PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 126, Number 2, February 1998, Pages 471–479 S 0002-9939(98)04098-2

CARDINAL INVARIANTS CONCERNING BOUNDED FAMILIES OF EXTENDABLE AND ALMOST CONTINUOUS FUNCTIONS

KRZYSZTOF CIESIELSKI AND ALEKSANDER MALISZEWSKI

(Communicated by J. Marshall Ash)

Abstract. In this paper we introduce and examine a cardinal invariant A_{b} closely connected to the addition of bounded functions from ℝ to ℝ. It is analogous to the invariant A defined earlier for arbitrary functions by T. Natkaniec. In particular, it is proved that each bounded function can be written as the sum of two bounded almost continuous functions, and an example is given that there is a bounded function which cannot be expressed as the sum of two bounded extendable functions.

1. Preliminaries

We will use the following terminology and notation. The letters ℕ, ℚ, and ℝ denote the set of positive integers, the set of rationals, and the real line, respectively. Functions will be identified with their graphs. The family of all functions from a set X into Y will be denoted by Y^{X}. The symbol |X| will stand for the cardinality of a set X. We consider cardinals as ordinals not in one-to-one correspondence with the smaller ordinals. For a cardinal number κ we will write cof(κ) for the cofinality of κ. A cardinal number κ is called regular, if κ = cof(κ). We define 𝔠 = |ℝ| and ω = |ℕ|. For a set A ⊆ ℝ its characteristic function is denoted by χ_{A}. The projection of a set A ⊆ ℝ^{2} onto the x-axis will be denoted by dom A. For a cardinal number κ ≤ 𝔠 and an open set U ⊆ ℝ^{n} we say that X ⊆ U is κ-dense in U if |X ∩ V | ≥ κ for every nonempty open set V ⊆ U.

The symbols Bd and BD will denote the class of all bounded functions from ℝ to ℝ and the class of all families 𝔄 ⊆ ℝ^{ℝ} of functions having common bound, respectively. For F ⊆ ℝ^{ℝ} we define two cardinal invariants (cf. [12] and [6]):

A(F) = min{|𝔄|: 𝔄 ⊆ ℝ^{ℝ} & ¬(∃g ∈ ℝ^{ℝ})(∀f ∈ 𝔄)(f + g ∈ F)} ∪ {(2^{𝔠})^{+}}
= min{|𝔄|: 𝔄 ⊆ ℝ^{ℝ} & (∀g ∈ ℝ^{ℝ})(∃f ∈ 𝔄)(f + g ∉ F)} ∪ {(2^{𝔠})^{+}},

and

A_{b}(F) = min{|𝔄|: 𝔄 ∈ BD & ¬(∃g ∈ Bd)(∀f ∈ 𝔄)(f + g ∈ F)} ∪ {(2^{𝔠})^{+}}
= min{|𝔄|: 𝔄 ∈ BD & (∀g ∈ Bd)(∃f ∈ 𝔄)(f + g ∉ F)} ∪ {(2^{𝔠})^{+}}.
\begin{matrix}c \\ ◯ \end{matrix}1998 American Mathematical Society

Received by the editors March 28, 1996 and, in revised form, August 11, 1996.

1991 Mathematics Subject Classification. Primary 26A21; Secondary 54C08.

Key words and phrases. Peripheral continuity, almost continuity, connectivity, extendability.

This work was partially supported by NSF Cooperative Research Grant INT-9600548.

The next proposition lists several basic properties of the function A_{b}. The similar properties for the function A can be found in [6]. They will be left without proofs. (Their easy proofs are analogous to those of Propositions 1.1 and 1.3 of [6].)

Proposition 1.1. Let F ⊆ ℝ^{ℝ}. Then (1): A_{b}(F) = A_{b}(F ∩ Bd); (2): A_{b}(F) ≥ 1; (3): if F ⊆ G ⊆ ℝ^{ℝ}, then A_{b}(F) ≤ A_{b}(G); (4): A_{b}(F) ≤ 2 if and only if {f_{1} − f_{2} : f_{1}, f_{2} ∈ F ∩ Bd} ≠ Bd; (5): A_{b}(F) ≥ 2 if and only if F ∩ Bd ≠ ∅; (6): if F ⊇ Bd, then A_{b}(F) = (2^{𝔠})^{+}.

We will find the value of the function A_{b} for the following classes of functions, where κ is a cardinal number ≤ 𝔠 and X is an arbitrary topological space.

PC_{κ}: of all functions f: ℝ → ℝ with the following property: for every x ∈ ℝ and every ε > 0 we have |f ∩ [(x − ε, x) × (f(x) − ε, f(x) + ε)]| ≥ κ and |f ∩ [(x, x + ε) × (f(x) − ε, f(x) + ε)]| ≥ κ. In particular, PC_{ω} is the class of peripherally continuous functions.

D: of all Darboux functions f: ℝ → ℝ, i.e., such that f[J] is an interval for

every interval J ⊆ ℝ. U_{κ}: of all functions f: ℝ → ℝ fulfilling the following condition: for all a < b and

each set A ⊆ [a, b] with |A| < κ+1, the set f[(a, b)\A] is dense in the interval (min{f(a), f(b)}, max{f(a), f(b)}). In particular, U_{𝔠} is the uniform closure of the class D [2], while the class U_{0} was examined earlier by T. Radakovič [14] and H. W. Ellis [8]. Moreover U_{𝔠} = PC_{𝔠} ∩ U_{0} [2, Theorem 3.2].

AC: of all almost continuous functions f: ℝ → ℝ, i.e., such that for every open

set U ⊆ ℝ^{2} containing f there is a continuous function h: ℝ → ℝ with h ⊆ U. Conn(X): of all connectivity functions f: X → ℝ, i.e., such that the restric-

tion f ↾ C (that is f ∩ [C × ℝ]) is connected in X × ℝ whenever C ⊆ X is connected.^{1}

Ext: of all extendable functions f: ℝ → ℝ, i.e., such that there exists a function

g ∈ Conn(ℝ × [0, 1]) with f(x) = g(x, 0) for every x ∈ ℝ. PR: of all functions f: ℝ → ℝ with perfect road, i.e., such that for every x ∈ ℝ

there exists a perfect set P ⊆ ℝ having x as a bilateral limit point for which the restriction f ↾ P is continuous at x.

For the above classes of functions we have the following proper inclusions, marked by arrows.

[figure: Ext; AC; Conn; D; U_{𝔠}; ⋯; U_{ω} = ⋯ = U_{0}; PR; PC_{𝔠}; ⋯; PC_{ω} =⋯=PC_{1}; PC_{0} =ℝ^{ℝ}]

For the upper row inclusions see, e.g., [1]. The inclusion Ext ⊊ PR follows from [9]. The other relations are more or less evident.

1Actually we will study only the class Conn = Conn(ℝ), but we need the class Conn(ℝ × [0, 1])

to define the class Ext.

The values of the function A for most of these classes have been established in several papers, as quoted below.

Theorem 1.2 ([6]). A(PC_{ω}) = 2^{𝔠} and A(Ext) = A(PR) = 𝔠^{+}.

Theorem 1.3 ([5]). 𝔠^{+} ≤ A(AC) = A(Conn) = A(D) ≤ 2^{𝔠}, cof(A(D)) > 𝔠, and it is pretty much all that can be shown in ZFC. More precisely, the assertion A(D) = 2^{𝔠} is independent of the cofinality of 2^{𝔠}, and for each regular cardinal λ between 𝔠^{+} and 2^{𝔠} it is consistent with ZFC that A(D) = λ.

The values of the function A for the other classes have not been considered so far. However, it is not difficult to find them.

Theorem 1.4. A(PC_{0}) = (2^{𝔠})^{+} and A(U_{κ}) = A(PC_{κ}) = 2^{𝔠} for every infinite cardinal number κ ≤ 𝔠.

The equality A(PC_{0}) = (2^{𝔠})^{+} is obvious. The other part will be proved in the next section.

It follows from Proposition 1.1(6) that A_{b}(PC_{0}) = (2^{𝔠})^{+}. It has been proved in [11] that A_{b}(D) = cof(𝔠). The goal of this paper is to establish the values of A_{b} for all other classes by proving the following theorem.

Theorem 1.5. (1): A_{b}(Ext) = A_{b}(PR) = 2.

(2): A_{b}(AC) = A_{b}(Conn) = A_{b}(D) = A_{b}(U_{𝔠}) = A_{b}(PC_{𝔠}) = cof(𝔠).

(3): A_{b}(U_{κ}) = A_{b}(PC_{κ}) = 𝔠 for every infinite cardinal number κ < 𝔠.

In particular, the equality A_{b}(AC) = cof(𝔠) solves Problem 2 of [10]. (This problem was restated in [11, p. 679].) Notice also that Theorem 1.5 and Proposition 1.1(4) imply immediately the following corollaries.

Corollary 1.6. Every bounded function f: ℝ → ℝ is the sum of two bounded almost continuous functions.

Corollary 1.7. There exists a bounded function f: ℝ → ℝ which is not the sum of two bounded functions with perfect road.

Notice that Corollary 1.6 generalizes the theorem of U. B. Darji and P. D. Humke that every bounded function is the sum of three bounded almost continuous functions [7]. On the other hand, by Corollary 1.7, T. Natkaniec’s result asserting that every bounded function is the sum of three bounded extendable functions [13, Theorem 1] cannot be improved.

In what follows we will use a generalization of [4, Lemma 4, p. 285] and [3, Lemma 1]. We will need the following notation. For every λ > ω and every function f: ℝ → ℝ let

A_{λ}^{−}(f) = {x ∈ ℝ: (∃ε > 0)(|f ∩ [(x − ε, x) × (f(x) − ε, f(x) + ε)]| < λ)},
A_{λ}^{+}(f) = {x ∈ ℝ: (∃ε > 0)(|f ∩ [(x, x + ε) × (f(x) − ε, f(x) + ε)]| < λ)},

and A_{λ}(f) = A_{λ}^{−}(f) ∪ A_{λ}^{+}(f).

Lemma 1.8. If λ is a cardinal number with cof(λ) > ω, then |A_{λ}(f)| < λ for every function f: ℝ → ℝ.

Proof. Assume, by way of contradiction, that |A_{λ}(f)| ≥ λ. Suppose that |A_{λ}^{−}(f)| ≥ λ, the case |A_{λ}^{+}(f)| ≥ λ being similar. Since cof(λ) > ω and

A_{λ}^{−}(f) = ⋃_{n∈ℕ}{x ∈ ℝ: |f ∩ [(x − n^{−1}, x) × (f(x) − n^{−1}, f(x) + n^{−1})]| < λ},

there exist an ε > 0 and a subset B of A_{λ}^{−}(f) of cardinality ≥ λ such that |f ∩[(x−ε, x)×(f(x)−ε, f(x)+ε)]| < λ for every x ∈ B. Choose an open interval J of length ε such that the set C = {x ∈ B: f(x) ∈ J} has cardinality greater than or equal to λ. Then, there is an x_{0} ∈ C with |(x_{0} − ε/2, x_{0}) ∩ C| ≥ λ. But f(x_{0}) ∈ J, so J ⊆ (f(x_{0}) − ε, f(x_{0}) + ε) and |f∩[(x_{0} − ε, x_{0}) × (f(x_{0}) − ε, f(x_{0}) + ε)]| ≥ λ, contradicting the fact that x_{0} ∈ B.

2. Proof of Theorem 1.4

Let κ ≤ 𝔠 be an infinite cardinal. It follows from the monotonicity of the function A and Theorem 1.2 that A(U_{𝔠}) ≤ A(U_{κ}) ≤ A(PC_{κ}) ≤ A(PC_{ω}) = 2^{𝔠}. Thus, it is enough to show that A(U_{𝔠}) ≥ 2^{𝔠}. This can be established by the following minor modification of the proof of [6, Theorem 1.7(3)].

Let F ⊆ ℝ^{ℝ} be such that |F| < 2^{𝔠}. We will find a function g: ℝ → ℝ such that f + g ∈ U_{𝔠} for every f ∈ F.

Let 𝔊 be the family of all pairs ⟨I, J⟩, where I and J are nonempty open intervals with rational end points. Let {⟨I_{ξ}, J_{ξ}⟩: ξ < 𝔠} be an enumeration of 𝔊 with each pair appearing 𝔠 many times. For each ξ < 𝔠 define a set B_{ξ} ⊆ I_{ξ} of cardinality 𝔠 such that B_{ξ} ∩ B_{η} = ∅ for any ξ < η < 𝔠. (See, e.g., [11, Lemma 5].)

Next, fix a ξ < 𝔠. For each f ∈ F construct a function h_{ξ,f}: B_{ξ} → ℚ such that

f(x) + h_{ξ,f}(x) ∈ J_{ξ} for every x ∈ B_{ξ}.

Then, by [6, Lemma 2.2] (used with B = B_{ξ} and H = {h_{ξ,f}: f ∈ F}), there exists a function g_{ξ}: B_{ξ} → ℚ such that h_{ξ,f} ∩ g_{ξ} ≠ ∅ for every f ∈ F.

Let g: ℝ → ℚ be a common extension of all functions g_{ξ}. Then for every ξ < 𝔠 and every f ∈ F there exists an x ∈ B_{ξ} such that f(x) + g(x) ∈ J_{ξ}. So, for every f ∈ F the graph of f + g is 𝔠-dense in ℝ^{2}. Thus, f + g ∈ U_{𝔠}.

3. Proof of Theorem 1.5(1)

By Proposition 1.1(5) and (3), we have 2 ≤ A_{b}(Ext) ≤ A_{b}(PR). Thus, it is enough to prove that A_{b}(PR) ≤ 2, i.e., by Proposition 1.1(4), that {f_{1}−f_{2} : f_{1}, f_{2} ∈ PR ∩ Bd} ≠ Bd. This follows immediately from the next example.

Example 3.1. Let B ⊆ ℝ be a Bernstein set and f = χ_{B} − χ_{ℝ\B}. If f = g_{0} − g_{1} and g_{0}, g_{1} ∈ PR, then both g_{0} and g_{1} are bilaterally unbounded.

Proof. By way of contradiction assume that there are g_{0}, g_{1} ∈ PR such that f = g_{0}− g_{1} and g_{0} is not bilaterally unbounded. We will assume that m_{0} = inf g_{0}[ℝ] > −∞. (The other case is analogous.) Put m_{1} = inf g_{1}[ℝ] and notice that m_{1} ≥ m_{0} − 1 > −∞.

For i < 2 choose a nonempty perfect set P_{i} such that g_{i} < m_{i} + 2^{−1} on P_{i}. Then

m_{0} ≤ inf g_{0}[P_{1}] = inf(g_{1} + f)[P_{1}] ≤ (m_{1} + 2^{−1} − 1) = m_{1} − 2^{−1} < m_{1}

and

m_{1} ≤ inf g_{1}[P_{0}] = inf(g_{0} − f)[P_{0}] ≤ (m_{0} + 2^{−1} − 1) = m_{0} − 2^{−1} < m_{0},

an impossibility. This completes the proof.

4. Proof of Theorem 1.5(2)

By Proposition 1.1(3), A_{b}(AC) ≤ A_{b}(Conn) ≤ A_{b}(D) ≤ A_{b}(U_{𝔠}) ≤ A_{b}(PC_{𝔠}). So, it is enough to prove that A_{b}(PC_{𝔠}) ≤ cof(𝔠) and A_{b}(AC) ≥ cof(𝔠). The first of these inequalities follows from the next example, and the second one follows immediately from Theorem 4.1.

Example 4.1. There is a family 𝔄 ∈ BD with |𝔄| = cof(𝔠) such that for every function g bounded above we have f + g ∉ PC_{𝔠} for some f ∈ 𝔄.

Proof. Let A be a family of sets such that |A| = cof(𝔠), |A| < 𝔠 for every A ∈ A, and |⋃ A| = 𝔠. Clearly we may assume that the sets in A are nonempty, pairwise disjoint, and that ⋃ A = ℝ. Put 𝔄 = {χ_{A}: A ∈ A}.

Let g: ℝ → ℝ be bounded above. Define M = sup g[ℝ] and choose an x ∈ ℝ with g(x) > M − 1. Then x ∈ A for some A ∈ A. Consequently, (g + χ_{A})(t) = g(t) ≤ M for every t ∉ A and (g + χ_{A})(x) > M. Thus

1 ≤ |{t ∈ ℝ: (g + χ_{A})(t) > M}| ≤ |A| < 𝔠

and g + χ_{A} ∉ PC_{𝔠}.

Theorem 4.1. If 𝔄 ⊆ (−1, 1)^{ℝ} and |𝔄| < cof(𝔠), then there is a function g ∈ (−2, 2)^{ℝ} such that f + g ∈ AC for every f ∈ 𝔄.

Proof. For each f ∈ 𝔄 let

U_{f} = {⟨x, y⟩: x ∈ ℝ & f(x) − 2 < y < f(x) + 2}
= ⋃_{x∈ℝ}[{x} × (f(x) − 2, f(x) + 2)]

and

K_{f} = {K ⊆ ℝ^{2} : K is closed & |dom(K ∩ U_{f})| = 𝔠}.

Clearly, |K_{f}| = 𝔠. Also, by Lemma 1.8, |A_{𝔠}(f)| < 𝔠 for every f ∈ 𝔄. So, since |𝔄| < cof(𝔠), we obtain

(1) the set A = ⋃_{f∈𝔄} A_{𝔠}(f) has cardinality less than 𝔠.

Let {⟨f_{ξ}, K_{ξ}⟩: ξ < 𝔠} be a transfinite sequence consisting of all pairs ⟨f, K⟩ such that f ∈ 𝔄 and K ∈ K_{f}. We will construct by transfinite induction a sequence {⟨x_{ξ}, y_{ξ}⟩: ξ < 𝔠} such that for every ξ < 𝔠 the following conditions hold:

(i) x_{ξ} ∉ {x_{ζ}: ζ < ξ} ∪ A; (ii) ⟨x_{ξ}, f_{ξ}(x_{ξ}) + y_{ξ}⟩ ∈ K_{ξ} ∩ U_{f_{ξ}}. So, assume that {⟨x_{ζ}, y_{ζ}⟩: ζ < ξ} have been already defined for some ξ < 𝔠. By

the definition of K_{f_{ξ}} and condition (1), we can choose

x_{ξ} ∈ dom(K_{ξ} ∩ U_{f_{ξ}}) \ ({x_{ζ}: ζ < ξ} ∪ A).

Let z_{ξ} be such that ⟨x_{ξ}, z_{ξ}⟩ ∈ K_{ξ} ∩ U_{f_{ξ}} and put y_{ξ} = z_{ξ} − f_{ξ}(x_{ξ}). Then (i) and (ii) are obviously satisfied. This completes the inductive construction.

Notice that y_{ξ} ∈ (−2, 2) for every ξ < 𝔠 since z_{ξ} ∈ (f_{ξ}(x_{ξ})−2, f_{ξ}(x_{ξ})+2). Define the function g: ℝ → (−2, 2) by

g(x) = \begin{cases}y_{ξ} & if x = x_{ξ} for some ξ < 𝔠, \\ 0 & otherwise. \end{cases}

Fix an f ∈ 𝔄. We will prove that f + g ∈ AC. Let V ⊆ ℝ^{2} be an open set containing f + g. First notice that if

E = dom(U_{f} \ V) = {x ∈ ℝ: [{x} × (f(x) − 2, f(x) + 2)] ⊈ V },

then |E| < 𝔠. Indeed, otherwise K = ℝ^{2} \ V ∈ K_{f}, so there exists a ξ < 𝔠 with ⟨f_{ξ}, K_{ξ}⟩ = ⟨f, K⟩. By (ii), we obtain

⟨x_{ξ}, (f + g)(x_{ξ})⟩ = ⟨x_{ξ}, f_{ξ}(x_{ξ}) + y_{ξ}⟩ ∈ K_{ξ} = K,

contradicting the fact that f + g ⊆ V.

Define

F = dom([ℝ × [−1, 1]] \ V ) = {x ∈ ℝ: [{x} × [−1, 1]] ⊈ V }.

Since [−1, 1] ⊆ (f(x) − 2, f(x) + 2) for every x ∈ ℝ, so [ℝ × [−1, 1]] ⊆ U_{f}. Hence F ⊆ E and |F | < 𝔠. But [ℝ × [−1, 1]] \ V is a closed subset of ℝ × [−1, 1]. Thus F is closed in ℝ and it is at most countable.

Let J be the family of all compact intervals J = [a, b] for which there exists a continuous function h: J → ℝ with h ⊆ V and h(a) = h(b) = 0. Moreover let G be the set of all x ∈ ℝ for which there exists a δ_{x} > 0 such that [a, b] ∈ J whenever a, b ∈ (x − δ_{x}, x + δ_{x}) \ F and a < b. The first claim is obvious.

Claim 1. If [a_{0}, a_{1}] ∈ J and [a_{1}, a_{2}] ∈ J, then [a_{0}, a_{2}] ∈ J.

Claim 2. If J = [a, b] ⊆ G and a, b ∉ F, then J ∈ J.

Indeed, the compactness of J and the relation J ⊆ ⋃_{x∈J}(x − δ_{x}, x + δ_{x}) imply that there exist x_{0}, …, x_{p} ∈ J such that J ⊆ ⋃_{i=0}^{p}(x_{i} − δ_{x_{i}}, x_{i} + δ_{x_{i}}). Hence we can find nonoverlapping compact intervals J_{0}, …, J_{l} ∈ J with J = ⋃_{j=0}^{l} J_{j}. By Claim 1, we obtain J ∈ J.

Claim 3. We have G = ℝ.

First notice that G is open and that ℝ\F ⊆ G. By way of contradiction suppose that the set P = ℝ \ G ⊆ F is nonempty. Thus, P is scattered and so, it contains an isolated point s. Let δ > 0 be such that [s − δ, s + δ] ∩ P = {s}. To get a contradiction it is enough to show that s ∈ G.

Let a, b ∈ (s − δ, s + δ) \ F and a < b. If a > s or b < s, then, by Claim 2, we obtain [a, b] ∈ J. So we may assume that a < s < b. Let ε > 0 be such that a < s − ε < s + ε < b and

R = [(s − ε, s + ε) × ((f + g)(s) − ε, (f + g)(s) + ε)] ⊆ V.

(Such an ε exists, since V is open and ⟨s, (f + g)(s)⟩ ∈ V.)

Notice that there exist ⟨q_{0}, y_{0}⟩, ⟨q_{1}, y_{1}⟩ ∈ U_{f} ∩ R such that q_{0} ∈ (s − ε, s) \ E and q_{1} ∈ (s, s + ε) \ E. We will prove it for i = 0, the other case being similar.

First suppose s ∈ A. Take an arbitrary q_{0} ∈ (s − ε, s) \ E. Then g(s) = 0, so (f + g)(s) = f(s) ∈ (−1, 1) ⊆ (f(q_{0}) − 2, f(q_{0}) + 2). Consequently, the interval

D = (f(q_{0}) − 2, f(q_{0}) + 2) ∩ ((f + g)(s) − ε, (f + g)(s) + ε)

is nonempty. So every point ⟨q_{0}, y_{0}⟩ with y_{0} ∈ D has the required properties.

Otherwise by the definition of A_{𝔠}^{−}(f), there is a q_{0} ∈ (s − ε, s) \ E with f(q_{0}) ∈ (f(s) − ε, f(s) + ε). Then |f(s) − f(q_{0})| < ε and

|(f + g)(s) − f(q_{0})| ≤ |g(s)| + |f(s) − f(q_{0})| < 2 + ε.

Define the set D as above. Observe that D ≠ ∅ and every point ⟨q_{0}, y_{0}⟩ with y_{0} ∈ D has the required properties.

Fix an i < 2. Put c_{i} = min{y_{i}, −1} and d_{i} = max{y_{i}, 1}. Since q_{i} ∉ E, we see that [{q_{i}}×[c_{i}, d_{i}]] ⊆ [{q_{i}}×(f(q_{i})−2, f(q_{i})+2)] ⊆ V. Since the set {q_{i}}×[c_{i}, d_{i}] is compact, we can find a_{i}, b_{i} ∈ (s − ε, s + ε) \ E such that q_{i} ∈ (a_{i}, b_{i}) and

R_{i} = [(a_{i}, b_{i}) × [c_{i}, d_{i}]] ⊆ V.

Construct continuous functions h_{0} : [a, a_{0}] → ℝ and h_{1} : [b_{1}, b] → ℝ such that h_{0} ∪ h_{1} ⊆ V and h_{0}(a) = h_{0}(a_{0}) = h_{1}(b_{1}) = h_{1}(b) = 0. (We use Claim 2.) Extend h_{0} ∪ h_{1} to h: [a, b] → ℝ by connecting the following pairs of points by straight line segments: ⟨a_{0}, 0⟩ with ⟨q_{0}, y_{0}⟩, ⟨q_{0}, y_{0}⟩ with ⟨q_{1}, y_{1}⟩, and ⟨q_{1}, y_{1}⟩ with ⟨b_{1}, 0⟩. Notice that these segments are contained in rectangles R_{0}, R, and R_{1}, respectively. Thus h ⊆ V. Clearly h(a) = h(b) = 0 and h is continuous, so [a, b] ∈ J. Hence s ∈ G, an impossibility. Claim 3 has been proved.

Using Claims 1–3 it is easy to show that there exists a continuous function h: ℝ → ℝ with h ⊆ V. So f + g ∈ AC.

5. Proof of Theorem 1.5(3)

Let κ < 𝔠 be an infinite cardinal number. By Proposition 1.1(3), we have A_{b}(U_{κ}) ≤ A_{b}(PC_{κ}) ≤ A_{b}(PC_{ω}). So, it is enough to prove that A_{b}(PC_{ω}) ≤ 𝔠 and A_{b}(U_{κ}) ≥ 𝔠. The first of these inequalities follows from Proposition 5.1, and the second one from Theorem 5.2.

Proposition 5.1. If a function g is bounded above, then g + χ_{{x}} ∉ PC_{ω} for some x ∈ ℝ.

Proof. Define M = sup g[ℝ] and choose an x ∈ ℝ with g(x) > M − 1. Then (g + χ_{{x}})(t) = g(t) ≤ M for t ≠ x and (g + χ_{{x}})(x) > M. Thus

|{t ∈ ℝ: (g + χ_{{x}})(t) > M}| = 1

and g + χ_{{x}} ∉ PC_{ω}.

Theorem 5.2. If κ < 𝔠, 𝔄 ⊆ (−1, 1)^{ℝ}, and |𝔄| < 𝔠, then there is a function g ∈ (−2, 2)^{ℝ} such that f + g ∈ U_{κ} for each f ∈ 𝔄.

Proof. The proof is a modification and simplification of that of Theorem 4.1.

Without loss of generality we may assume that κ ≥ |𝔄| + ω. Let D be a κ-dense subset of ℝ with |D| = κ, and let

K = {(p, q) × {d}: p, q, d ∈ D & p < q}.

For each f ∈ 𝔄 let

U_{f} = {⟨x, y⟩: x ∈ ℝ & f(x) − 2 < y < f(x) + 2}

and

K_{f} = {K ∈ K: |dom(K ∩ U_{f})| > κ}.

Clearly, |K_{f}| = κ. Also, by Lemma 1.8,

the set A = ⋃_{f∈𝔄} A_{κ^{+}}(f) has cardinality ≤ κ.

Let {⟨f_{ξ}, K_{ξ}⟩: ξ < κ} be a transfinite sequence consisting of all pairs ⟨f, K⟩ such that f ∈ 𝔄 and K ∈ K_{f}. As in Theorem 4.1 we can construct by transfinite induction a sequence {⟨x_{ξ}, y_{ξ}⟩: ξ < κ} such that for every ξ < κ the following conditions hold:

(i) x_{ξ} ∉ {x_{ζ}: ζ < ξ} ∪ A; (ii) ⟨x_{ξ}, f_{ξ}(x_{ξ}) + y_{ξ}⟩ ∈ K_{ξ} ∩ U_{f_{ξ}}. Notice that y_{ξ} ∈ (−2, 2) for every ξ < κ. Define g: ℝ → (−2, 2) by

g(x) = \begin{cases}y_{ξ} & if x = x_{ξ} for some ξ < κ, \\ 0 & otherwise. \end{cases}

Fix an f ∈ 𝔄. We will prove that f +g ∈ U_{κ}. Let a < b and define c = (f +g)(a) and d = (f + g)(b). Clearly we may assume that c ≠ d. We will assume that c < d, the case d < c being similar.

Let y ∈ (c, d) ∩ D. We will find p, q ∈ (a, b) ∩ D such that p < q and

K = (p, q) × {y} ∈ K_{f}. (2)

Indeed, if y ∈ (−1, 1), then [ℝ × {y}] ⊆ U_{f} and the relation (2) holds for any p, q ∈ (a, b) ∩ D with p < q. So, suppose that y ∉ (−1, 1). We will assume that y ≥ 1, the case y ≤ −1 being essentially the same.

Then f(b) + g(b) = d > y ≥ 1 > f(b), so g(b) ≠ 0. In particular, b ∉ A_{f}^{−}. So, there exist an ε ∈ (0, d − y) and p, q ∈ (a, b) ∩ D such that the set S = {x ∈ (p, q): f(x) > f(b) − ε} has cardinality greater than κ. But for every x ∈ S we have

f(x) + 2 > f(x) + g(b) > f(b) − ε + g(b) = d − ε > y ≥ 1 > f(x) − 2.

Thus S × {y} ⊆ U_{f} and (p, q) × {y} ∈ K_{f}.

Let ξ < κ be such that ⟨f, K⟩ = ⟨f_{ξ}, K_{ξ}⟩. Then x_{ξ} ∈ (p, q) ⊆ (a, b) and (f + g)(x_{ξ}) = f_{ξ}(x_{ξ}) + y_{ξ} = y. It follows that (c, d) ∩ D ⊆ (f + g)[(a, b)], i.e., that (f + g)[(a, b)] is κ-dense in (c, d). This completes the proof.

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Department of Mathematics, West Virginia University, Morgantown, West Virginia 26506–6310

   E-mail address: kcies@wvnvms.wvnet.edu

Department of Mathematics, Pedagogical University, Arciszewskiego 22, 76–200 Słupsk, Poland

   E-mail address: wspb05@pltumk11.bitnet