Wetzel families and the continuum

We provide answers to a question brought up by Erdős about the construction of Wetzel families in the absence of the continuum hypothesis: A Wetzel family is a family F$\mathcal {F}$ of entire functions on the complex plane which pointwise assumes fewer than |F|$\vert \mathcal {F} \vert$ values. To be more precise, we show that the existence of a Wetzel family is consistent with all possible values κ$\kappa$ of the continuum and, if κ$\kappa$ is regular, also with Martin's Axiom. In the particular case of κ=ℵ2$\kappa = \aleph _2$ this answers the main open question asked by Kumar and Shelah [Fund. Math. 239 (2017) no. 3, 279–288]. In the buildup to this result, we are also solving an open question of Zapletal on strongly almost disjoint functions from Zapletal [Israel J. Math. 97 (1997) no. 1, 101–111]. We also study a strongly related notion of sets exhibiting a universality property via mappings by entire functions and show that these consistently exist while the continuum equals ℵ2$\aleph _2$ .


Introduction
This paper is an investigation related to Wetzel's problem which comes from analysis yet has surprising set-theoretical aspects.While the subjects of analysis and set theory might nowadays be conceived as somewhat distant to each other, there are some examples of topics belonging to them both.Among the most prominent one is the problem of sets of uniqueness, cf.[21], which led Cantor to investigate sets of real numbers and subsequently to the founding of set theory.
One of the greatest contributions of Cantor was the discovery of ordinals and cardinals and the distinction between countable and uncountable sets of reals.This led him to the formulation of the continuum hypothesis, CH, which states that the cardinality of the set of all real numbers is the smallest one conceivable in light of this, the smallest uncountable cardinal ℵ 1 , cf. [9].In 1904 Ernst Zermelo axiomatised set theory in a way conforming to mathematician's practise hitherto, cf.[34].Subsequently, A. Fraenkel added the replacement scheme, cf.[15], thus yielding the system ZFC.Somewhat later, Kurt Gödel showed that the continuum hypothesis cannot be refuted within this system (provided that there is anything which cannot be derived within it), cf.[17].Gödel conjectured, cf.[19, Section 4], that neither can it be proved in it but only several decades later, Paul Cohen, at the origin an analyst just like Cantor, developed the method of forcing and could prove that this is indeed the case, cf.[12].
Nowadays it is common to subdivide analysis into real and complex analysis.The latter's theorems about its objects of study, holomorphic functions, revealed deep connections between analysis and geometry and found applications in various areas, among them number theory.One striking feature of the family of functions which are holomorphic on some domain of complex numbers opposite the family of those which are merely smooth on an interval of real numbers is the intertwinement of the local and global behaviour of holomorphic functions.Whereas two distinct functions may be both identical and infinitely often differentiable on an interval of real numbers, the situation in the complex domain is quite different, due to the famous "identity theorem".According to it, for any two distinct functions holomorphic on some complex domain, the set of points where they agree is discrete (see Proposition 2.1).
As the complex plane is separable, no uncountable set of complex numbers is discrete.Therefore, any two distinct holomorphic functions can only agree on a countable set of points.Subsequently further theorems underscored the difference between the realms of entire functions on the one hand and smooth functions on the real number line on the other.In the middle of the nineteenth century, it emerged from work of Liouville, cf.[18,Chapter 11], that all bounded entire functions are constant.Later, Nevanlinna developed the theory named after him, cf.[10], one upshot of which is that for any two distinct entire functions f and g there can be at most four complex values a for which the preimages of {a} are equal.Another interesting property that is purely of combinatorial nature and can be stated irrespective of the topological and algebraic structure of C, is Picard's Little Theorem, namely the fact that any non-constant entire function can avoid at most one single value (see e.g.[27,Theorem 16.22]).
The emerging picture of complex analysis in general and of entire functions in particular was one of strong general principles governing their behaviour.Against this backdrop, while writing his dissertation during the sixties of the last century, John E. Wetzel asked (cum grano salis) whether any family F of entire functions such that for each complex number z the set {f (z) : f ∈ F } is countable, must be countable itself.Dixon showed that, assuming the failure of the continuum hypothesis, this is indeed the case -in fact, this is a corollary of the aforementioned identity theorem, cf.[16].Shortly thereafter, Erdős proved that not only does the negation of the continuum hypothesis imply this statement, it is equivalent to it, [13].In fact this result is one among many statements in various areas of mathematics proved to be equivalent to the continuum hypothesis, cf.[31,6].Towards the end of his paper, Erdős asked whether the analogue statement resulting from replacing "countable" by "fewer than continuum many" can be proved without assuming the continuum hypothesis.Following a suggestion by Martin Goldstern we subsequently refer to a family of entire functions whose members everywhere assume fewer values than the family has members altogether as a Wetzel family (see Definition 3.1).In this terminology, Erdős asked whether the existence of a Wetzel family is provable from ZFC.One might also ask whether the continuum hypothesis is equivalent to the existence of a Wetzel family.
Not long after the development of forcing by Cohen, Solovay and Tennenbaum instigated the theory of iterated forcing and Tony Martin stated what became known as Martin's Axiom, or MA for short, a weakening of the continuum hypothesis.It does not prescribe a particular value for the cardinality of the continuum but it does for instance imply that its cardinality is regular.In many cases when something can be proved for countable sets within ZFC, Martin's Axiom allows us to generalise this to sets with fewer than 2 ℵ0 elements.Meanwhile Erdős' question remained unanswered.
The threads regarding Wetzel's problem were only picked up again in 2017 by Ashutosh Kumar and Saharon Shelah who answered both questions above in the negative, albeit in a slightly non-satisfactory way.They showed that there is no Wetzel family in the side-by-side Cohen model and provided a model with a Wetzel family (and hence a continuum, cf.Lemma 3.2) of cardinality ℵ ω1 .The singularity of ℵ ω1 is quite crucial in their argument and the result could not be generalized to other cardinals.Moreover, their model necessarily fails to satisfy Martin's Axiom.
It seems that the paper by Kumar and Shelah resuscitated interest in Wetzel's problem.We are aware of two more papers dealing with it since then, a formalisation of Erdős' proof, [26], and a proof that the continuum hypothesis implies the existence of sparse analytic systems, [11].But no one yet addressed the open question which Kumar and Shelah ask at the end of [22], of whether the existence of a Wetzel family is consistent with a continuum of cardinality ℵ 2 .Erdős' proof relied on the fact that any countable dense set of complex numbers is universal for countable sets via entire functions, that is to say that any countable set may be mapped into it via a non-constant entire function (see Definition 3.4 and Proposition 3.5).In fact one finds a few papers from the sixties and seventies of the last century studying similar yet slightly stronger mapping properties, cf.[24,2,3,28,25].Kumar and Shelah observed that a Wetzel family would exist in a model of 2 ℵ0 = ℵ 2 in which there is a set of cardinality ℵ 1 universal in the sense above for sets of cardinality ℵ 1 of complex numbers.
The main result of our paper is Theorem 5.13, that shows that starting from a model of the generalised continuum hypothesis and any cardinal κ of uncountable cofinality, there is a cardinal and cofinality preserving forcing extension with a Wetzel family of size κ.In particular this completely solves Kumar and Shelah's open problem by showing that Wetzel families put no further restriction on the size of the continuum.Moreover, for regular κ we can also force Martin's axiom.We also study the notion of universality from above and show that while MA precludes the existence of sufficiently universal sets, they can consistently exist while 2 ℵ0 = ℵ 2 .
While some basic knowledge of set theory is needed to understand the main results, a large part of the arguments (with an exception of those in Section 4 and 7) is more analytic than set theoretic.
The paper is organized as follows.In the following first section we review some of the preliminaries in complex analysis and forcing that are used in the paper.In the next section, Section 3, we introduce Wetzel families and universal sets and prove some ZFC results about them.Among other things, we show that Wetzel families must have cardinality 2 ℵ0 , we provide a proof of Erdős' result on countable dense sets and we show that universal sets imply the existence of Wetzel families.In Section 4, we show how to force a certain family of strongly almost disjoint functions that serves as a basic (and somewhat necessary) ingredient in the proof of the main result.In fact, it turns out that this solves [33,Question 22].We also obtain some interesting additional results related to MA that shine some light on one of our open questions.This section can be read completely independently from the rest the paper.Section 5 is the longest and contains the proof the main result, Theorem 5.13.In Section 6, we show that universal sets do not exist under MA + ¬CH.As a corollary we obtain that the converse of Proposition 3.7, namely the statement that Wetzel families imply the existence of a universal set, does not hold.In the next Section, we then show that a universal set, as suggested by Kumar and Shelah, can consistently exist with continuum ℵ 2 .This uses a proper forcing, based on some of the previous arguments, with pairs of models as side-conditions.We finish the paper with a list of open problems.

Preliminaries
2.1.Complex analysis.Throughout the paper, C denotes the set of complex numbers.A function f : C → C is entire if it is holomorphic on the domain C, in other words, its complex derivative f ′ exists in every point z ∈ C. The set of entire functions will be denoted H(C).
For z ∈ C and δ a positive real number, we let be the ball of radius δ around z.We also define the semi-norms Recall that a sequence of functions f = f n : n ∈ ω on C is said to converge uniformly on compact sets, if for every δ > 0, f converges uniformily on B δ (0), i.e. for every ε > 0, there is N ∈ ω so that for all n 0 , n 1 > N , Among the most useful facts about entire functions that we will use is the following.Proposition 2.2 (see [27,Theorem 10.28]).Let f n : n ∈ ω be a sequence of entire functions that converges uniformily on every compact set.Then the pointwise limit f of f n : n ∈ ω is entire.
Moreover, the sequence of derivatives f ′ n : n ∈ ω converges uniformily on every compact set to f ′ .

2.2.
Forcing.Here we review a few standard facts about forcing that we will find useful.We use standard forcing notation as used in the reference books [20] or [23].
Lemma 2.3 ([23, Lemma V.3.9,V.3.10]).Let P * Q be a two step iteration.Then P * Q is ccc iff P is ccc and P Q is ccc.Definition 2.4.Let P, Q be forcing notions.Then P is a subforcing of Q if P ⊆ Q and the extension as well as the incompatibility relations agree.Definition 2.5.Let M ⊆ V be a transitive model of ZF − (possibly a proper class). 1 Let P ∈ M be a subforcing of Q.Then we write P ⋖ M Q to say that every predense set E ∈ M of P is predense in Q.We write P ⋖ Q for P ⋖ V Q and say that P is a complete subforcing of Q. Lemma 2.6 ([32, Theorem 6.3]).The iterative direct limit of ccc forcings is ccc.To be more precise, suppose P δ : δ ≤ α is a sequence of posets, α limit, so that (1) for all γ ≤ δ ≤ α, P γ ⋖ P δ , (2) for every limit δ ≤ α, γ<δ P γ is dense in P δ , (3) and for all δ < α, P δ is ccc.Then also P α is ccc.
Lemma 2.7.Let P be a complete subforcing of Q, Ȧ a P-name and Ḃ a Q-name for a forcing notion.Then Proof.It suffices to notice that a predense subset of P * Ȧ (and respectively of Q * Ḃ), is precisely the same as a P-name (or Q-name) for a predense subset of Ȧ (respectively Ḃ).For the direction from left to right, see also for example [7,Lemma 13].Proof.Suppose towards a contradiction that |F | < 2 ℵ0 .For any distinct f, g ∈ H(C), X f,g = {z ∈ C : f (z) = g(z)} does not have an accumulation point by Proposition 2.1 and thus must be countable.Thus, if X = f =g∈F X f,g , then In particular, there is some z ∈ C \ X.But then for any distinct f, g ∈ F , f (z) = g(z) and so |{f (z) : This lemma in fact shows that a Wetzel family induces a quite non-trivial combinatorial object, especially when 2 ℵ0 > ℵ 2 .Namely, consider an enumeration z α : α < κ of C. For each α < κ, there is a bijection e α : {f (z α ) : f ∈ F } → µ α for some cardinal µ α < 2 ℵ0 .In this way, we can think of each f ∈ F as the function σ f ∈ α<κ µ α , where σ f (α) = e α (f (z α )).At the same time, the elements of F and thus of {σ f : f ∈ F } have pairwise countable intersections.
Under CH, this type of almost disjoint family of functions can be obtained quite easily.For instance, for any α < ω 1 , simply let σ α be constantly 0 below α and constantly equal to α above α.
Also this is not particularily hard when 2 ℵ0 = ℵ 2 .Whenever we have constructed σ β : β < α for some α < ω 2 , we can find a single function σ : α → ω 1 that has countable intersection with all σ β using a standard diagonalization argument.Then we simply let σ α equal σ below α and constantly equal to α above α.
For larger continuum, the existence of such families becomes much less clear, as a consequence of the larger gap between the countable size of the pairwise intersections and the size of the family and their elements.In fact, in Section 5 we will show how to force these types of families, based on a techinque by Baumgartner.This will be a key starting point for the construction of a Wetzel family by forcing.
As a small observation of independent interest let us mention: Lemma 3.3.A Wetzel family cannot consist only of polynomials.
Proof.Suppose F is such a family.Since |F | = 2 ℵ0 and 2 ℵ0 has uncountable cofinality, we can assume that all polynomials in F have the same degree n.Then pick any n + 1-many points a 0 , . . ., a n ∈ C. The set {f ↾ {a 0 , . . ., a n } : f ∈ F } has size less than 2 ℵ0 , because F is Wetzel.But each f ∈ F is uniquely determined by f ↾ {a 0 , . . ., a n } and so F has small size as well, contradicting Lemma 3.2.
Let us write the argument for sake of completeness.In a way, the forcing notions we will use later mimic this construction.
Proof.Let Y ⊆ C be countable dense and X ⊆ C be arbitrary countable and enumerated as z n : n ∈ ω .We recursively construct a sequence f n : n ∈ ω of entire functions that converges uniformily on compact sets.Start by simply letting f 0 be constantly 0. Next, if f n has been defined, consider Note that the set of zeros of g n,ξ is exactly {z m : m < n}, when |ξ| > 0. Then there is δ > 0, so that Then let f n+1 = f n + g n,ξ .Finally, let f be the limit of Proposition 3.6.Let Y be a universal set.Then |Y | + = 2 ℵ0 and in particular the continuum is a successor cardinal.
Proof.Suppose that there is and f ′′ X ⊆ Y , by the pigeonhole principle, there is y ∈ Y such that {z ∈ X : f (z) = y} is uncountable.But then this set has an accumulation point and by Proposition 2.1, f is constant.Proposition 3.7.If there is a universal set there is also a Wetzel family.
Proof.Let Y be universal and let First of all, we note that |F | = κ.By Proposition 3.6, κ is a successor and in particular regular.Thus, if |F | < κ, there is an unbounded subset S ⊆ κ so that f α = f β , for all α, β ∈ S.But then for any such α ∈ S, f α (C) ⊆ Y .This would imply that f α is constant, as in the proof of Proposition 3.6.Now let z α ∈ C be arbitrary.Then {f (z α ) : f ∈ F } ⊆ {f β (z α ) : β ≤ α} ∪ Y and the right-hand-side has size less than κ.

Strongly almost disjoint functions and Baumgartner's thinning-out forcing
This section can be skipped entirely if one wants to pass directly to the proof of the main result.The main purpose is to prove Proposition 4.1 below which is used in the setup of our main forcing construction.It is an adaptation of Baumgartner's "thining-out" technique to obtain certain types of almost disjoint families (see [5]).To be more precise, we show how to obtain the type of family of functions necessitated by a Wetzel family, as shown in Section 3, where pairwise intersections are finite.Interestingly, a different type of strongly almost disjoint family was also used in [22] to obtain a Wetzel family with continuum ℵ ω1 .It is unclear to us how this relates to our argument.Proposition 4.1.(GCH) Let κ be an infinite cardinal of uncountable cofinality and for every α < κ, let µ α = max(|α|, ℵ 0 ).Then there is a cardinal and cofinality preserving forcing extension of V where 2 ℵ0 = κ and there is σ α : α < κ , such that for all α < β < κ, (1) for all α, β ∈ dom p, proj p(α) = proj p(β), i.e. p(α) and p(β) have the same projection to the first coordinate, (3) and for all α ∈ dom p and ξ ∈ proj p(α) ∩ λ, |p(α) ∩ ({ξ} × µ ξ )| = µ ξ .We will simply write proj p to denote proj p(α), for any α ∈ dom p, and if dom p = ∅, we let proj p = ∅.
For any p, q ∈ P λ , q ≤ p iff dom p ⊆ dom q and for any distinct α, β ∈ dom p, Whenever G is P λ -generic, let S α = p∈G p(α) for every α < κ.Then it is not hard to check that every vertical section of S α is non-empty (see more below) and for any α < β, In particular, when λ is a successor cardinal, In fact, for any ξ ∈ [ω, λ − ), the section with index ξ of S α equals µ ξ .The reason why we include these sections is purely notational.
It is clear that P λ is < λ-closed.Ideally we would like to force with P ω , since if we then choose σ α , such that (ω + ξ, σ α (ξ)) ∈ S α , for every ξ < κ, (1) and ( 2) of the proposition are satisfied.But P ω is far from being ccc.To circumvent this we use Baumgartner's thinning out trick.
Let P ⊆ λ∈K P λ consist of all p = p λ : λ ∈ K such that for λ ′ < λ, dom p λ ′ ⊆ dom p λ and for any α ∈ dom p λ ′ , p λ ′ (α) ⊆ p λ (α).When G is P-generic, we obtain the sets S λ α = p∈G p λ (α) for every λ ∈ K and it is very easy to see again that Let us check that all vertical sections of S λ α are non-empty.To this end let p ∈ P be arbitrary.
Let P 1 consist of all p ↾ λ + , for p ∈ P, where for each α ∈ dom p λ .Then it is easy to verify that if ] and suppose that pi : i < λ + is an antichain in P 1 .
Note that for any i < λ + , there is δ < λ so that dom p i λ ′ as well as p i λ ′ (α) are the same for all λ ′ ∈ [δ, λ) ∩ K and fixed α ∈ dom p i λ , since these sets grow and are subsets of dom p i λ and p i λ (α) respectively which have size strictly less than λ. 3 We may assume that this δ is the same for all i < λ + and let 4Since λ <λ = λ, a ∆-system argument lets us assume that there are sets D and E such that D = D i ∩ D j and dom p i λ ∩ dom p j λ = E, for all i < j < λ + .Note that we can also do the same for all λ ′ ∈ δ ∩ K simultaneously, as δ < λ.To be more precise we can assume D λ ′ = dom p i λ ′ ∩ dom p j λ ′ , for all λ ′ ∈ δ ∩ K and i < j < λ + .Moreover, using another ∆-system argument we can assume that R i λ ∩ R j λ = R, for some fixed R and all i < j < λ + , where R i λ = α∈dom p i λ p i λ (α).Now comes the heart of the thinning out argument.Let i < λ + be arbitrary, then note that for any distinct α, β and any λ ′ , p i λ ′ (α) ∩ S(β) is a < λ sized subset of S(α) ∩ S(β), which has size at most λ.Also R has size < λ.Thus we find i < j < λ + such that , for all λ ′ ∈ δ∩K and distinct α, β ∈ D λ ′ , also for any, equivalently all, λ ′ ∈ [δ, λ)∩K and distinct α, β ∈ D, and for λ ′ = λ and distinct α, β ∈ E. It is straightforward to check that pi and pj are compatible.Now let us check that all regular cardinals θ are preserved.Suppose that in V P , cf(θ) = µ < θ.If µ ≥ κ, then either κ is regular and we have shown above that P is κ + -cc, so in particular θ-cc.Or κ is singular, µ ≥ κ + and |P| ≤ κ + , so P is κ ++ -cc and also θ-cc.If µ < κ, P is the iteration of a µ + -closed and a µ + -cc forcing and θ ≥ µ + .

Wetzel families with arbitrary continuum and MA
In this section we prove our main result that Wetzel families can coexist with arbitrary values of the continuum and in combination with Martin's Axiom.where a p ∈ [C] <ω , f p ∈ H(C), ε p is a positive rational number and m p ∈ ω.A condition q extends p iff a p ⊆ a q , f q ↾ a p = f p ↾ a p , ε q ≤ ε p , m p ≤ m q and f q − f p mp ≤ ε p − ε q .
Definition 5.2.Let H : X → P(C), for some X ⊆ C. Then we define Note that the notion of incompatibility of conditions p, q ∈ Q(H) isn't dependent on H. Namely, if r extends p and q in Q, then (a p ∪ a q , f r , ε r , m r ) ∈ Q(H) and also extends p and q.In other words, Q(H) is a subforcing of Q (see Definition 2.4).For most considerations it is also not relevant in which transitive model of set theory M we evaluate the definition of Q(H), as long as H ∈ M .If we restrict the functions appearing as f p to those that are polynomials in coefficients in the field generated by dom H ∪ z∈dom H H(z), we obtain a dense subforcing of Q(H).Note that f ξ (y) = f p (y), for every y ∈ a p .Let z ∈ dom H \ a p be arbitrary.Since H(z) is dense, we can easily find a small enough ξ so that dense open.We claim that for any sequences p n : n ∈ ω and q n : n ∈ ω , with p n , q n ∈ D n ∩ G, f pn : n ∈ ω and f qn : n ∈ ω converge uniformly on compact sets to the same function f ∈ H(C).To see this, use Proposition 2.2 and notice that when p, q ∈ D n ∩ G are arbitrary, there is r ≤ p, q and thus For any z ∈ dom H, we can find a decreasing sequence p n : n ∈ ω such that Let us make the following interesting observation that will somewhat elucidate the necessity of the approach taken in the proof of the main result.
Lemma 5.4.Let dom H be uncountable and suppose that there is an entire f such that f (z) ∈ H(z), for every z ∈ dom H. Then Q(H) is not ccc.
It would be interesting to obtain some sort of converse to Lemma 5.4.For instance, suppose that H only maps to countable sets.Does the non-ccc of Q(H) imply at least that there is a Borel function f , with f (z) ∈ H(z) for uncountably many z ∈ dom H(z)?Corollary 5.5.Let dom H be uncountable and suppose that H(z) is dense in C, for every z ∈ dom H. Then Q(H) forces that Q(H) is not ccc.
Proof.Either Q(H) is already not ccc and stays so in any extension, or dom H is preserved to be uncountable and by Lemma 5.3 we are in the situation of the previous lemma.
Thus the forcings Q(H) can generally not be recycled in a ccc construction.If one wants to add another entire function, one has to pass to a new H.

Tools for the successor step.
Lemma 5.6.Let l, m ∈ ω and K ⊆ C l+1 be compact, such that every element of K is one-to-one.Then there is L > 0 such that for any z ∈ K, there is g ∈ H(C) with g m < L, g(z i ) = 1 for every i < l and g(z k ) = 0.
Proof.Consider an interpolation formula such as and simply note that z → g(z, •) is a continuous map from K to H(C) in the norm • m .The claim follows from the compactness of K.
Lemma 5.7.Let H 0 , . . ., H n be such that 5 For instance, this follows easily from Proposition 2.2.This part of the argument strongly depends on the special geometry of holomorphic functions.The statement is clearly not true for functions that are merely infinitely often differentiable.
Note that by a simple inductive argument, the previous lemma implies that we can extend simultaneously each H i in countably many arbitrary points with arbitrary countable sets of values and preserve the ccc: 6 When l = 0, then C l contains one element, namely the empty sequence, which all zα then equal to.
5.3.Tools for the limit step.
Lemma 5.9.Let M ⊆ V be a transitive model of ZF − (possibly a proper class). where Whenever f ∈ M , we can assume that f = f .
Before we continue to the proof let us note that g ξ ∈ H(C) is simply a function such that g ξ ↾ K is constantly ξ and g ξ (z) = 0, for z ∈ a.Also, the smaller the absolute value of ξ is, the smaller g ξ m gets.
For a set F of finite partial functions on C, let are a particular type of forcings of the form Q 0 (F ).As with Q(H), the interpretation of Q 0 (F ) in any transitive model that contains F is easily seen to be a dense subset of Q 0 (F ) as interpreted in V .The following lemma will be formulated for this more general type of forcing, since that will be needed in Section 7. The proof of this core lemma originates in ideas fleshed out in Burke's [8].For instance, Claim 2.8 in the aforementioned paper corresponds roughly to Claim 5.11 below.Burke remarks that this is a version of an argument by Shelah from [29].
Lemma 5.10.Let M ⊆ V be as in Lemma 5.9, F ∈ M , z ∈ C ∩ (M \ h∈F dom h) and c a Cohen real over M . 7Furthermore, let F ′ = F ∪ {h ∪ {(z, c)} : h ∈ F } and P ∈ M be a forcing notion that is dense in a forcing P ′ ∈ V .Then Proof.It is easy to see that Q 0 (F ) × P is a subforcing of Q 0 (F ′ ) × P ′ (the incompatibility relation is preserved).Now let E ∈ M , E ⊆ Q 0 (F ) × P be predense in Q 0 (F ) × P and suppose towards a contradiction there exists p = (p 0 , p 1 ) ∈ Q 0 (F ′ ) × P ′ , such that p ⊥ E, where z ∈ a p0 and so f p0 (z) = c.By extending p, we may assume without loss of generality that p 1 ∈ P and that a p0 ⊆ B mp 0 (0).
Once we prove the claim we are done.Namely, as c ∈ O is Cohen generic over M , c ∈ U .Then, according to the claim, there is q ≤ E, (p 0 , p 1 ) such that f q0 (z) = c.Letting r 0 := (a r0 , f q0 , ε q0 , m q0 ), where a r0 := a q0 ∪ {z}, we clearly have that r = (r 0 , q 1 ) ≤ q and r ∈ Q 0 (F ′ ) × P ′ .Moreover, we have that r 0 ≤ p 0 and thus r ≤ p, E: This contradicts the assumption that p ⊥ E.
Proof of Claim.Work in M .Let O 0 ⊆ O be an arbitrary non-empty open set.We will find a non-empty open set O 1 ⊆ O 0 that will be included in U .Let e ∈ O 0 be an arbitrary rational complex number.Then we find ξ ∈ B δ (0) such that f ξ (z) = e, where f ξ = f + g ξ and g ξ is the function from the statement of Lemma 5.9.Then, by (1) of the lemma, In particular, (q ′ 0 , p 1 ) ≤ (p 0 , p 1 ).Now let q′′ ∈ Q 0 (F ) × P be such that q′′ = (q ′′ 0 , q ′′ 1 ) ≤ E, (q ′ 0 , p 1 ).Let γ ∈ (0, ε q ′ 0 ) be small enough so that for any v ∈ B γ (0) there is an entire function h v with h v m q ′′ 0 < ε q ′′ 0 , h v (z) = v and h v ↾ a q ′′ 0 constantly equals 0 (e.g. using a similar formula as in Lemma 5.9).
As f q ′′ 0 − f ξ m q ′ 0 < ε q ′ 0 , we have that for any v ∈ B γ (0) and h v as above, (f . U is constructed in M as the union of all sets O 1 that we obtain in this way. Let us check that this works.So working in V , let d ∈ O 1 be arbitrary.Then there is v ∈ B γ (0) and h v as before so that f q ′′ 0 (z) + h v (z) = d.Let q 0 = (a q0 , f q0 , ε q0 , m q0 ), where a q0 = a q ′′ 0 , 0 is constantly 0. So if we let q = (q 0 , q ′′ 1 ), then q ∈ Q 0 (F ) × P and q ≤ q′′ ≤ E, (p 0 , p 1 ): This finishes the proof of the claim.Proposition 5.12.Let M ⊆ V be as in Lemma 5.9, H 0 , . . ., H n ∈ M and for any pairwise distinct pairs (i j , z j ), j ≤ l, where i j ≤ n, z j ∈ dom(H ′ ij ) \ dom(H ij ), and any sequence c j : j ≤ l , c j ∈ H ′ ij (z j ), we have that c j : Proof.This is an inductive argument using Lemma 5.10.Let E ∈ M be predense in } is enumerated by a sequence (i j , z j ) : j ≤ l .Then according to (2), c j : j ≤ l = f pi j (z j ) : j < l is mutually Cohen generic over M .In M [c 0 , . . ., c l−1 ], consider . Now apply Lemma 5.10 once to accommodate the full condition p and reach a contradiction.5.4.The main theorem.Theorem 5.13.(GCH) Let κ be an infinite cardinal of uncountable cofinality.Then there is a cofinality and cardinal preserving forcing extension in which (1) there is a Wetzel family, (3) if κ is regular, MA holds.
Suppose B(α) is a P α -name Ȧ for a ccc poset of size < κ.In V Pα , either there are In this case, we force with Q(H σ ξ 0 ↾α ) × • • • × Q(H σ ξn ↾α ) and kill the ccc of A. Otherwise we force with A. In V , using the ccc of P α , we let X − α be a countable set that contains any such ξ 0 , . . ., ξ n that we might choose in the former case.If B(α) is not a name for a ccc forcing, we simply skip this step.In any case, let P + α be the forcing that results from this.Note that ( * ) still holds if we replace P α by P + α and if ξ 0 < • • • < ξ n are in κ \ (X − α ∪ δ<α X δ ).Next we let żα = B(γ), where γ is least such that B(γ) is a P α -name for a complex number distinct from any żδ , δ < α.Let C µα be the forcing for adding mutually generic Cohen reals c α,ξ,i : ξ < µ α , i ∈ ω .Let P ++ α = P + α × C µα and Ċα,ξ be a P ++ α -name for {c α,ξ,i : i ∈ ω}.For any ), and ḟα be a name for the entire function added by Q(H ση α ↾α ).Now, for any ) is ccc and by Proposition 5.8, 9 Q(∅) has a countable dense subset, e.g.consisting of those conditions p, where fp is a polynomial in rational coefficients.
This finishes the construction.The bookkeeping function B is supposed to enumerate all P-names for complex numbers, and in case κ is regular, also all P-names for ccc forcings on ordinals < κ unboundedly often.This is a standard argument.When κ is regular, it suffices to let B enumerate all elements of H(κ) unboundedly often.A standard argument then shows that MA holds after forcing with P. When κ is not regular, let B enumerate all elements of P κ ω (κ), where ] ≤ω , and we take unions at limits.Since 2 ℵ0 = κ, |P κ ω (κ)| = κ as well.It is then standard to see that P ⊆ P κ ω (κ), using the ccc and choosing appropriate names for Qα , α < κ.
Finally, after forcing with P, we have that z α : α < κ enumerates the complex numbers and for every δ, α < κ, Let us note that it is not very important that the σ α 's had finite pairwise intersections and we could easily get by with assuming only countable intersections.In that case, we would just have to split up the limit stages into countable and uncountable cofinalities.The proof for uncountable cofinality stays the same and for countable cofinality the ccc follows easily from the previous steps.
Also, some interesting modifications can be made to the forcing construction above.For example, instead of taking care of all complex numbers along the iteration, we can leave out some values.For example, we may leave out exactly 0. The resulting family F is then a Wetzel family on the modified domain C {0}, while all values f (0), for f ∈ F , are pairwise distinct.To see this, note that if z ∈ dom H, then the function added by Q(H) maps to a generic complex number at z.
More generally, for any given infinite Ω ⊆ C, we can construct a family of entire functions that is Wetzel on Ω, while attaining 2 ℵ0 -many values at any point outside of Ω.For κ regular, we can force ♦ κ over the model from Proposition 4.1 without collapsing cardinals or changing cofinalities, by adding a κ-Cohen real in the standard way.This doesn't affect the result of Proposition 4.1.Then, using a standard guessing argument in the iteration of Theorem 5.13, it should not be hard to modify the construction in order to obtain a model where such families exist for every infinite subset Ω ⊆ C. Whenever Ω α ⊆ z β : β < α is guessed at step α, we may force with Q(H ση α ↾α ↾ Ω α ) instead.Here note that if Q(H) is ccc then for any restriction H ′ ⊆ H, Q(H ′ ) is a subforcing (not necessarily complete) of Q(H) and thus also ccc.

MA and universal sets
In this section, we show that under MA + ¬CH there is no universal set.Recall that MA is saying that for any ccc partial order P and any family D of less than 2 ℵ0 -many dense subsets of P, there is a filter G ⊆ P such that G ∩ D = ∅, for every D ∈ D.
We begin by introducing the ccc poset that we will use.The forcing S shall consist of conditions of the form p = (w, s) = (w p , s p ), where w ∈ [C] <ω and s is a finite sequence of open intervals with rational endpoints in (0, 1 A condition q extends p iff w p ⊆ w q and s p ⊆ s q .Lemma 6.1.S is ccc.
Proof.Suppose that p α ∈ S, for α < ω 1 are pairwise incompatible.Then we may assume without loss of generality that s pα = s and |w pα | = n is the same for all α < ω 1 .Let us write w pα = {z α i : i < n}, for every α and consider the function Thus, by continuity, there is k large enough so that In other words, for all z 0 , z , where b is strictly bigger than the maximal distance between points in w pα ∪ w pα k that lies in (0, c 8 ).Similarly, let 0 < a < b, where a is strictly smaller than the minimal such distance.Letting I be the interval (a, b), (w pα ∪ w pα k , s ⌢ I) is a condition extending both p α and p α k , while α = α k .This is a contradiction to the assumption that p α ⊥ p α k .
In the following, Q is the set of rational numbers.Lemma 6.2.For every z ∈ C and every n ∈ ω, the sets D z = {q ∈ S : z ∈ w q + (Q + iQ)} and E n = {q ∈ S : |s q | ≥ n} are dense in S.
Proof.Let p ∈ S be arbitrary.If w p = ∅, then clearly q = ({z}, s p ) extends p and lies in D z .Otherwise, there is z 0 ∈ w p and a small open neighborhood O of z 0 so that for any z 1 ∈ O, there is an extension q of p with z 1 ∈ w q .This is similar to the argument in the proof of Lemma Then there is no non-constant entire function f such that f ′′ X ⊆ Y .
Proof.Suppose there is such f .Since f is non-constant we can find an accumulation point x ∈ X of X such that f ′ (x) = 0. Let x n → x, where x n ∈ X for every n ∈ ω.
Then there is n k : k ∈ ω such that for all k, which does not converge, and from the latter we follow that which converges to 0. This contradicts that f ′ (x) = 0.
Let a n be the left-endpoint of s(n), for every n ∈ ω and consider O = n∈ω (a 4 n , a 2 n ).Then note that for every x ∈ U , y ∈ O, min(x, y) < max(x, y) 2 .
Finally apply Lemma 6.3 to find a set X ⊆ C of size ℵ 1 such that {|z 0 − z 1 | : z 0 , z 1 ∈ X, z 0 = z 1 } ⊆ O.We claim that there is no entire f such that f ′′ X ⊆ Y .
Then note that Q 0 ∈ N 0 is a forcing of the form Q 0 (F ) for F ∈ N 0 . 11Furthermore, let E = {w ′ ∈ Q : ∃s ′ ((w ′ , s ′ ) ∈ A)}.Then E ⊆ Q 0 , E ∈ N 0 and by the inductive hypothesis, it is predense in Q 0 .Now use Lemma 5.10, for P, P ′ trivial forcings, and a similar argument as in Proposition 5.12 to finish the inductive step.
Finally, if (w, s) ∈ K is an arbitrary condition, then we have shown that (w, s ∪ {(M, N )}) is a master condition over K.This proves the lemma.Lemma 7.4.Let p, Y ∈ K H(θ), K countable, for large θ, and let c be a Cohen real over K. Then there is a master condition q ≤ p over K so that q c is Cohen over K[ Ġ].This is known as "almost preserving ⊑ Cohen " in [4, Section 6.3.C] and implies the preservation of non-meager sets in countable support iterations.H(θ) be countable with K, H, c ∈ K + .Let p = (w, s) and N = K + ∩ H(ω 1 ).Then q = (w, s ∪ {(M, N )}) is a master condition over K, as in the proof of Lemma 7.2.Let G ∋ q be P(Y )-generic over V .According to Claim 7. Proof.Let (w, s) ∈ P(Y ) and z ∈ C. Extending (w, s) further, we can assume that z ∈ M 0 for some (M 0 , N 0 ) ∈ s, where M 0 is minimal with this property and there is a successor (M 1 , N 1 ) ∈ s of (M 0 , N 0 ).Since Y is everywhere non-meager, M 1 knows this and since N 0 is countable in M 1 , there is c ∈ Y ∩ M 1 that is arbitrarily close to f w (z) and Cohen generic over N 0 [c 0 , . . ., c n ], where c 0 , . . ., c n enumerates the mutual Cohen generics f w (x), for x ∈ a w that first appear in M 0 .The rest then follows as in the proof of Lemma 5.3.
Proof of Theorem 7.1.Let Y = C V and iterate P(Y ) in a countable support iteration of length ω 2 .By Lemma 7.4 and [4, Lemma 6.3.17,6.3.20],Y stays everywhere non-meager along the iteration.Everything else follows from standard counting of names arguments and the fact that P(Y ) has size ℵ 1 under CH.
It seems that it would also suffice to consider nodes where M, N are merely countable transitive models of ZF − , (M, ∈, Y ∩ M ) is countable in N and Y ∩ M is non-meager in M .This has the slight advantage that P(Y )∩M is already a definable subclass of M and not just definable in N .Also in the proof of Lemma 7.4 we could directly let 11 To see that Q 0 , F ∈ N 0 , note that the subset of M 0 consisting of countable elementary submodels of (H(ω 1 ), ∈, Y ) can be defined in N 0 as the elements of M 0 that are elementary submodels of (M 0 , ∈, Y ∩ M 0 ).
There is a also a ccc way to do this this starting from a model of ♦.This is a modification of the construction presented in [8], from which our result draws its main inspiration.Instead of generically adding an ∈-increasing sequence of nodes using side conditions, we start already with a given sequence (M α , N α ) : α < ω 1 , where N α : α < ω 1 is an "oracle" (see [30]).It can then be shown that the resulting forcing, consisting of those w ∈ Q such that (w, s) ∈ P(Y ) for every s ⊆ (M α , N α ) : α < ω 1 , is "oracle-cc".In fact, our proper forcing is built directly from this construction.The advantage is that it has a much easier setup and does not depend on a particular chosen sequence of nodes.Also there might be a bigger potential of generalising it to continuum higher than ℵ 2 , although this is not very clear to us.

Corollary 4 . 5 .
The answer to[33, Question 22]  is positive.Namely, assuming GCH, it possible to add arbitrarily large strongly almost disjoint families of functions in ℵ ℵω+1 ω without collapsing cardinals.

Lemma 5 . 3 .
Thus it suffices in the definition of the forcing Q(H) to only consider functions that lie dense enough in relation to the values of H. Suppose that H(z) is dense in C, for every z ∈ dom H. Then Q(H) generically adds an entire function f such that f (z) ∈ H(z) for every z ∈ dom H. Proof.Let G be Q(H)-generic over V .For any n ∈ ω, the set D n = {p ∈ Q(H) : ε p < 1 n ∧ m p > n} is clearly dense open.Moreover, for any ξ ∈ C and any p ∈ Q(H) consider f ξ (z) = f p (z) + ξ y∈ap (z − y).
Proof.Let M = K ∩ H(ω 1 ) and H be a Coll(ω, λ)-generic overK[c], where λ = |H(ω 1 )| K .Then c is still a Cohen real over K[H] and K[H] |= |M | = ω.Moreover, P(Y ) ∩ K = P(Y ) ∩ M ∈ K[H], since P(Y ) ∩ M is definable from M, Y ∩ M ∈ K[H].Now let K + 3, for any predense subsetA ∈ K + of P(Y ) ∩ K, A ∩ G = ∅.Thus G ∩ K is P(Y ) ∩ Kgeneric over K + and in particular over K[H][c] ⊆ K + .But P(Y ) ∩ K is a countable forcing in K[H]and thus equivalent to Cohen forcing (or, in the simplest case, a trivial forcing) witnessed through an isomorphism in K[H].So K[H][c][G ∩ K] is a Cohen extension of K[H][c], and c is still Cohen generic over K[H][G ∩ K].In particular c is still Cohen generic over K[G] = K[G ∩ K] ⊆ K[H][G ∩ K].Lemma 7.5.Let Y ⊆ C be everywhere non-meager.Then P(Y ) generically adds a non-constant entire function f such that f (C V ) ⊆ Y .

8 . 5 ? 8 . 3 .
Open questions Question 8.1.Does MA or PFA imply that there is a Wetzel family?Is MA+2 ℵ0 = ℵ 2 sufficient?Recall that non(M) is the least size of a non-meager set.Question 8.2.Is every universal set non-meager under ¬CH?In particular, can we replace MA with non(M) = 2 ℵ0 in Theorem 6.Question Is the existence of a universal set consistent with 2 ℵ0 = ℵ 3 ?With 2 ℵ0 = κ for arbitrary successor cardinal κ? Recall that a domain Ω ⊆ C is any open connected subset of C. We may then define the analoguous notion of Wetzel families on Ω for functions that are holomorphic on Ω. Question 8.4.Let Ω ⊂ C be any domain and suppose that there is a Wetzel family on Ω.Does there exist a Wetzel family on the whole of C? What about Ω = C \ {0}?Question 8.5.Can we characterize when Q(H) is ccc?