PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 130, Number 7, Pages 2007–2013 S 0002-9939(01)06254-2 Article electronically published on December 27, 2001

ON SUMS OF DARBOUX AND NOWHERE CONSTANT CONTINUOUS FUNCTIONS

KRZYSZTOF CIESIELSKI AND JANUSZ PAWLIKOWSKI

(Communicated by Alan Dow)

Abstract. We show that the property (P) for every Darboux function g: ℝ → ℝ there exists a continuous nowhere

constant function f: ℝ → ℝ such that f + g is Darboux follows from the following two propositions: (A) for every subset S of ℝ of cardinality 𝔠 there exists a uniformly continuous

function f: ℝ → [0, 1] such that f[S] = [0, 1], (B) for an arbitrary function h: ℝ → ℝ whose image h[ℝ] contains a nontrivial interval there exists an A ⊂ ℝ of cardinality 𝔠 such that the restriction h ↾ A of h to A is uniformly continuous,

which hold in the iterated perfect set model.

Our set theoretic terminology is standard and follows that of [3]. In particular, functions are identified with their graphs and |X| stands for the cardinality of a set X. The symbol 𝔠 stands for |ℝ| and C(ℝ) denotes the family of all continuous functions from ℝ into ℝ. A function f: ℝ → ℝ is Darboux if a conclusion of the intermediate value theorem holds for f or, equivalently, when f maps every interval onto an interval; f is a Sierpiński-Zygmund function if its restriction f ↾ Y is discontinuous for every subset Y of ℝ of cardinality 𝔠; f is nowhere constant if it is not constant on any non-trivial interval.

The class of Darboux functions has been studied for a long time as one of possible generalizations of the class of continuous functions. However, it has some peculiar properties. For example, it is not closed under addition. In fact, in 1927 Lindenbaum [8] noticed that every function f: ℝ → ℝ can be written as a sum of two Darboux functions (see also [9]) while Erdős [5] showed that if additionally f is measurable, then both of the summands can be chosen to be measurable. In 1990 Kirchheim and Natkaniec showed that there are some truly bad Darboux functions by proving the following.

Proposition 1 (Kirchheim, Natkaniec [6]). If the union of less than 𝔠 many meager subsets of ℝ is meager (thus under CH or MA), then there exists a Darboux function g: ℝ → ℝ such that f + g is not Darboux for every continuous nowhere constant function f: ℝ → ℝ.

\begin{matrix}c \\ ◯ \end{matrix}2001 American Mathematical Society

Received by the editors November 13, 2000 and, in revised form, January 24, 2001.

1991 Mathematics Subject Classification. Primary 26A15; Secondary 03E35.

Key words and phrases. Darboux, nowhere constant, images of continuous functions.

The work of the second author was partially supported by KBN Grant 2 P03A 031 14.

The problem of whether the additional set-theoretic assumptions are necessary in this theorem was investigated in 1992 by Komjáth [7] and was settled in 1995 by Steprāns.

Proposition 2 (Steprāns [11]). It is consistent with ZFC that (P) for every Darboux function g: ℝ → ℝ there exists a continuous nowhere con-

stant function f: ℝ → ℝ such that f + g is Darboux.

In his paper Steprāns showed that (P) holds in the iterated perfect set model. His direct forcing argument is quite complicated and difficult to follow. We will show that (P) follows from the principles (A) and (B) defined in the abstract. Since the proofs that (A) and (B) hold in the the iterated perfect set model are much simpler than the argument given in [11], our argument for (P) is essentially more accessible.

The proof that (A) holds in the iterated perfect set model is due to Miller [10]. The proof that (B) holds in this model is due to Balcerzak, Ciesielski, and Natkaniec [1] and consists of a page-long argument based on the results from [10]. In addition, properties (A) and (B) follow quite easily from the Covering Property Axiom (CPA) as shown by the authors in [4]. For more on the Darboux property see also the survey article [2].

Note also that condition (B) is equivalent to its following seemingly stronger version: (B^{∗}) for an arbitrary S ⊂ ℝ and a function h: S → ℝ whose image h[S] contains

a non-trivial interval there exists an A ⊂ S of cardinality 𝔠 such that the restriction h ↾ A of h to A is uniformly continuous.

Clearly (B^{∗}) implies (B). To see the other implication let h be as in (B^{∗}) and f: ℝ → ℝ be Sierpiński-Zygmund. Put h^{∗} = f ↾ (ℝ \ S) ∪ h and apply (B) to find A^{∗} ⊂ ℝ of cardinality 𝔠 such that h^{∗} ↾ A^{∗} is uniformly continuous. Then A = A^{∗}∩S is as in (B^{∗}).

Our main theorem is the following.

Theorem 3. Assume that (A) and (B) hold. Then for every Darboux function d: ℝ → ℝ there exist a complete metric ρ on C(ℝ) and a dense G_{δ} subset G of ⟨C(ℝ), ρ⟩ of nowhere constant functions such that d + g is Darboux for every g ∈ G.

In particular (A)&(B) implies (P).

Our proof will be based on the following two lemmas.

Lemma 4. Let d: ℝ → ℝ be Darboux and D ⊂ ℝ be such that d ↾ D is dense in d. If g: ℝ → ℝ is continuous and such that

[(d + g)(α), (d + g)(β)] ⊂ (d + g)[[α, β]] for all α, β ∈ D, (1)

then d + g is Darboux.

Proof. An easy proof can be found in [11, lemma 4.1].

Lemma 5. Assume that (A) holds and for every n < ω let A_{n} ∈ [ℝ]^{𝔠}. Then for every n < ω there exists a B_{n} ∈ [A_{n}]^{𝔠} such that the closures of B_{n}’s are pairwise disjoint.

Proof. First note that (A) implies

∀⟨C_{n} ∈ [ℝ]^{𝔠}: n < ω⟩ ∃C ∈ [C_{0}]^{𝔠} ∀n < ω |C_{n} \ cl(C)| = 𝔠. (2)

Indeed, by condition (A), we can find a continuous function f: ℝ → [0, 1] such that |C_{0} ∩ f^{−1}(c)| = 𝔠 for every c ∈ [0, 1]. (Just take a composition of a function from (A) with the Peano curve followed by a projection.) Identify 2^{ω} with a subset of [0, 1] and by induction on n < ω choose an increasing sequence s_{n} ∈ 2^{n+1} such that |C_{n} \ f^{−1}({c ∈ 2^{ω}: s_{n} ⊂ c})| = 𝔠. Put s = ⋃_{n<ω} s_{n}. Then C = C_{0} ∩ f^{−1}(s) satisfies (2).

Next note that there exists a sequence ⟨D_{n}: n < ω⟩ of closed subsets of ℝ such that |A_{n} ∩ (D_{n} \ ⋃_{i<n} D_{i})| = 𝔠. This sequence is constructed by using induction on n < ω with each set D_{n} chosen by applying (2) to the sequence ⟨A_{k} \ ⋃_{i<n} D_{i}: n ≤ k < ω⟩.

Finally, since each set D_{n} \ ⋃_{i<n} D_{i} is an F_{σ}-set, we can find closed subsets of them, E_{n}, with B_{n} = E_{n} ∩ A_{n} having cardinality 𝔠. It is easy to see that the sets B_{n} are as required.

Remark 6. Note that Lemma 5 cannot be proved in ZFC. Indeed, if we assume that there exists a 𝔠-Luzin set^{1} and ⟨A_{n} ∈ [ℝ]^{𝔠}: n < ω⟩ is a sequence of 𝔠-Luzin sets such that every non-empty open interval contains one of A_{n}’s, then for such a sequence there are no B_{n}’s as in the lemma.

Lemma 7. Assume that (A) and (B) hold and d: ℝ → ℝ is a Darboux function. If D is a countable subset of ℝ such that (∗) for every α < β from D for which d(α) = d(β) there exist p, q ∈ ℝ such that

α < p < q < β and d[(p, q)] = {f(α)}, then there exists a countable family A ⊂ [ℝ]^{𝔠} such that

(a) different elements of A have disjoint closures,

(b) d ↾ A is uniformly continuous for every A ∈ A, and

(c) for every α, β ∈ D there exists an A ∈ A with the property that d ↾ A ⊂ [α, β] × [d(α), d(β)].

Proof. First for every α < β from D we choose A_{α}^{β} ∈ [ℝ]^{𝔠} such that the family A_{0} = {A_{α}^{β}: α, β ∈ D & α < β} satisfies (b) and (c). For this fix α < β from D.

If d(α) = d(β), then it is enough to put A_{α}^{β} = (p, q), where p and q are from (∗). So, assume that d(α) ≠ d(β). Then I = [d(α), d(β)] is a non-trivial interval, which is a subset of d[[α, β]], since d is Darboux. Let S = [α, β] ∩ d^{−1}(I) and notice that h = d ↾ S maps S onto I. So, by (B^{∗}), we can find an A_{α}^{β} ∈ [S]^{𝔠} such that d ↾ A_{α}^{β} is uniformly continuous.

It is easy to see that A_{0} satisfies (b) and (c) so it is enough to decrease its elements to get condition (a), while assuring that they still have cardinality 𝔠. This can be done by applying Lemma 5.

Proof of Theorem 3. Take a Darboux function d: ℝ → ℝ and choose a countable dense set D ⊂ ℝ satisfying condition (∗) from Lemma 7 such that d ↾ D is dense in d. Let A ⊂ [ℝ]^{𝔠} be the family from Lemma 7 and let {A_{n}: n < ω} be an enumeration of A ∪ {{d}: d ∈ D \ ⋃ A}. Let ρ_{0} be the uniform convergence metric on C(ℝ), that is,

ρ_{0}(f, g) = min{1, sup{|f(x) − g(x)|: x ∈ ℝ}}

1A set L ⊂ ℝ is a 𝔠-Luzin set if |L| = 𝔠 but |L ∩ N| < 𝔠 for every nowhere dense subset N of ℝ. It is well known (see, e.g., [10, sec. 2]) and easy to see that no 𝔠-Luzin set can be mapped continuously onto [0, 1]. Thus (A) implies that there is no 𝔠-Luzin set.

and, for f, g ∈ C(ℝ), let

ρ_{1}(f, g) = 2^{− min{n<ω: f↾A_{n}≠g↾A_{n}}}.

(If {n < ω: f ↾ A_{n} ≠ g ↾ A_{n}} = ∅ we assume that ρ_{1}(f, g) = 0.) Then ρ_{1} is a pseudometric on C(ℝ). Consider C(ℝ) with the following metric ρ:

ρ(f, g) = max{ρ_{0}(f, g), ρ_{1}(f, g)}

and notice that ⟨C(ℝ), ρ⟩ forms a complete metric space. We will prove that if G is the set of all nowhere constant continuous functions g: ℝ → ℝ for which d + g satisfies (1) of Lemma 4, then G contains a dense G_{δ} subset of ⟨C(ℝ), ρ⟩. For this we will show that for every α < β from D the following two types of sets contain dense open subsets of ⟨C(ℝ), ρ⟩:

H_{α}^{β} = {g ∈ C(ℝ): g is not constant on (α, β)}

and

G_{α}^{β} = {g ∈ C(ℝ): [(d + g)(α), (d + g)(β)] ⊂ (d + g)[[α, β]]}.

Then

⋂ {H_{α}^{β} ∩ G_{α}^{β}: α, β ∈ D & α < β} ⊂ G

will contain a dense G_{δ} subset of ⟨C(ℝ), ρ⟩.

To see that H_{α}^{β} contains a dense open subset first note that, by (a) and (c) of Lemma 7, elements of A are nowhere dense. Next, take an f ∈ C(ℝ), and fix an ε > 0. Let B(f, ε) be the open ρ-ball centered at f and of radius ε. We will find g ∈ B(f, ε) and δ > 0 such that B(g, δ) ⊂ B(f, ε)∩H_{α}^{β}. So take an n < ω such that 2^{−n} < ε/4, choose a non-empty open interval J ⊂ (α, β)\⋃_{i=1}^{n} A_{i}, and pick different x, y ∈ J. It is easy to find g ∈ C(ℝ) such that g(x) ≠ g(y), g ↾ (ℝ \ J) = f ↾ (ℝ \ J), and ρ_{0}(f, g) < 2^{−n}. Then, by the choice of J we also have ρ_{1}(f, g) < 2^{−n}, so g ∈ B(f, ε/4). Now, if δ = min{|g(x) − g(y)|/4, ε/4}, then B(g, δ) ⊂ B(f, ε) ∩ H_{α}^{β}.

To see that each G_{α}^{β} contains a dense open subset take f ∈ C(ℝ) and ε > 0. As previously we will find g ∈ B(f, ε) and δ > 0 such that B(g, δ) ⊂ B(f, ε)∩G_{α}^{β}. Find α = x_{0} < x_{1} < ⋯ < x_{m} = β, x_{i} ∈ D, such that the variation of f on each interval [x_{i}, x_{i+1}] is less than ε/8. Also, since d ↾ [x_{i}, x_{i+1}] is Darboux, we can partition each [x_{i}, x_{i+1}] even further to also get that |d(x_{i}) − d(x_{i+1})| < ε/8 for all i < m.

Pick an n < ω such that 2^{−n} < ε/8 and {x_{i}: i ≤ m} ⊂ ⋃_{i=1}^{n} A_{i}. For every i < m choose an index k_{i} > n such that d ↾ A_{k_{i}} ⊂ (x_{i}, x_{i+1}) × [d(x_{i}), d(x_{i+1})]. By (A) for every i < m we can also pick a uniformly continuous function h_{i} from A_{k_{i}} onto [(d + f)(x_{i}), (d + f)(x_{i+1})]. For each i < m define h ↾ A_{k_{i}} = (h_{i} − d − f) ↾ A_{k_{i}} and note that

h[A_{k_{i}}] ⊂ h_{i}[A_{k_{i}}] − d[A_{k_{i}}] − f[A_{k_{i}}] ⊂ [−ε/2, ε/2]

since [(d+f)(x_{i}), (d+f)(x_{i+1})] ⊂ [d(x_{i})+f(x_{i})−ε/4, d(x_{i})+f(x_{i})+ε/4], d[A_{k_{i}}] ⊂ [d(x_{i}), d(x_{i+1})] ⊂ [d(x_{i}) − ε/8, d(x_{i}) + ε/8], and f[A_{k_{i}}] ⊂ (f(x_{i}) − ε/8, f(x_{i}) + ε/8). Also define h as 0 on ⋃_{i=1}^{n} A_{i} and extend it to a uniformly continuous function from ℝ into [−ε/2, ε/2]. Put g = f + h and note that ρ(f, g) ≤ ε/2. Also let k = max{n, k_{0}, …, k_{m−1}} and δ ∈ (0, 2^{−k}). We claim that B(g, δ) ⊂ B(f, ε) ∩ G_{α}^{β}.

Indeed, it is easy to see that B(g, δ) ⊂ B(f, ε). To see that B(g, δ) ⊂ G_{α}^{β} take a g_{0} ∈ B(g, δ). By the choice of δ, h, and g for every i ≤ m we have

(d + g_{0})(x_{i}) = (d + g)(x_{i}) = (d + f + h)(x_{i}) = (d + f)(x_{i})

and, for A = ⋃_{j<m} A_{k_{j}},

g_{0} ↾ A = g ↾ A = (f + h) ↾ A = (f + (h_{i} − d − f)) ↾ A = (h_{i} − d) ↾ A.

So,

[(d + g_{0})(α), (d + g_{0})(β)] = [(d + f)(x_{0}), (d + f)(x_{m})]
⊂ ⋃_{j<m} [(d + f)(x_{j}), (d + f)(x_{j+1})]
= ⋃_{j<m} h_{j}[A_{k_{j}}]
= ⋃_{j<m}(d + (h_{j} − d))[A_{k_{j}}]
= ⋃_{j<m}(d + g_{0})[A_{k_{j}}]
⊂ (d + g_{0})[[α, β]]

proving that g_{0} ∈ G_{α}^{β}. This finishes the proof of the theorem.

We say that a function f: ℝ → ℝ is a D^{1} function if it is differentiable with its derivative f′(x) finite at every x ∈ ℝ; f is in the class “D^{1}” provided its derivative f′(x) exists at every point, but it can have an infinite value. We say that f belongs to C^{1} (to “C^{1}”) if it belongs to D^{1} (to “D^{1}”, respectively) and its derivative f′ is continuous.

Let us also notice that in (P) we cannot require that function g is “C^{1}”. This follows from the following fact which, for the functions from the class C^{1}, was first noticed by Steprāns [11, p. 118]. Since Steprāns leaves his statement without any comments concerning its proof, we include here a missing argument.

Proposition 8. There exists, in ZFC, a Darboux function d: ℝ → ℝ such that d + f is not Darboux for every non-constant “C^{1}” function f.

Proof. Let ℝ = {x_{ξ}: ξ < 𝔠} and let {f_{ξ}: ξ < 𝔠} be an enumeration of all nonconstant “C^{1}” functions. Notice that for every ξ < 𝔠 there exists a non-empty open interval I_{ξ} such that f_{ξ} ↾ I_{ξ} is strictly monotone. (Just take an x ∈ ℝ such that f′_{ξ}(x) ≠ 0, which exists since f_{ξ} is non-constant. Then I_{ξ} is chosen as a neighborhood of x on which f′_{ξ} is non-zero.)

The function d that we construct will be strongly Darboux in a sense that d^{−1}(r) is dense in ℝ for every r ∈ ℝ. For such d in order to show that d+f_{ξ} is not Darboux it is enough to show that (d + f_{ξ})^{−1}(y_{ξ}) is not dense in ℝ for some y_{ξ} ∈ ℝ. (See, e.g., [3, prop. 7.2.4].)

By induction we construct a sequence {⟨Q_{ξ}, d_{ξ}, y_{ξ}⟩: ξ < 𝔠} such that for every ξ < 𝔠 we have:

(i) Sets {Q_{η}: η ≤ ξ} are countable and pairwise disjoint, and x_{ξ} ∈ ⋃_{η≤ξ} Q_{η}. (ii) d_{ξ}: Q_{ξ} → ℝ and d_{ξ}^{−1}(x_{ξ}) is dense in ℝ. (iii) y_{ζ} ∉ (d_{η} + f_{ζ})[Q_{η} ∩ I_{ζ}] for every ζ, η ≤ ξ.

Notice that if we have such a sequence, then, by (i), ℝ = ⋃_{ξ<𝔠} Q_{ξ} so d = ⋃_{ξ<μ} d_{ξ}: ℝ → ℝ. It is strongly Darboux by condition (ii), while, by (iii), for every ζ < 𝔠 the set (d + f_{ζ})^{−1}(z_{ζ}) is not dense in ℝ, since it misses I_{ζ}. Thus, d is as desired.

To make an inductive ξ-th step first choose a countable dense subset Q_{ξ}^{0} of ℝ disjoint with

⋃_{η<ξ} Q_{η} ∪ ⋃_{ζ<ξ} (I_{ζ} ∩ f_{ζ}^{−1}(y_{ζ} − x_{ξ})) .

This can be done since, by the choice of of the intervals I_{ζ}’s, each of the sets I_{ζ} ∩ f_{ζ}^{−1}(y_{ζ} − x_{ξ}) has at most one element.

Define d_{ξ} ↾ Q_{ξ}^{0} as constantly equal to x_{ξ}. This guarantees (ii), while condition (iii) is satisfied for the part defined so far: for every ζ < ξ and x ∈ I_{ζ} ∩ Q_{ξ}^{0} we have d_{ξ}(x)+f_{ζ}(x) = f_{ζ}(x)+x_{ξ} ≠ y_{ζ}, since otherwise we would have x ∈ I_{ζ} ∩f_{ζ}^{−1}(y_{ζ} −x_{ξ}) contradicting the choice of Q_{ξ}^{0}.

If x_{ξ} ∈ ⋃_{η<ξ} Q_{η} we define Q_{ξ} = Q_{ξ}^{0}. Otherwise we put Q_{ξ} = Q_{ξ}^{0} ∪ {x_{ξ}} and if d_{ξ}(x_{ξ}) is not defined yet (i.e., if x_{ξ} ∉ Q_{ξ}^{0}) we define d_{ξ}(x_{ξ}) by choosing d_{ξ}(x_{ξ}) ∉ {y_{ζ} − f_{ζ}(x_{ξ}): ζ < ξ}. This guarantees that (i) is satisfied, while (iii) is preserved.

Finally, we choose y_{ξ} to have

y_{ξ} ∈ ℝ \ ⋃_{η≤ξ}(d_{η} + f_{ξ})[Q_{η}].

This will guarantee that (iii) also holds for ζ = ξ.

Let us also note that the following question, for the case of D^{1} due to Steprāns [11, Question 5.1], remains open.

Problem 9. Does there exist, in ZFC, a Darboux function d: ℝ → ℝ such that d + f is not Darboux for every nowhere constant “D^{1}” function f? What if we restrict the choice of f to D^{1}?

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

   E-mail address: K Cies@math.wvu.edu
   URL: http://www.math.wvu.edu/~kcies

Department of Mathematics, University of Wrocław, pl. Grunwaldzki 2/4, 50-384 Wrocław, Poland – and – Department of Mathematics, West Virginia University, Morgantown, West Virginia 26506-6310

   E-mail address: pawlikow@math.uni.wroc.pl