 1},

a closed subset of C

(X). The space E can be identified with F

= {U ∩A | U is clopen and a ∈ U }. Let F be the filter on A \ {a} generated by F

, i.e.,

F = {B ⊂ A \ {a} | ∃(f ∈ E) (f

({1}) ⊂ B ∪ {a})}.

By an argument of [DMM, Corollary 3.6], F is an analytic filter and con- sequently it belongs to the σ-field generated by the open sets and the first category sets of 2

. By 4.3, C

is a Z

-space. Since C

(X) is a dense linear subspace of C

× R ⊂ R

× R = R

, C

(X) is a Z

-space.

2

Assume A is discrete. For every x ∈ X \A, let F

= {U ∩A | x ∈ Int U } be a filter of neighborhoods of x.

Claim. There exists x

∈ X \ A such that F

is of the first category.