Bounding 2D Functions by Products of 1D Functions 3

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f(x, y) ≤ max{g1(x),g2(y)}. We will mostly consider the case that X and Y are cardinals (which are ordinals). When this happens, without loss of generality λ1,λ2 ≥ µ. In this section we will consider arbitrary µ, but for the rest of the paper we will focus on the µ = ω case. Theorem 2.2. Let µ be a regular cardinal. Let X be a set. Then Φ(X,µ,µ) holds iﬀ each family hfx : x ∈ Xi of functions from µ to µ is eventually dominated (mod <µ) by some single function. Proof. First assume every such family indexed by X can be eventually dominated by a single function from µ to µ. Fix f : X × µ → µ. Let g2 : µ → µ be such that for each x ∈ X, the set Sx := {y<µ : f(x, y) > g2(y)} has size < µ. Let g1 : X → µ be such that for each x ∈ X, g1(x) ≥ sup{f(x, y): y ∈ Sx}. The functions g1 and g2 work. BOUNDING 2D FUNCTIONS BY PRODUCTS OF 1D FUNCTIONS 3

For the other direction, assume there is a family indexed by X of functions from µ to µ that cannot be eventually dominated by a single function. Using such a family, let f : X × µ → µ be such that the functions of the form y 7→ f(x, y) cannot be eventually dominated by a single function from µ to µ. By possibly increasing the values of the function f, we may assume that whenever x ∈ X and Y ∈ [µ]µ, the set f“{x} × Y is unbounded below µ. Now, consider any g2 : µ → µ. By hypothesis, ﬁx some x ∈ X such that g2 does not eventually dominate the function y 7→ f(x, y). That is, Sx := {y ∈ µ : f(x, y) >g2(y)} is in µ [µ] , so f“{x} × Sx is unbounded below µ. Hence, there is no α<µ such that for all y<µ, f(x, y) ≤ max{α,g2(y)}, so there can be no function g1 which works (because no value of α = g1(x) works). Every family of size µ of functions from µ to µ is eventually dominated by a single function. On the other hand, there exists a family indexed by µ2 of functions from µ to µ that cannot be eventually dominated by any single function. This gives us the following: Corollary 2.3. (ZF) Let µ be a regular cardinal. Then Φ(µ,µ,µ) holds but Φ(µ2,µ,µ) does not. Corollary 2.4. (ZF) Let µ be a regular cardinal. Assume there is no injection of µ+ into µ2. Then for any cardinal λ (that is an ordinal), Φ(λ,µ,µ) holds. Proof. Any size λ family of functions from µ to µ has at most µ distinct members, and hence the family can be eventually dominated by a single function. Recall that in ZFC, b(µ) is the cardinality of the smallest family of functions from µ to µ that cannot be eventually dominated by a single function. See [1] for more on b(ω). Corollary 2.5. (ZFC) Let µ be a regular cardinal. Let λ be a cardinal. Then Φ(λ,µ,µ) holds iﬀ λ< b(µ). 3. First Countable Topologies

By Corollary 2.3, Φ(R,ω,ω) fails in ZF, and hence so does Φ(R,ω1,ω) and Φ(R, R,ω). In this section, we will show in ZF that if ω1 is regular and there is an injection of ω1 into R, then Φ(ω1,ω1,ω) fails as well. Given a topological space X and a point x ∈ X, we call B ⊆ P(X)a neighborhood basis for x iff 1) every U ∈ B is open and contains x and 2) for every open W ∋ x, there is some U ∈ B such that U ⊆ W . We call X first-countable iff every x ∈ X has a countable neighborhood basis. 4 FRANÇOISDORAISANDDANHATHAWAY

Definition 3.1. Let X be a topological space. We call X uniformly first-countable iff there is a function F with domain X such that for each x ∈ X, F (x) is a sequence hUx,n : n<ωi such that {Ux,n : n<ω} is a neighborhood basis for x. Assuming the Axiom of Choice, a topological space is first-countable iff it is uniformly first-countable. Recall that a topological space X is T1 iff for any distinct x, y ∈ X, there is an open set which contains x but not y.

Theorem 3.2. (ZF) Let X be a T1 and uniformly first-countable topological space. Let Y ⊆ X and assume Φ(X,Y,ω). Then we may write Y = n<ω Yn where each Yn is a closed discrete subset of Y and Y0 ⊆ Y1 ⊆ .... S Proof. Let the function F witness that Y is uniformly first-countable, where F (x) = hUx,n : n<ωi. Assume without loss of generality that for each x ∈ X, we have Ux,0 ⊇ Ux,1 ⊇ .... Let f : X × Y → ω be defined by min{n<ω : y 6∈ U } if x =6 y, f(x, y) := x,n (0 if x = y. Thus for all distinct x ∈ X and y ∈ Y ,

y 6∈ Ux,f(x,y).

By Φ(X,Y,ω), there are functions g1 : X → ω and g2 : Y → ω such that f(x, y) ≤ max{g1(x),g2(y)} for all distinct x ∈ X and y ∈ Y . Thus

y 6∈ Ux,max{g1(x),g2(y)} for all distinct x ∈ X and y ∈ Y . Deﬁne Yn := {y ∈ Y : g2(y) ≤ n}. Then

y 6∈ Ux,max{g1(x),n} =: N(x, n) for any distinct x ∈ X and y ∈ Yn. Since N(x, n) is independent of y ∈ Yn, we have that N(x, n) is a neighborhood of x whose only possible point in Yn that it contains is x itself. That is,

∅ if x 6∈ Yn, N(x, n) ∩ Yn = ({x} if x ∈ Yn.

It follows that Yn is a discrete space. Also note that since each point in X−Yn has a neighborhood disjoint from Yn, we have X−Yn is open, and hence Yn is closed. Also note that Y = n<ω Yn and Y0 ⊆ Y1 ⊆ .... Corollary 3.3. (ZF) Assume ω1 is regularS and there is an injection of ω1 into R. Then Φ(ω1,ω1,ω) is false. BOUNDING 2D FUNCTIONS BY PRODUCTS OF 1D FUNCTIONS 5

Proof. Assume, towards a contradiction, that Φ(ω1,ω1,ω) holds. By ω hypothesis, let f : ω1 → ω be an injection. Let Y be the range of ω f. We have that Φ(Y,Y,ω) holds. Since ω is T1 and uniformly ﬁrst- countable, so is Y . By Theorem 3.2, Y is the union of a countable chain of discrete spaces. Since ωω is second countable, so is Y , and hence every discrete subspace of Y is countable. Hence Y is the union of countably many countable sets. Using the bijection f : ω1 → Y , we see that ω1 is a countable union of countable sets. This contradicts ω1 being regular.

Corollary 3.4. (ZFC) Φ(ω1,ω1,ω) is false. We will use the following later: Lemma 3.5. (ZFC) Let S be a set of ordinals such that each α ∈ S is either a successor ordinal or has countable coﬁnality. Then S, with its usual order topology, is T1 and uniformly ﬁrst-countable.

Proof. We will deﬁne the appropriate function F (β)= hUβ,n : n<ωi. For each successor ordinal β ∈ S, deﬁne F (β) to be the sequence h[β], [β], [β], ...i. For each limit ordinal β ∈ S, use the Axiom of Choice to choose an ω-sequence β0 < β1 < ... with limit β. Let F (β) be the sequence h(β0, β], (β1, β], ...i. One can check that this works.

4. Consistency Strength Upper Bound The main result of this section is that

(1) ZF+DC+ ω1 is a measurable cardinal implies Φ(ω1,ω1,ω). Using the Solovay model, we have that the consistency of ZFC + there is a measurable cardinal implies the consistency of (1). We will sketch that proof now: Proposition 4.1. (ZFC) Let κ be a measurable cardinal. Let G be generic over V for the forcing Col(ω, <κ). Then in the L(R) of V [G], we have DC + ω1 is a measurable cardinal. Proof. Because V [G] satisfies DC, so does L(R)V [G] (see [7]). To show V [G] that ω1 is measurable in L(R) , let U be a normal ultrafilter on V [G] R V [G] κ. We want to define a normal ultrafilter on κ = ω1 in L( ) . Whenever x is generic over V by a poset of size < κ, let Ux refer to the ultrafilter generated by U. That is, given A ⊆ κ in V [x], define that A ∈ Ux iff (∃B ∈ U) B ⊆ A. The set Ux is a normal ultrafilter in V [x] (see [10]). 6 FRANÇOISDORAISANDDANHATHAWAY

Define V [G] UG := {Ux : x ∈ R }. V [G] Note that by the definition[ of the Ux’s, for any x ∈ R , we have UG ∩ V [x]= Ux. V [G] We will first show that UG is an ultra. Let A ⊆ κ be in L(R) . The set A is ordinal definable in V [G] using a real number x ∈ RV [G] as a parameter. The model V [G] is generic over V [x] by the forcing Col(ω, <κ) (see [6]). Since this forcing is homogeneous, we have A ∈ V [x]. Since V [x] |=(Ux is an ultrafilter), either A ∈ Ux or κ − A ∈ Ux (but not both). This establishes that UG is ultra. V [G] Note also that whenever A ⊆ B ⊆ κ are in L(R) , if A ∈ UG, then V [G] B ∈ UG (consider Ux where x ∈ R is such that A, B ∈ V [x]). A similar argument shows that UG is a filter. To see that UG is normal, let A = hAβ ∈ UG : β < κi be in the L(R) of V [G]. Let B = {(α, β): α ∈ Aβi ⊆ κ × κ. We have B is ordinal definable in V [G] from a real x, and since V [G] is a homogeneous extension of V [x], we have B ∈ V [x], and so A ∈ V [x]. Now (∀β < κ) Aβ ∈ Ux. Using the normality of Ux, the diagonal intersection of A is in Ux, and hence is in UG. Note that a similar proof to the one above shows that if κ is measurable and if we force using Col(ω, <κ), then the N := HOD(R) of the forcing extension also thinks that ω1 is measurable. N also satisfies DC (see [14]). For clarification, here are equivalents of ω1 being measurable assuming DC. We will use the third one to show Φ(ω1,ω1,ω). Proposition 4.2. (ZF) Assume DC. The following are equivalent:

1) The club ﬁlter on ω1 is a normal ultraﬁlter. 2) ω1 is a measurable cardinal. 3) 3a) (∀r ∈ R) r# exists, 3b) (∀A ⊆ ω1)(∃r ∈ R) A ∈ L[r].

Proof. 1) ⇒ 2): if there is a non-principal normal ultrafilter on ω1, then ω1 is measurable by definition. 2) ⇒ 3): this is the only direction where we use DC. From Theorem 5 and Corollary 6 in [15], we see that if ZF + DC + ω1 is almost <ω1- Erdös, then every subset of ω1 is constructible from a real. Hence since ω1 is measurable and we are assuming DC, we have every subset of ω1 is constructible from a real. Also since ω1 is measurable, every real has a sharp. 3) ⇒ 1): first we will show that the club filter is ultra. Let A ⊆ ω1. Let r ∈ R be such that A ∈ L[r]. Using that r# exists, for a final BOUNDING 2D FUNCTIONS BY PRODUCTS OF 1D FUNCTIONS 7 segment of L[r]-indiscernibles α<ω1, either α ∈ A or α 6∈ A. A final segment of L[r]-indiscernibles is a club, so this shows that the club filter is ultra. Note for later that such a final segment is in L[r#]. To show that the club filter is normal, let A := hAβ : β<ω1i be an ω1-sequence of sets in the club filter. Let B = {(α, β): α ∈ Aβ}. Now B can be coded by a subset of ω1, so fix a real r ∈ R such that B ∈ L[r]. Hence A ∈ L[r]. Each Aβ contains some club that is in # L[r ]. This is because if not, then ω1 − Aβ would contain a club in # L[r ], but this is impossible because Aβ contains a club in V . Since # # L[r ] satisfies the Axiom of Choice, let hCβ : β<ω1i ∈ L[r ] be such that (∀β<ω1) Cβ is a club subset of Aβ. Let ∆β<ω1 Cβ and ∆β<ω1 Aβ be the diagonal intersection of the Cβ’s and Aβ’s respectfully. The diagonal intersection of clubs is club (see [6]), hence ∆β<ω1 Cβ is club.

Also ∆β<ω1 Cβ ⊆ ∆β<ω1 Aβ. So ∆β<ω1 Aβ contains a club, which is what we wanted to show

Here is the proof of the consistency of Φ(ω1,ω1,ω): Theorem 4.3. (ZF) Assume • (∀r ∈ R) r# exists, • (∀A ⊆ ω1)(∃r ∈ R) A ∈ L[r].

Then Φ(ω1,ω1,ω) holds.

Proof. Let f : ω1 × ω1 → ω. By hypothesis, fix r ∈ R such that f ∈ L[r]. Let I ⊆ Ord be the club of Silver indiscernibles for L[r]. Let a1, ..., an ∈ I be the indiscernibles used in the definition of f in L[r] L[r]. That is, there is a Skolem term t such that t(a1, ..., an) = f. # α α In L[r ], for each α<ω1 choose Cα := (b1 , ..., bmα , tα) to be some α α sequence of L[r]-indiscernibles b1 , ..., bmα ∈ I and a Skolem term tα such that α α L[r] α = tα(b1 , ..., bmα ) . # The function α 7→ Cα is in L[r ]. Let Type(α) := (mα, tα), and let T be the collection of all possible types. Note that T is countable. The value of f(α1,α2), since it is a natural number, depends only on the Skolem term for α1, the Skolem term for α2, and the relative order- ing of the ordinals in Cα1 and Cα2 with each other and with a1, ..., an. Knowing just Type(α1) and Type(α2), only finitely many such order- ings of the indiscernibles are possible, and so only finitely many values ′′ of f(α1,α2) are possible. Given any T1, T2 ∈ T , define f (T1, T2) to be the maximum of these possible values. We have f ′′ : T×T → ω. ′ ′ ′′ Define f : ω1 ×ω1 → ω by f (α1,α2)= f (Type(α1), Type(α2)). We have that f ≤ f ′ (f ′ everywhere dominates f). If we can find functions ′ ′ ′ ′ ′ g1,g2 : ω1 → ω such that (∀x, y ∈ ω1) f (x, y) ≤ max{g1(x),g2(y)}, 8 FRANÇOISDORAISANDDANHATHAWAY then we will be done. By the definitions of f ′ and f ′′, it suffices to find ′′ ′′ ′′ ′′ ′′ g1 ,g2 : T → ω such that (∀T1, T2 ∈ T ) f (T1, T2) ≤ max{g1 (T1),g2 (T2)}, ′ ′′ ′ ′′ because then we can define g1(x) := g1 (Type(x)) and g2(y) := g2 (Type(y)). ′′ ′′ There must be such functions g1 and g2 because |T | = ω.

Corollary 4.4. (ZF) Assume DC + ω1 is measurable. Then Φ(ω1,ω1,ω) holds. It is also true that the club ﬁlter on ω1 is normal.

Solovay showed that AD implies ω1 is measurable (see Theorem 33.12 in [6]). Also, Solovay proved under AD that every subset of ω1 is constructible from a real (see Lemma 2.8 of [9]). These two facts give us the following:

Corollary 4.5. (ZF) Assume AD. Then Φ(ω1,ω1,ω) holds.

Proof. Now fix f : ω1 × ω1 → ω. We must find the g1,g2 : ω1 → ω to satisfy the definition of Φ(ω1,ω1,ω). By AD, every subset of ω1 is constructible from a real. So f ∈ L[r] for some real r, and so in L(R) particular f ∈ L(R). So if we show Φ(ω1,ω1,ω) , then we will be done. R Note that ADL( ) holds, because for each A ⊆ R in L(R), A is determined in L(R) iff it is determined in V . Now ADL(R) implies L(R) both DC by [8], and that ω1 is measurable in L(R). Hence by L(R) Proposition 4.2 and Theorem 4.3, Φ(ω1,ω1,ω) holds, and we are done.

Just as we observed in Corollary 2.4, note that if Φ(ω1,ω1,ω) holds and there is no injection of ω2 into P(ω1), then Φ(λ,ω1,ω) holds for each ordinal λ. It is natural to ask what fragments of the Axiom of Choice are consistent with Φ(ω1,ω1,ω). One such fragment consistent with Φ(ω1,ω1,ω) is the existence of a non-principal ultrafilter on ω, which we will prove ω now. Recall that ω → (ω)2 is that statement that given any coloring of the infinite subsets of ω using two colors, there is an infinite subset of ω all of whose infinite subsets have the same color. Theorem 4.6. Assume ZFC + “there exists a measurable cardinal” is consistent. Then there is a model of ZF in which Φ(ω1,ω1,ω) holds but there is a non-principal ultrafilter on ω.

Proof. By hypothesis, there must be a model M1 of ZFC + “there exists a measurable cardinal”. Let G be generic for the Levy collapse M1[G] of κ to ω1. Let M2 := L(R) . By Proposition 4.1, M2 satisfies both DC and that ω1 is measurable. Thus M2 satisfies Φ(ω1,ω1,ω). Since κ is inaccessible in M1, by a result of Mathias [11], we have that M2 ω satisfies ω → (ω)2 . BOUNDING 2D FUNCTIONS BY PRODUCTS OF 1D FUNCTIONS 9

Let P be the P (ω)/Fin forcing of M2. Let H be P-generic over M2 ω P and let M3 := M2[H]. Since M2 |= ω → (ω)2 , forcing with will add no new sets of ordinals (see [4]). Hence, M3 |= Φ(ω1,ω1,ω). On the other hand, M3 |= “there is a non-principal ultraﬁlter on ω”, by the nature of the P forcing.

5. Consistency Strength Lower Bound

To show that Φ(ω1,ω1,ω) implies there is an inner model with a large cardinal, we will additionally assume some fragment of the Axiom of Choice. The ﬁrst such fragment we will consider is that ω1 is a regular cardinal:

Proposition 5.1. Assume ω1 is regular and that Φ(ω1,ω1,ω) holds. Let M be an inner model of ZFC. Then ω1 is strongly inaccessible in M.

Proof. By Corollary 3.3, there is no injection of ω1 into R. By a well- known argument, this implies that ω1 is strongly inaccessible in M. For this section, we will use the following deﬁnition: Deﬁnition 5.2. Let M be an inner model. Then M S := {α<ω1 : M |= cf(α) ≤ ω}, M R := {α<ω1 : M |= α is a regular cardinal}.

Lemma 5.3. (ZF) Assume ω1 is regular and that Φ(ω1,ω1,ω) holds. Let M be an inner model of ZFC. Then SM is a nonstationary subset of ω1. M Proof. By Lemma 3.5, S is T1 and uniformly first-countable in M. It M follows from Theorem 3.2 that we can write S = n<ω Sn where each M Sn is a closed and discrete subset of S , and where S0 ⊆ S1 ⊆ .... Since M M S Sn is a closed subset of S and S has the subspace topology induced M by ω1, we have Sn = Cn ∩ S for some closed Cn ⊆ ω1. Without loss ′ of generality, let Cn be the closure of Sn in ω1. Let Cn be the set of ′ M limit points of Cn. We claim that Cn ∩ S = ∅. ′ To see why, let α ∈ Cn. Now α is a limit of points β in Cn, where each such β is either in Sn or is the limit of points in Sn. Hence α is M a limit of points in Sn. Suppose towards a contradiction that α ∈ S . Fix m ≥ n such that α ∈ Sm. Then α ∈ Sm is the limit of a sequence of points in Sm, which contradicts Sm being discrete. m Now since S = n<ω Sn and ω1 is regular in V , fix n ∈ ω such that ′ Sn is unbounded below ω1. Then Cn is unbounded below ω1 and is also M S ′ M closed. So S is disjoint from the club Cn, so S is nonstationary. 10 FRANÇOISDORAISANDDANHATHAWAY

Now let us strengthen the hypothesis “ω1 is regular” to “the club ﬁlter on ω1 is normal”. Note this is indeed a strengthening because if ω1 was singular, then ω1 would be the union of countably many countable sets (which are therefore each nonstationary), so the nonstationary ideal on ω1 would not be closed under countable unions. Using this stronger fragment of the Axiom of Choice, we can show that Φ(ω1,ω1,ω) implies that every real has a sharp, and that there is an inner model with a measurable cardinal.

Lemma 5.4. (ZF) Assume the club ﬁlter on ω1 is normal and that M Φ(ω1,ω1,ω) holds. Let M be an inner model of ZFC. Then R contains a club subset of ω1.

Proof. For every β<ω1,

Xβ := {α<ω1 : M |= cf(α) ≤ β}

M[s] is nonstationary, since Xβ ⊆ S where s ∈ R is such that M[s] |=(β is countable). Let C be the diagonal intersection of the sets ω1 − Xβ for β<ω1. Since each ω1 − Xβ contains a club and since the club ﬁlter on ω1 is normal, C itself contains a club. Now

C = {α<ω1 :(∀β<α) α 6∈ Xβ}. That is,

C = {α<ω1 :(∀β<α) M |= cf(α) > β}. So C = RM . Deﬁnition 5.5. Given an inner model M, we say M computes singulars correctly iﬀ whenever κ is a singular cardinal in V , then κ is a singular cardinal in M. By a result of Jensen (see [12]), for any real r ∈ R, if L[r] does not compute singulars correctly, then r# exists.

Proposition 5.6. (ZF) Assume the club ﬁlter on ω1 is normal and # that Φ(ω1,ω1,ω) holds. Then for each r ∈ R, r exists. Proof. Fix r ∈ R. By the previous lemma, RL[r] contains a club C ⊆ ω1 (which is not in L[r]). Now L[r][C] is an inner model of ZFC, by Proposition 1.3. By Proposition 5.1, ω1 is strongly inaccessible in L[r][C]. So we may ﬁnd a κ ∈ C that is a singular cardinal in L[r][C]. Since κ ∈ C, κ is regular in L[r]. So L[r] is not correct about the singulars of L[r][C]. Hence L[r][C] |= r# exists. Any inner model that thinks it contains r# is right about this, so V |= r# exists. BOUNDING 2D FUNCTIONS BY PRODUCTS OF 1D FUNCTIONS 11

For the next result, let K refer to KDJ , the Dodd-Jensen core model. By a result of of Dodd and Jensen (see [12]), if there is no inner model with a measurable cardinal, then 1) K computes singulars correctly, 2) if N is any inner model such that N ⊇ K, then KN = K.

Proposition 5.7. (ZF) Assume the club filter on ω1 is normal and that Φ(ω1,ω1,ω) holds. Then there is an inner model with a measurable cardinal. Proof. Suppose, towards a contradiction, that there is no inner model with a measurable cardinal. Let M be the Dodd-Jensen core model K M of V . By the previous lemma, R contains a club C ⊆ ω1 (which is not in M). Now M[C] is an inner model of ZFC by Proposition 1.3. By Proposition 5.1, ω1 is strongly inaccessible in M[C]. So we may find a κ ∈ C that is a singular cardinal in M[C]. Now κ is regular in M but singular in M[C]. Thus M[C] thinks that M does not compute singulars correctly. But M[C] has no inner model with a measurable cardinal (because V has none). So M[C] thinks KM[C] computes singulars correctly. But since M[C] contains M = K, we have KM[C] = K = M. So M[C] thinks M both does and does not compute singulars correctly, which is impossible. Combining this with Corollary 4.4 and Proposition 4.1, we have this: Corollary 5.8. The following theories are equiconsistent: • ZFC + “there is a measurable cardinal”, • ZF + “the club filter on ω1 is normal” + Φ(ω1,ω1,ω).

Note that the club ﬁlter on ω1 being normal is a theorem of ZFC.

6. Questions

We ﬁrst ask what are the consequences of Φ(ω1,ω1,ω). We know that ZF + “the club ﬁlter on ω1 is normal” + Φ(ω1,ω1,ω) implies there is an inner model with a measurable cardinal, but we ask this:

Question 6.1. Does ZF+ Φ(ω1,ω1,ω) imply there is an inner model with a measurable cardinal?

Question 6.2. Does ZF + “the club ﬁlter on ω1 is normal” + Φ(ω1,ω1,ω) imply that every subset of ω1 is constructible from a real? Also given any inner model M, we can ask whether for every function f : ω1 × ω1 → ω in M there exists functions g1,g2 : ω1 → ω in V such 12 FRANC¸OISDORAISANDDANHATHAWAY that (∀x, y ∈ ω1) f(x, y) ≤ max{g1(x),g2(y)}. But the most intriguing question is whether ω1 can be enlarged:

Question 6.3. Is ZF + Φ(ω2,ω2,ω) consistent? Does it follow from AD?

And in general, we can ask about the status of Φ(λ1,λ2, κ) for var- ious λ1, λ2, and κ. It would be interesting to know if there is any instance of Φ(λ1,λ2, κ), with λ1 and λ2 being ordinals, which fails in L(R) assuming AD. References [1] Blass, A., 2010. Combinatorial cardinal characteristics of the continuum, in Foreman, M. and Kanamori A. (Eds.) Handbook of Set Theory Volume 1, Springer, New York, pp. 395-489. [2] Dance, C., 2016. Contributions to Descriptive Set Theory. Doctoral disserta- tion. [3] Feng, Q., Magidor, M., Woodin, H., 1992. Universally Baire sets of reals, in Woodin, H., Judah, H., Just, W. (Eds.), Set Theory of the Continuum. Mathematical Sciences Research Institute Publications, vol 26, North-Holland pp. 203-242. [4] Henle, J., Mathias, A., Woodin, H., 1985. A Barren Extension. Di Prisco C.A. (Ed.) in Methods in Mathematical Logic. Lecture Notes in Mathematics, vol 1130. Springer, Berlin, Heidelberg. [5] Grigorieff, S., 1975. Intermediate Submodels and Generic Extensions in Set Theory, Annals of Mathematics, Second Series, Vol. 101, No. 3, pp. 447-490. [6] Jech, T., 2006. Set Theory, third ed. Springer, New York. [7] Kanamori., A., 1994. The Higher Infinite. Springer-Verlag, Berlin. [8] Kechris, A., 1984. The Axiom of Determinancy Implies Dependent Choices in L(R), Journal of Symbolic Logic 49, pp. 161-173. [9] Kleinberg., E., 1977. Infinitary Combinatorics and the Axiom of Determinate- ness. Springer-Verlag, Berlin. [10] Levy, A., Solovay, R. 1967. Measurable cardinals and the continuum hypothesis. Israel Journal of Mathematics 5, pp. 234-248. [11] Mathias, A., 1977. Happy families, Annals of Mathematical Logic 12, pp. 59111. [12] Mitchell, W., 2010. The Covering Lemma, in Foreman, M. and Kanamori A. (Eds.) Handbook of Set Theory Volume 1, Springer, New York, pp. 1497-1594. [13] Solovay, R., 1970. A Model of Set-Theory in Which Every Set of Reals is Lebesgue Measurable, Annals of Mathematics 92, no.1, pp. 1-56. [14] Unger, S., 2015. A Brief Account of Solovay’s Model, http://math.huji.ac.il/∼sunger/solovay-model.pdf. [15] Welch, P., 1994. Characterising Subsets of omega1 Constructible from a Real, Journal of Symbolic Logic 59, pp. 1420-1432.

Department of Mathematics, University of Vermont, Innovation Hall, 82 University Place, Burlington, VT 05405 U.S.A. E-mail address: [email protected] BOUNDING 2D FUNCTIONS BY PRODUCTS OF 1D FUNCTIONS 13

Department of Mathematics, University of Vermont, Innovation Hall, 82 University Place, Burlington, VT 05405 U.S.A. E-mail address: [email protected] URL: http://mysite.du.edu/~dhathaw2/