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SUMS OF DARBOUX AND CONTINUOUS

arXiv:math/9304205v1 [math.LO] 27 Apr 1993SUMS OF DARBOUX AND CONTINUOUSFUNCTIONSJURIS STEPR¯ANSAbstract. It is shown that that for every Darboux function Fthere is a non-constant continuous function f such that F + f isstill Darboux.

It is shown to be consistent — the model used isiterated Sacks forcing — that for every Darboux function F thereis a nowhere constant continuous function f such that F + f isstill Darboux.This answers questions raised in [4] where it isshown that in various models of set theory there are universallybad Darboux functions, Darboux functions whose sum with anynowhere constant, continuous function fails to be Darboux.1. IntroductionA function which maps any connected set to a connected set is knownas a Darboux function.

This paper will be concerned with functionsfrom R to R and, in this context, Darboux simply means that the imageof any interval is an interval. While there are various results establish-ing similarities between continuous functions and Darboux functions offirst Baire class, the fact that it is possible to construct Darboux func-tions by transfinite induction allows all sorts of pathologies to exist.For example, transfinite induction can be used to construct a Darbouxfunction F such that the function F(x) + x is not Darboux [7].

In [5]it is shown that if G is a family of functions such that | G |+ < 2ℵ0then there is a Darboux function F such that F + g is not Darbouxfor all g ∈G. This result is extended in [4] where it is established, as-suming certain set theoretic hypotheses, that there exists a universallybad Darboux function f : R →R which means that, for every nowhereconstant continuous g : R →R, f +g does not have the Darboux prop-erty.

In unpublished work W. Weiss has shown that a universally badDarboux function can be constructed assuming only the existence of a2ℵ0 additive ideal I on B, the Borel subsets fo R, such that the Booleanalgebra B/I has the 2ℵ0 chain condition; in other words, there do notexists 2ℵ0 elements of B whose pairwise intersections belong to I.This research was partially supported by NSERC.1

2JURIS STEPR¯ANSIn this paper it will be shown that some form of set theoretic hypoth-esis is necessary for such a result because there is a model of set theorywhere for every Darboux function F there is a nowhere constant con-tinuous function F such that F + f is also Darboux. The significanceof the adjective “nowhere constant” in this statement requires somecomment because it might seem a minor point.

An indication that thisis not so is given by the fact that, in spite of having shown that thereis a Darboux function f such for every nowhere constant continuousg : R →R, f + g does have the Darboux property, the authors of [4]pose the following question at the end of their paper.Question 1.1. Does there exist a Darboux function F : R →R suchthat F +g does not have the Darboux property whenever g is continuousbut not constant?Section 2 provides a negative answer to this problem.Section 3contains some technical material on Sacks forcing and Section 4 makesuse of this material in proving the main consistency result.

The finalsection contains some open questions.2. Sums with Non-constant FunctionsThe following lemma is easy but the proof is included anyway.

It isessentially the Sen-Massera condition which, in the case of Baire class1 functions, is equivalent to being Darboux [2].Lemma 2.1. If F is Darboux and not continuous at x then there isan interval (a, b) ̸= ∅such that for each y ∈(a, b) there is a sequence{xn | n ∈ω} such that limn→∞xn = x and F(xn) = y for all n ∈ω.Proof:Because F is not continuous at x there are sequences{yan}n∈ω and {ybn}n∈ω such thatlimn→∞yan = x = limn→∞ybn and limn→∞F(yan) = a < b limn→∞F(ybn)Given n ∈ω and y ∈(a, b) let k be such that | yak −x |< 1n, | ybk−x |< 1nand F(ybk) > y > F(yak).

Then use the Darboux property of F to findxn between yak and ybk such that F(xn) = y.■Corollary 2.1. If F is a Darboux function which is finite-to-one thenF is continuous.Lemma 2.2.

If F : R →R is a Darboux function which is continuousat only countably many points then there is a non-constant, continuousfunction f such that F + f is Darboux.

SUMS OF DARBOUX AND CONTINUOUS FUNCTIONS3Proof: To each real x at which F is not continuous, use Lemma 2.1to assign an interval (ax, bx) such that for each y ∈(ax, bx) there is asequence {xn | n ∈ω} such that limn→∞xn = x and F(xn) = y forall n. For rationals p and q let X(p, q) be the set of all x such thatax < p < q < bx and note that X(p, q) is a closed set. Because F iscontinuous at only countably many points, it can not be the case thatX(p, q) is nowhere dense for each pair of rationals p and q.

Therefore let[s, t] and [a, b] be intervals such that [ax, bx] ⊇[a, b] for each x ∈[s, t]and, furthermore F(s) = a and F(t) = b. Observe that F −1{y} isdense in [s, t] for each y ∈[a, b].Next, choose a family of open intervals I such that• I = ∪n∈ωIn where In = {(uni , vni ) | i ∈2n −1}• vn+1i< uni < vni < un+1i+1 for each n and i• ∪I is dense in [s, t]• F(uni ) = a and F(vni ) = b for each n and i• sup F ↾[vni−1, un+1i] ∪[vn+1i, uni ] < sup F ↾[un+1i, vn+1i] +1n+1 whereit is understood that, in the case i = 0, vn−1 = s and, in the casei = n, unn = t• inf F ↾[vni−1, un+1i] ∪[vn+1i, uni ] > inf F ↾[un+1i, vn+1i] −1n+1 whereit is understood that, in the case i = 0, vn−1 = s and, in the casei = n, unn = tNow let g be a nondecreasing, continuous function such that g ↾I isconstant for each I ∈I and such that g(s) = 0 and g(t) = (b −a) andsuch that g is constant on (−∞, s] and [t, ∞).

The reader who insistson concreteness may verify thatg(x) = sup{ 2i+12n | uni ≤x}b −asatisfies these requirements.To see that F +g is Darboux suppose that x < y and w lies betweenF(x)+g(x) and F(y)+g(y). First of all observe that it may be assumedthat s ≤x < y ≤t.

The reason it may be assumed that s ≤x is thatif x < s then either w lies between F(x) + g(x) and F(s) + g(s) or elseit lies between F(s) + g(s) and F(y) + g(y). In order to eliminate thefirst case use the fact that F is Darboux and g is constant on [x, s] tofind z ∈[x, s] such that F(z) + g(z) = w. In the second case it may, ofcourse, be assumed that x = s. A similar argument can be applied toshow that, without loss of generality, y ≤t.First consider the case that there is some (u, v) ∈I such that ghas constant value c on (u, v) and w ∈(a + c, b + c) and such thatx ≤u < v ≤y.

It is possible to use the fact that g is constant on

4JURIS STEPR¯ANS[u, v], F is Darboux, F(u) = a and F(v) = b to find z ∈(u, v) suchthat F(z) + g(z) = w.In the remaining case it follows from the fact that ∪I is dense andthe continuity of g that either w ≥b + g(r) for every r ∈∪I ∩(x, y)or w ≤a + g(r) for every r ∈∪I ∩(x, y). Only the first case willbe considered since the other one is dealt with similarly.

Furthermore,it will be assumed that F(x) + g(x) < F(y) + g(y) since the othercase is also similar.To begin, suppose that y ∈(u, v) ∈I.SinceF(u) = a < b < w −g(y) it follows that F(u) < w −g(y) < F(y)and so it is possible to appeal to the Darboux property of F and theconstantness of g on [u, v].On the other hand, if y /∈∪I then there must be some m ∈ω suchthat F(y) + g(y) −w > 1/m. Choose δ > 0 such that y −x > δ andg(y) −g(r) >12m if y −r < δ.

Then there is some k ≥2m such that(uki , vki ) ∈Ik and y −δ ≤uki < vki < y ≤uk−1ifor some i ∈2n −1. Itfollows that F(y) ≤sup F ↾[vki , uk−1i] < sup F ↾[uki , vki ]+1/k and hencesup{F(r) + g(r) | r ∈[uki , vki ]} > F(y) + g(y) −1/m.

Therefore w a + g(uki ) = F(uki ) + g(uki ); in otherwords there is some r ∈[uki , vki ] such thatF(uki ) + g(uki ) < w < F(r) + g(r)Because g is constant on [uki , vki ] it now follows from the Darboux prop-erty of F that there is z ∈[uki , vki ] such that F(z) + g(z) = w.■Lemma 2.3.

If F is a Darboux function which is continuous on anuncountable set then there is a continuous, non-constant function gsuch that F + g is DarbouxProof:Because the set of points where F is continuous is Borel,it is possible to find a perfect, nowhere dense set P such that F iscontinuous at each point of P. Because P is a perfect, nowhere denseset it follows that R \ P = ∪I where I is a disjoint family of openintervals of order type the rationals. Let g be any continuous, non-decreasing function which is not constant yet, g has constant value gIon each interval I ∈I.To see that F + g is Darboux suppose that x < y and that F(x) +g(x) < w < F(y) + g(y) — a similar proof works if F(x) + g(x) >w > F(y) + g(y).

If there is some interval I ∈I such that I ⊆[x, y]and sup F ↾I ≥w −gI ≥inf F ↾I then the Darboux property of Fguarantees that there is some z ∈I such that F(z) + g(z) = w.

SUMS OF DARBOUX AND CONTINUOUS FUNCTIONS5If there is no such I then consider first the case that there are I andJ in I such that I ⊆[x, y] and J ⊆[x, y] and sup F ↾I < w −gIand inf F ↾J > w −gJ and suppose that sup I < inf J. Let z bethe infimum of all intervals J′ such that sup I < inf J′ and such thatw −gJ′ < inf F ↾J′.

First observe that z /∈∪I and so F is continuousat z and hencew −g(z) ≤lim infr→z+ F(r) = limr→z+ F(r) = F(z)On the other hand, since g is also continuous at z it follows from thedefining property of z thatw −g(z) ≥lim infr→z−F(r) = limr→z−F(r) = F(z)and so F(z)+g(z) = w. Similar arguments in the other cases establishthat one the following two possiblities holds• if I ⊆[x, y] then sup F ↾I < w −gI• if I ⊆[x, y] then inf F ↾I > w −gIConsider the first alternative. If y /∈∪I then F is continuous at y andsolims→y−F(s) + g(s) = F(y) + g(y) ≤wand this is impossible becasue F(y) + g(y) > w. If y ∈(a, b) ∈I andF(a) + g(a) > w then the same argument applies because a /∈∪I .On the other hand, if F(a) + g(a) ≤w then the Darboux property ofF and the fact that g is constant on [a, b] yields z ∈[a, y) such thatF(z) + g(z) = w. The other alternative is dealt with similarly.■Theorem 2.1.

If F is a Darboux function then there is a non-constantcontinuous function g such that F + g is Darboux.Proof:Either F is continuous on an uncountable set or it is not.If it is, use Lemma 2.3 and if it is not then use Lemma 2.2.■3. Sacks RealsThe Sacks partial order of perfect trees will be denoted by S andthe iteration, of length ξ, of this partial order will be denoted by Sξ —so S1 = S and S0 = ∅.

For other notation and definitions concerningSacks reals see [6] as well as [1]. For any p ∈Sξ definep∗= {θ : ξ×ω →2 | (∀F ∈[ξ]<ℵ0)(∀m ∈ω)(θ ↾F×m is consistent with p)}

6JURIS STEPR¯ANSIt is easy to see p∗⊆2ξ×ω is a closed set; but there is no reason tobelieve that it should be non-empty. However, if p is determined —see page 580 of [6] for a definition — then p∗is a reasonably accuratereflection of p. In [6] a notion very similar to p∗is defined and denotedby Ep.

The only difference is that Ep ⊆(2ω)A where A is the domainof p. The projection function from 2ξ×ω to 2γ×ω will be denoted byΠξ,γ.Lemma 3.1. If p ∈Sξ is (E, k)-determined and p ⊩Sα “x ∈R \ V ”then for each E ∈[α]<ℵ0 and k ∈ω there is q such that (q, k) ≤E (p, k)and a function Z : q∗→R such that1.

q ⊩“x = Z(G)”2. Z(x) ̸= Z(y) unless Πω2,1(x) = Πω2,1(y)Proof:This is essentially Lemma 6 on page 580 of [6].

The onlydifference is that it is now required that (q, k) ≤E (p, k) whereas Miller’sLemma 6 only asserts that q ≤p. On the other hand, the assertionrequired here is only that Z(x) ̸= Z(y) unless Πω2,1(x) = Πω2,1(y);whereas a canonical condition for x, in Miller’s terminology, actuallyyields a one-to-one function Z.

The way around this is to choose foreach σ : E × k →2 a condition qσ and a one-to-one function Zσ : (qσ ↾β(σ))∗→R such that qσ ⊩“x = Zσ(Πω2,β(σ)(G))”. The point to noticeis that the domain of Zσ depends on β(σ) and so there may not be asingle ordinal which works for all σ.

Nevertheless, β(σ) ≥1 for eachσ and so it is possible to define Z = ∪σZσ ◦Πω2,β(σ). It follows thatZ(x) ̸= Z(y) unless Πω2,1(x) = Πω2,1(y).■Lemma 3.2.

If p ∈Sξ is (E, k)-determined and F : p∗→R andG : p∗→R are continuous functions such that F ↾q∗̸= G ↾q∗foreach q ≤p then there is some q such that (q, k)

SUMS OF DARBOUX AND CONTINUOUS FUNCTIONS74. Darboux Functions and the Sacks ModelLemma 4.1.

If H : I →R is Darboux then there is a countable set Dsuch that, for any continuous function F, if for every a ∈D and b ∈Dand t such thatH(a) + F(a) < t < H(b) + F(b)there is some c between a and b such that H(c) + F(c) = t then H + Fis also Darboux.Proof:Let D be any countable set such that (H) ↾D is dense inthe graph of H and suppose that F is continuous. If F(x)+H(x) < t 0 suchthat F(z) + H(x) < t if | z −x |< ǫ and F(z) + H(y) > t if | z −y |< ǫ.Because H is Darboux and H ↾D is dense in the graph of H thereare dx ∈D and dy ∈D, between x and y, such that | dx −x |< ǫand | dy −y |< ǫ and H(dx) < t −H(x) and H(dy) > tH(y).

HenceF(dx) + H(dx) < t and F(dy) + H(dy) > t. Hence, if there is some zbetween dx and dy such that H(z) + F(z) = t then z also lies betweenx and y.■For the rest of this section by a condition in Sξ will be meant adetermined condition. Real valued functions will be considered to haveas their domain the unit interval I.

This is merely a convenience thatallows the use of the complete metric space of al continuous real valuedfunctions on the unit interval using the sup metric. This space will bedenoted by C(I, R) and its metric will be ρ(f, g) = sup{| f(x) −g(x) |:x ∈I}.Theorem 4.1.

Let V be a model of 2ℵ0 = ℵ1and V [G] be obtained byadding ω2 Sacks reals with countable support iteration. If H : I →Ris a Darboux function in V [G] then there is a second category set ofcontinuous functions f such that H + f is also Darboux.Proof:If the theorem fails then, in V [G], let H be a Darbouxfunction and X be a comeagre subset of C(I, R) which provide a coun-terexample.

Let D be a countable set, provided by Lemma 4.1, suchthat for any continuous function F, if for every a ∈D and b ∈D andt such that H(a) + F(a) < t < H(b) + F(b) there is some c betweena and b such that H(c) + F(c) = t then H + F is also Darboux. Itmust be true that, for each continuous function g ∈X there is aninterval N(g) = [a, b], with endpoints in D, and a real T(g) betweenH(a) + g(a) and H(b) + g(b) such that there is no z ∈[a, b] such thatH(z) + g(z) = T(g).

8JURIS STEPR¯ANSBy a closure argument, there must exist α ∈ω2 such that• D ∈V [G ∩Sα]• T(f) ∈V [G ∩Sα] for every f ∈V [G ∩Sα]• if x is in V [G ∩Sα] then so is H(x)• X = ∩n∈ωUn where each Un is a dense open set belonging toV [G ∩Sα].To simplify notation it may be assumed that V = V [G ∩Sα]. In V ,let {di | i ∈ω} enumerate D, let G denote the generic function fromω2 × ω to 2 which is obtained from an Sω2 generic set and, let p0 ∈Sω2be a determined condition.Let M be a countable elementary submodel of (H(ω3), ∈) containingthe functions T and N and the name H. Let {En | n ∈ω} be an in-creasing sequence of finite sets such that ∪n∈ωEn = M∩ω2.

(The use ofthe elementary submodel is only a convenience that allows the finitesset En to be chosen before beginning the fusion argument, therebyavoiding some bookkeeping.) Construct, by induction on n ∈ω, func-tions fn, as well as conditions pn ∈Sω2, reals ǫn > 0 and integers kn,all in M, such that:IH(0) fn ∈C(I, R) and f0 is chosen arbitrarilyIH(1) the neighbourhood of fn of radius ǫn in C(I, R) is contained in∩i≤nUnIH(2) ρ(fn −fn+1) < ǫn · 2−n−1IH(3) pn is (En, kn) determinedFor each n, an integer Jn and a sequence Cn = {cnj | j ≤Jn} such thatcn0 = 0 < cn1 < cn2 · · · < cnJn = 1 will be chosen so thatIH(4) dn ∈Cn and Ci ⊆Cn if i ∈nIH(5) if i ∈n and c ∈Ci then fn(c) = fi(c)For each n and each j ≤Jn a continuous function Φn,j : p∗n →R willbe found so that there is a name zn,j such thatIH(6) pn ⊩Sω2 “H(zn,j) = Φn,j(G)” for each j ∈JnA function Zn,j : 2ω2×ω →R will also be constructed so thatIH(7) pn ⊩Sω2 “zn,j = Zn,j(G)”IH(8) if Zn,j(x) = Zn,j(y) then Πω2,1(x) = Πω2,1(y)Let Cn,m,j denote the image of pn under the mapping Zm,j.IH(9) if m < k ≤n , j ∈Jm and i ∈Jk then Cn,m,j ∩Cn,k,i = ∅By [Am,j, Bm,j] will be denoted the interval whose endpoints are the twopoints (fm + H)(cmj ) and (fm + H)(cmj+1).

Observe that IH(5) impliesthat the definition of [Am,j, Bm,j] does not change at later stages of theinduction.

SUMS OF DARBOUX AND CONTINUOUS FUNCTIONS9IH(10) the image of pn under fn ◦Zn,j + Φn,j ◦Πω2,1 contains the interval[An,j, Bn,j] for each j ∈JnFor x ∈[0, 1] let pxn,m,j be the join of all conditions pn | σ such thatσ : En × kn →2 is consistent with pn and x belongs to the image of(pn | σ)∗under the mapping fn ◦Zm,j + Φm,j ◦Πω2,1. The following isthe key inductive requirement.IH(11) if x ∈[Am,j, Bm,j] then (pxn+1,m,j, kn+1)

The integers m and j mustexist because the endpoints of N(f) belong to D and so N(f) = [cmi , cmk ]for some m, i and k. Furthermore, from IH(5) it follows that [H +f(cmi ), H + f(cmk )] = [H + fm(cmi ), H + fm(cmk )] = Si≤v

It follows that T(f) belongs to the image of p∗ωunder the mapping f ◦Zm,j + Φm,j ◦Πω2,1. Furthermore, because thediameters of the images of (pT(f)n,m,j)∗under the mappings fn◦Zm,j +Φm,japproach 0 as n increases, it follows that f ◦Zm,j + Φm,j ◦Πω2,1 hasconstant value T(f) on p∗ω.It follows that pω ⊩“f ◦Zm,j + Φm,j ◦Πω2,1(G) = T(f)”.

FromIH(6) and the fact that pω ≤pm it follows that pω ⊩“H(zm,j) =Φm,j◦Πω2,1(G)” and from IH(7) that pω ⊩“zm,j = Zm,j(G)”. Thereforepω ⊩“f(zm,j) + H(zm,j) = T(f)” and this is a contradiction becausezm,j ∈[cmj , cmj+1] ⊆N(f) by definition.To carry out the induction suppose that fn, {Φn,j | j ∈Jn} and{Zn,j | j ∈Jn} as well as conditions pn ∈Sω2 have all been defined forn ≤K.

To begin, let 0 = cK+10< cK+11< cK+12< . .

. < cK+1JK+1 = 1 besuch that:• {dK} ∪CK ⊆CK+1 = {cK+1i| i ≤JK+1}• the diameter of the image of [cK+1i, cK+1i+1 ] under fK is less thanǫK · 2−K−4

10JURIS STEPR¯ANS• 0 <| H(cK+1i) −H(cK+1i+1 ) |< ǫK · 2−K−3The first condition ensures that IH(4) is satisfied. The second is easilyarranged using uniform continuity.

The last condition can be satisfiedby a further refinement using the Darboux property of H.Note that Πω2,1(p∗K), the image of p∗K under Πω2,1, is perfect and so,for each i ∈JK+1 it is possible to find ΦK+1,i : Πω2,1(p∗K) →R suchthat• ΦK+1,i is a continuous mapping• the image of p∗K under ΦK+1,i◦Πω2,1 is the interval whose endpointsare H(cK+1i) and H(cK+1i+1 )• if (m, j) ̸= (K + 1, i) then ΦK+1,i ◦Πω2,1(x) ̸= Φm,j ◦Πω2,1(x) forevery x ∈p∗K• ΦK+1,i is finite-to-oneObserve that the last point implies that ΦK+1,i ◦Πω2,1(G) does notbelong to the ground model V .In any generic extension there must be a real between cK+1iand cK+1i+1at which H takes on the value ΦK+1,i ◦Πω2,1(G) because H is assumedto be Darboux. Let zK+1,i be a name for such a real.

It follows fromthe choice of ΦK+1,i that 1 ⊩“zm,j ̸= zK+1,i” for each m ≤K + 1 andj ∈Jm such that (K + 1, i) ̸= (m, j).Now find k and p such that• (p, kK)

SUMS OF DARBOUX AND CONTINUOUS FUNCTIONS11(q, kK+1) ≤EK+1 (¯p, kK+1) and for each i ∈JK+1 there is a functionZK+1,i : q∗→[cK+1i, cK+1i+1 ] such that• q ⊩“zK+1,i = ZK+1,i(G)”• ZK+1,i(x) ̸= ZK+1,i(y) unless Πω2,1(x) = Πω2,1(y)Now observe that if (m, j) ̸= (K + 1, i) then there can not be ˆq ≤qsuch that Zm,j ↾ˆq∗= ZK+1,i ↾ˆq∗because it has already been remarkedthat 1 ⊩“zm,j ̸= zK+1,i” for each m ≤K + 1 and j ∈Jm such that(K+1, i) ̸= (m, j). It is therefore possible to use Lemma 3.2 repeatedlyto find a single condition pK+1 such that (pK+1, kK+1)

Observe that pK+1 is (EK+1, kK+1) de-termined because ¯p is. Hence IH(3) is satisfied.

Now define CK+1,K+1,ito be the range of ZK+1,i. This, along with the induction hypothesis,will guarantee that IH(6), IH(7), IH(8) and IH(9) are all satisfied.For integers m ≤K + 1, j ∈Jm let {[u0m,j,v, u1m,j,v] | v ∈Lm,j} bea partition of [Am,j, Bm,j] into intervals of length ǫK · 2−K−2.

Now, foreach σ : EK+1 × kK+1 →2 and for each pair of integers m ≤K + 1,j ∈Jm and for each v ∈Lm,j let W[σ, m, j, v] be a perfect, nowheredense subset of(fK + Φm,j ◦(Πω2,1 ↾(pK+1 | σ)) ◦Z−1m,j)−1[u0m,j,v −ǫK2K+3, u1m,j,v + ǫK2K+3]if this is possible. By choosing smaller sets, if necessary, it may be as-sumed that the sets W[σ, m, j, v] are pairwise disjoint and that W[σ, m, j, v]∩CK+1 = ∅.

Then define Fσ,m,j,v : W[σ, m, j, v] →[u0m,j,v, u1m,j,v] to beany continuous surjection and let fσ,m,j.v = Fσ,m,j,v −Φm,j ◦Πω2,1 ◦Z−1m,j.Note that IH(8) implies that Πω2,1◦Z−1m,j is a function even though Zm,jis not one-to-one.Similarly, for each i ≤JK+1 let Wi be a perfect, nowhere densesubset of [cK+1i, cK+1i+1 ] disjoint from each W[σ, m, j, v] and define Fi :Wi →[AK+1,i, BK+1,i] to be a continuous surjection. Then let f i =Fi −ΦK+1,i◦Πω2,1◦Z−1K+1,i.

Notice that the domains of all the functionsfσ,m,j,v and f i are pairwise disjoint. Hence it is possible to find fK+1 :I →R extending each of these functions in such a way that ρ(fK+1, fK)does not exceedmax{| fK+1(y) −fK(y) |: y ∈(∪iW i) ∪(∪σ,m,j,vW[σ, m, j, v])}and, moreover, because W[σ, m, j, v] ∩CK+1 = ∅and Wi ∩CK+1 = ∅, itmay also be arranged that fK+1(c) = fK(c) if c ∈CK.

Therefore IH(5)is satisfied as well as IH(0). Observe that IH(10) is satisfied becausethe choice of Fj ensured that it maps Wj onto [AK+1,j, BK+1,j].

Since

12JURIS STEPR¯ANSfK+1 ↾Wj = Fj −ΦK+1,j ◦Πω2,1 ◦Z−1K+1,j it follows that fK+1 ◦ZK+1,j +ΦK+1,j ◦Πω2,1 maps pK+1 onto [AK+1,j, BK+1,j].To see that IH(2) holds it suffices to consider onlyfK+1 ↾(∪iW i) ∪(∪σ,m,j,vW[σ, m, j, v])because fK+1 was defined not to exceed this bound.Consider firsty ∈W[σ, m, j, v]. Then | fK+1(y) −fK(y) | is equal to| fσ,m,j,v(y) −fK(y) |=| Fσ,m,j,v(y) −(Φm,j ◦Πω2,1 ◦Z−1m,j(y) + fK(y)) |Next, it is possible to use the definition of Fσ,m,j,v to conclude thatFσ,m,j,v(y) ∈[u0m,j,v, u1m,j,v].

Using the definition of W[σ, m, j, v] it ispossible to conclude that(fK + Φm,j ◦Πω2,1 ◦Z−1m,j)(y) ∈[u0m,j,v −ǫK2−K−3, u1m,j,v +ǫK2−K−3]because y ∈W[σ, m, j, v].Consequently, | fK+1(y) −fK(y) | is nogreater than the diameter of[u0m,j,v −ǫK2−K−3, u1m,j,v +ǫK2−K−3]which is ǫK · 2−K−3 + ǫK · 2−K−2 + ǫK · 2−K−3 = ǫK · 2−K−1.On the other hand, if y ∈Wi then, as before,| fK+1(y) −fK(y) |=| Fi(y) −(Φm,j ◦Πω2,1 ◦Z−1m,j(y) + fK(y)) |Recall that ΦK+1,i is chosen to map onto [H(cK+1i), H(cK+1i+1 )]; moreover,because y ∈[cK+1i, cK+1i+1 ] it follows from the choice of CK+1 thatfK(y) ∈[fK(cK+1i) −ǫK2−k−4, fK(cK+1i+ǫK2−k−4]and so Φm,j ◦Πω2,1◦Z−1m,j(y)+fK(y) belongs to [AK+1,i−ǫK2−K−4, BK+1,i+ǫK2−K−4]. Furthermore, Fi(y) belongs to [AK+1,i, Bk+1,i] by design.

By thechoice of CK+1 the diameter of [AK+1,i, Bk+1,i] is less thanǫK2−K−4 +ǫK2−K−3and so the diameter of[AK+1,i −ǫK2−K−4, BK+1,i +ǫK2−K−4]is no greater thanǫK2−K−1 and so it follows that | fK+1(y) −fK(y) |<ǫK · 2−K−1.Now all of the induction hypotheses have been shown to be satisfiedexcept for IH(1) and IH(11). To verify IH(11) suppose that m ≤K,j ∈Jm and x ∈[Am,j, Bm,j].

It follows that there is some v ∈Lm,jsuch that x ∈[u0m,j,v, u1m,j,v]. Suppose also that σ : EK+1 × kK →2 isconsistent with pxK,m,j.

It follows that there is some σ′ : EK+1 × k →2such that• σ′ is consistent with p

SUMS OF DARBOUX AND CONTINUOUS FUNCTIONS13• σ ⊆σ′• the distance from x to the image of (p | σ′)∗under the mappingfK ◦Zm,j + Φm,j ◦Πω2,1 is less than ǫK · 2−K−4It suffices to show that τ is consistent with pxK+1,m,j for each τ : EK+1×kK+1 →2 such that σ′ ⊆τ; the reason for this is that kK+1 was cho-sen so that (¯p, kK+1) ≤EK+1 (pK, k) and (pK+1, kK+1) ≤EK+1 (¯p, kK+1).Recall that the diameter of the image of (p | σ′)∗under the mappingfK ◦Zm,j +Φm,j ◦Πω2,1 is less than ǫK ·2−K−4 because σ′ : EK+1×k →2is consistent with p.Because the distance from x to to the imageof (p | σ′)∗under the mapping fK ◦Zm,j + Φm,j ◦Πω2,1 is less thanǫK · 2−K−4 it must be that this image is contained in [u0m,j,v −ǫK ·2−K−3, u1m,j,v −ǫK · 2−K−3]. Because pK+1 < p it follows that the imageof (pK+1 | τ)∗under the mapping fK ◦Zm,j + Φm,j ◦Πω2,1 is containedin [u0m,jv −ǫK · 2−K−3, u1m,jv −ǫK · 2−K−3] and so W[τ, m, j, v] ̸= ∅.

Thechoice of Fτ,m,j,v ensures that it maps W[τ, m, j, v] onto [u0m,j,v, u1m,j,v]and therefore fK+1+Φm,j ◦(Πω2,1 ↾(pK+1 | τ))◦Z−1m,j maps W[τ, m, j, v]onto [u0m,j,v, u1m,j,v]. Hence τ is consistent with pxK+1,m,j.Finally, choose ǫK+1 so that the neighbourhood of fK+1 of radiusǫK+1 is contained in UK+1.■Corollary 4.1.

If set theory is consistent then it is consistent thatfor every Darboux function F there is a nowhere constant continuousfunction f such that F + f is also Darboux.Proof:The model to use is the one for Theorem 4.1.Given aDarboux F to obtain a nowhere constant continuous f use the factthat the set of nowhere constant function is comeagre in C(I, R).■5. Further RemarksIt should be observed that the function f in Corollary 4.1 has veryfew nice properties other than continuity.It is natural to ask thefollowing question.Question 5.1.

Is there a Darboux function H : I →R such that H+fis not Darboux for every non-constant, differentiable function f?The answer to Question 5.1 for functions with continuous derivativeis positive. The same question can be asked with absolutely continu-ous in the place of differentiable.

One should recall that differentiablefunctions satisfy the property T1 of Banach [8].

14JURIS STEPR¯ANSDefinition 5.1. A function F : R →R satisfies T1 if and only if theset of all x such that f −1{x} is infinite has measure zero.Banach showed that differentiable functions satisfy T1.

Question 5.1is of interest for differentiable functions because Corollary 2.1 showsthat a strengthening of T1 yields a positive theorem.Another potentially interesting direction to pursue would be to askwhether the size of the set of continuous functions in Theorem 4.1 canbe increased.Question 5.2. Is there a Darboux function F such that the set of con-tinuous functions f such that F + f is Darboux is comeagre?Question 5.3.

Is there a Darboux function F such that the set of con-tinuous functions f such that F + f is Darboux has measure one?In [4] the authors consider not only sums of a Darboux functionand a continuous function but also products and other algebraic con-structions. It is not difficult to check that everything that has beenestablished in this paper for sums also holds for products, but it is notclear that this must always be so.Question 5.4.

If there is a Darboux function F such that F + g isnot Darboux for every nowhere constant function g must it also be thecase that there is a Darboux function F such that F · g is not Dar-boux for every nowhere constant function g? What about the oppositeimplication?References1.

J. E. Baumgartner and R. Laver, Iterated perfect set forcing, Ann. Math.

Logic17 (1979), 271–288.2. A. Bruckner, Differentiation of real functions, Lecture Notes in Mathematics,vol.

659, Springer-Verlag, Berlin, 1978.3. A. Bruckner and J. Ceder, On the sums of Darboux functions., Proc.

Amer.Math. Soc.

51 (1975), 97–102.4. B. Kirchiem and T. Natkaniec, On universally bad Darboux functions., RealAnal.

Exchange 16 (1990-91), 481–486.5. P. Komjath, A note on Darboux functions, Real Anal.

Exchange 18 (1992-93),no. 1, 249–252.6.

A. Miller, Mapping a set of reals onto the reals, J. Symbolic Logic 48 (1983),no.

3, 575–584.7. T. Radakoviˇc, Uber Darbouxsche und stetige Funktionen, Monatshefte Math.Phys.

38 (1931), 117–122.8. S.Saks, Theory of the integral, Hafner, New York, 1937.Department of Mathematics, York University, Toronto, CanadaM3J 1P3, stepransnexus.yorku.ca

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