How to have more things by forgetting how to count them

Cohen's first model is a model where there is a Dedekind-finite set of real numbers, and it is perhaps the most famous model where the Axiom of Choice fails. We force over this model to add a function from this Dedekind-finite set and some infinite ordinal $\kappa$. In the case we force the function to be injective, it turns out that the resulting model is the same as adding $\kappa$ Cohen reals to the ground model, and that we only enumerated the canonical Dedekind-finite set. In the case where the function is merely surjective it turns out that we do not add any reals, sets of ordinals, or collapse any Dedekind-finite sets. This motivates the question if there is any combinatorial condition on a Dedekind-finite set $A$ such that a forcing will preserve its Dedekind-finiteness or not add new sets of ordinals. We answer this question in the case of"Adding a Cohen subset"by presenting a varied list of equivalent conditions to the preservation of Dedekind-finiteness. For example, $2^A$ is extremally disconnected, or $[A]^{<\omega}$ is Dedekind-finite.


Introduction
Cohen developed the method of forcing to prove that Cantor's Continuum Hypothesis is not provable from the axioms of Zermelo-Fraenkel and the Axiom of Choice. He then quickly adapted the known techniques for producing models where the Axiom of Choice fails using atoms (or urelements) to match the method of forcing. His models, therefore, proved that the Axiom of Choice does not follow from the rest of the Zermelo-Fraenkel axioms of set theory. In this model there is a canonical set of real numbers which is Dedekind-finite, that is infinite but without a countably infinite subset.
Over the years we see time and time again how rich and interesting is the theory of Cohen's first model. Recently, for example, Beriashvili, Schindler, Wu, and Yu proved in [1] that in Cohen's first model there is a Hamel basis for the real numbers as a vector space over the rational numbers.
From a modern perspective, Cohen's first model is constructed by adding a countable sequence of Cohen reals, and then "forgetting the enumeration, while remembering the set" using a method called symmetric extensions that extends the method of forcing and is the main tool in proving consistency results related to the Axiom of Choice. We give an overview of this technique in section 2 and an overview of Cohen's first model in section 3.
In this paper we show that forcing over Cohen's first model can have some counterintuitive results. Our two main results to that effect are Theorem 4.2, which shows that we can introduce an arbitrary enumeration of the canonical Dedekindfinite set and the resulting model is itself the appropriate Cohen extension of the ground model; and Theorem 5.1 where we show that an analogue of the Levy collapse adds a surjection from the canonical Dedekind-finite set onto any fixed ordinal, but does not add new sets of ordinals. In particular, this forcing preserves Dedekind-finiteness of sets in Cohen's first model.
We then characterise, in section 6, those Dedekind-finite sets which remain Dedekind-finite after adding a Cohen subset to them. We give 10 different conditions, and we show that if the Dedekind-finite set is a set of real numbers, like in Cohen's first model, then these conditions are satisfied.
Acknowledgements. The authors would like to thank Matti Rubin for inviting the second author to Israel in March 2015 where this project began, and to the Hausdorff Center for Mathematics in Bonn and to the Ben-Gurion University of the Negev for supporting him during this visit. Additional thanks go to the organisers of the HIF programme for extending an invitation to Cambridge in October 2015, where work on this project continued. Final acknowledgement goes to Mike Oliver who gave a talk titled "How to have more things by forgetting where you put them", which inspired the title of this work.

Preliminaries
Since we are dealing with models of ZF and with cardinals, it is worth clarifying what we mean by cardinals. We say that a set X can be well-ordered, or that it is well-orderable, if there is an ordinal α and a bijection f : X → α. If X can be wellordered, the cardinal of X is the smallest ordinal in bijection with X. If, however, X cannot be well-ordered, we utilise Scott's trick and define its cardinal as the set {Y ∈ V α | ∃f : X → Y a bijection}, where α is the least ordinal for which the set is not empty. The letters κ and λ will always denote well-ordered cardinals.
We say that a set X is Dedekind-finite if it has no countably infinite subset, although we will use the term Dedekind-finite exclusively to mean that it is also infinite, as finite sets already have a much shorter name. It is a simple exercise to prove that X is Dedekind-finite if and only if every injection f : X → X is a bijection. We note that if X is a Dedekind-finite set which can be linearly ordered, e.g. a subset of the real numbers, then [X] <ω = {a ⊆ X | x is finite} is Dedekindfinite as well. This is because every finite subset of X can be uniformly enumerated, so the union of countably many finite subsets will be a countable subset of X. Note that it is possible that X is a Dedekind-finite set while [X] <ω is Dedekind-infinite, for example {P n | n < ω} where each P n is a pair, and no infinite family of pairs admits a choice function (such set is sometimes referred to as a Russell set, or a socks set).
Our forcing terminology is standard. We say that P is a notion of forcing if it is a partial ordered set with a maximum, 1 P . We call the elements of P conditions, and we say that a condition q is stronger, or that it extends a condition p if q ≤ p. We also follow Martin Goldstern's alphabet convention which dictates that if p, q are both conditions in P, then p will not denote a stronger condition than q. If two conditions have a joint extension we say that they are compatible, otherwise they are incompatible. P-names are denoted byẋ, and canonical names for ground model objects are denoted byx when x is the object in the ground model.
If X is a set, Add(ω, X) is the forcing whose conditions are finite partial functions from p : X × ω → 2. We denote by supp(p) the support of p, which is the maximal (finite) subset of X such that p : supp(p) × ω → 2. In the context of ZF, if A is a set, we denote by Add(A, X) the set of partial functions p : X × A → 2 such that dom p can be well-ordered and |p| < |A|.
Finally, given a family of P-names, {ẋ i | i ∈ I}, we denote by {ẋ i | i ∈ I} • the obvious name this family defines, that is { 1 P ,ẋ i | i ∈ I}. This notation extends to ordered pairs and to sequences as well. Using this notation,x = {y | y ∈ x} • .
2.1. Symmetric extensions. The method of forcing, albeit very useful, preserves the Axiom of Choice in the ground model. In order to accommodate consistency proofs related to the failure of the Axiom of Choice we need to extend the method of forcing. Let P be a notion of forcing, and let π be an automorphism of P. Then π acts on P-names via this recursive definition: Let G be a group of automorphisms of P. We say that F is a filter of subgroups over G if it is a filter on the lattice of subgroups, namely it is a nonempty family of subgroups of G closed under finite intersections and supergroups. We say that F is normal if whenever H ∈ F and π ∈ G , πHπ −1 ∈ F as well.
We call P, G , F a symmetric system if P is a notion of forcing, G is a subgroup of Aut(P), and F is a normal filter of subgroups over G . In all cases it is enough to require that F is a basis for a normal filter instead of a filter.
Let us fix a symmetric system for the rest of the section. We denote by sym G (ẋ) the group {π ∈ G | πẋ =ẋ}, also called the stabiliser ofẋ. We say thatẋ is F -symmetric if sym G (ẋ) ∈ F . Whenẋ is F -symmetric, and this condition holds hereditarily to the names inẋ, we say thatẋ is hereditarily F -symmetric. The class HS F denotes the class of all hereditarily F -symmetric names. When the symmetric system is clear from the context, and here it will always be clear from context, we omit the subscripts.
The forcing relation can be relativised to HS, and we use HS to denote this relativisation.
The first part of the proof appears as Lemma 14.37 in [4], and the last sentence is an easy consequence of the fact thatẋ ∈ HS if and only if πẋ ∈ HS.
Theorem. Let G ⊆ P be a V -generic filter and let M = HS G = {ẋ G |ẋ ∈ HS}.
The model M in the theorem is called a symmetric extension (of V ). The theorem appears in [4] as Theorem 15.51.

Cohen's first model
Cohen's first model is the classical example of a model of set theory where the Axiom of Choice fails. This model was investigated by Halpern and Levy in [3] where they prove that every set in the model can be linearly ordered, and much more. (The model is sometimes referred to as the Halpern-Levy, or the Cohen-Halpern-Levy model). This model has many presentations through the literature (see Chapter 5 in [5] for a comprehensive analysis of the construction, for example). For convenience of the reader we give a brief overview of the construction here as well.
We assume that V satisfies ZFC, 1 and we take P to be Add(ω, ω). Our group of automorphisms is given by the group of finitary permutations of ω acting on P in the natural way: πp(πn, m) = p(n, m), or equivalently πp(n, m) = p(π −1 n, m). 2 And finally, F is the filter of subgroups For each n, define the nameȧ n as the canonical name for the nth Cohen real, i.e. { p,m | p(n, m) = 1}, and letȦ = {ȧ n | n < ω} • . Claim 3.1. If π ∈ G , then πȧ n =ȧ πn , therefore πȦ =Ȧ. Consequently,Ȧ ∈ HS.
Proof. Suppose thatḟ ∈ HS and p HSḟ :ω →Ȧ. Let E be a support forḟ , and without loss of generality supp(p) ⊆ E as well.
Let n / ∈ E be some natural number, and assume towards contradiction that q ≤ p is a condition such that q HSḟ (m) =ȧ n for some m < ω. Let j < ω be some natural number such that j / ∈ E ∪ supp(q), and let π be the 2-cycle (n j). Then the following hold: (1) π ∈ fix(E) and therefore πp = p and πḟ =ḟ .
(3) πq is compatible with q, since the only coordinates changed between πq and q are j and n, but these are mutually exclusive to the conditions. (4) πq HS πḟ (πm) = πȧ n which is to say, by the above, πq HSḟ (m) =ȧ j . Therefore q ∪ πq HS "ȧ n =ḟ (m) =ȧ j andȧ n =ȧ j ". This is impossible, therefore the assumption that there are such q and m must be false. Therefore, p forces that the range ofḟ is finite, and in fact a subset of {ȧ n | n ∈ E} • , so in particular, p must force thatḟ is not injective.
In the following two sections we work in the Cohen model. We fix a V -generic filter G ⊆ P and denote by M the symmetric extension obtained by it and the symmetric system defined here. We will write a n and A to denoteȧ G n andȦ G respectively. The idea that V (A) is the symmetric extension is relatively straightforward, and it is worth sketching the argument behind it. On the one hand, since V ⊆ M , and A ∈ M we immediately have V (A) ⊆ M . On the other hand, by analysing the proof of Lemma 5.25 and Lemma 5.26 in [5], we see that if x ∈ M , then we can assign to it a minimal finite subset, A 0 , of A and a name in V such that A 0 is the copy of a support of the name, and from this we can define x using A 0 , A and the name from V as parameters. By induction on rank x we get that M ⊆ V (A).

Injective collapse
For two sets X and Y , define Col inj (X, Y ) as the partial order given by wellorderable partial injections p : X → Y . We note that Col inj (X, Y ) is isomorphic to Col inj (Y, X). In this section we will focus on Col inj (A, κ) when κ is an infinite ordinal, 4 and since A is Dedekind-finite, the conditions are finite. It will be easier to work with Col inj (κ, A) instead, and as it is isomorphic to Col inj (A, κ), we can do that without a problem.
Let f : κ → ω be a finite partial injection, and letq f denote the following name: , and all cardinals are preserved.
Proof. For a pair p ∈ P and a nameq f , let r p,f be the condition in Add(ω, κ) If D * is a dense subset of Add(ω, κ), then we define a name for a subset of Col inj (κ, A): We claim thatḊ is a name for a dense set. Suppose thatq f ′ is any condition. Let p ′ be some condition such that supp(p ′ ) = rng(f ′ ) (we may extend f ′ if necessary, thus strengtheningq f ), and let r ′ = r p ′ ,f ′ . By density there is some r ∈ D * such that r ≤ r ′ , then we can extend f ′ to any injective f and define p by In other words, for every nameq f ′ for a condition in Col inj (κ, A) we showed that every condition p can be extended to one which forces some extension ofq f ′ is inḊ.
Suppose now that F is an M -generic function for Col inj (κ, A), and let D * ∈ V be a dense open subset of Add(ω, κ). LetḊ be the name obtained from D * as above, since F is M -generic, there is some p ∈ G andq f such that for some r ∈ D, r = r p,f andq G f ⊆ F . But this means that r ⊆ F , when seen F as a function from κ × ω → 2, replacing each Cohen real in a by its characteristics function. Therefore F is V -generic for Add(ω, κ) as wanted. Proof. This is true since M = V (A), and the argument for this equality is the same even when using Add(ω, κ), this is easy to see from analysing the same proofs as in the case κ = ω. 5 Remark 4.4. The corollary means that the process works in reverse as well, namely, starting with Add(ω, λ) with finitary permutations of λ and a filter of subgroups generated by pointwise stabilisers of finite subsets of λ, we end up with an analogue of the Cohen model where we have a set of Cohen reals which is Dedekind-finite. Using finite injective functions f : κ → λ provides us with the same proof as above.
This result is odd. Indeed, upon first reading, it makes no sense. It quite literally implies that there is a bijection between ω and κ. But we should point out that all it implies inside V is that there is a generic bijection between them, i.e. we can 4 We can of course assume it is a cardinal, but the assumption is never used. 5 We can also obtain the same by applying Feferman's theorem appearing as Problem 23 in Chapter 5 of [5] instead of the proof analysis.
generically collapse κ to be countable. Moreover, the generic objects that we always refer to are not guaranteed to exist "out of the blue", rather we have a working assumption that V is some countable transitive model in a larger universe. And of course, under this assumption, κ is in fact a countable ordinal.
In the case where κ is singular of countable cofinality (recall that we only required that κ is infinite). It is well-known that adding ℵ ω Cohen reals (to a model of CH, at least), adds ℵ ω+1 of them. When we move from A having order-type ω 1 to ω ω , we seemingly add a lot more reals, which will soon disappear as we move again, say to ω 2 . This is the place to point out, of course, that the additional reals are the consequence of being able to define new reals from certain infinite subsequences of the generic, none of which is symmetric enough to enter the Cohen model itself.
And finally, a question. One should make the obvious, and immediate, observation that taking the above question at face value the answer is no. Split A into two infinite parts, A 0 and A 1 (e.g., those reals which include 0 and those that omit it), and force with Col inj (κ, A 0 ) instead. It is not hard to see that we do not add any enumeration of A 1 , which therefore remains Dedekind-finite. However over the symmetric model that is V (A 0 ), the result was indeed the same as above.

"Levy collapse" without adding reals (or sets of ordinals)
For two sets X and Y , denote by Col(X, Y ) the set of partial functions p : X → Y such that |p| is well-ordered and |p| < |X|, ordered by reverse inclusion. This coincides with the standard definition when X can be well-ordered, but we will focus on the case where X = A in Cohen's first model, meaning the conditions are, as before, finite functions.
In a manner similar to Theorem 4.1, if q ∈ Col(A, κ) is a condition, where κ is some well-ordered cardinal, then there is some f : ω → κ in V such that q =q G f , withq f defined as before. Proof. LetẊ ∈ M be a Col(A, κ)-name for a set of ordinals. It is easier to consideṙ X as a name in the iteration P * Col(Ȧ,κ) • , whose projection to a P-name of a Col(A, κ)-name, denoted by [Ẋ], is in HS. 6 Moreover, since we have such canonical names for conditions in Col(A, κ), and we are only interested in this iteration of two steps, we may assume that the conditions in this iteration have the form p,q f . Note that if π ∈ G , then π acts on P * Col(Ȧ,κ) • in the obvious way: Let p,q f be a condition which forces thatẊ is a name for a set of ordinals, and let E be a support forẊ, i.e. a finite subset of ω for which fix(E) ⊆ sym([Ẋ]). We may assume that supp(p) = E = dom f . Suppose that p 0 ,q f0 and p 1 ,q f1 are two extensions of p,q f . Again, we may assume that supp(p i ) = dom f i for i < 2.
If p 1 ↾ E = p 2 ↾ E, then the two must agree on any statement of the form α ∈Ẋ. This is because there is an automorphism in fix(E) moving supp(p 0 ) \ E to be disjoint of supp(p 1 ), which means that πp 0 , πq f0 is compatible with p 1 ,q f1 while πα =α and πẊ =Ẋ (here we used the fact that dom f = E).
In particular this means that if p ′ ,q f ′ ≤ p,q f and p ′ ,q f ′ α ∈Ẋ, then p ′ ↾ E,q f ′ ↾E = p ′ ↾ E,q f already forced this statement, and the same forα / ∈Ẋ, of course. Therefore the conclusion follows, and thereforeẊ is a name for a set in M , given by the nameẊ This provides another proof to the known fact (see Problem 16 in Chapter 5 of [5]) that two models of ZF with the same sets of ordinals are not necessarily equal. 7 Corollary 5.2. Forcing with Col(A, κ) over M preserves cofinalities. , κ), and f can be coded as a real, namely a subset of ω. However, by Theorem 5.1 no new reals are added, and therefore f ∈ M . This is a contradiction since A is Dedekind-finite in M .

Corollary 5.4. Every Dedekind-finite set remains Dedekind-finite after forcing with Col(A, κ).
Proof. There is an injection from every set in M into [A] <ω × η for some ordinal η, therefore for a Dedekind-finite set in M we can take η = ω. But this means that if A remains Dedekind-finite, so must every [A] <ω , as A is a set of real numbers, and therefore every other Dedekind-finite set.
Question 5.5. In the previous section and in this one, the proofs involved in a fairly meaningful way the Cohen forcing itself. What happens when we consider a similar symmetric extension obtained by using a different kind of real numbers, e.g. random reals, Sacks reals, etc., or even a mixture of these? On its face it seems that the proof uses more of the fact that we take a finite-support product of infinitely many copies of the same forcing, rather than the specific properties of the Cohen forcing. To what extent can this be pushed? 6. Adding a Cohen subset to a Dedekind-finite set Corollary 5.3 shows that the Dedekind-finite set A in Cohen's first model is still Dedekind-finite after forcing with Col(A, κ), and this leads to the problem of finding more general condition that ensure this. In this section, we provide characterisations, in ZF, of those Dedekind-finite sets A that remain Dedekindfinite after forcing with Add(A, 1) = Col(A, 2). The combined results can already be stated in the following result, although some of the notions in the theorem are only defined below. Some of these conditions (e.g., (1)) already imply that A is Dedekind-finite, but others do not (e.g., (7) holds for A = ω 1 when assuming ZFC, and (6) holds for A = ω even in ZF).
This theorem admits an easy corollary, which is applicable to Cohen's first model.

Corollary 6.2. If
A is a Dedekind-finite which can be linearly ordered, in particular a set of real numbers, then all the conditions of Theorem 6.1 hold.
In the proofs, we will use the sunflower lemma, a finite version of the ∆-system lemma. Before we state the lemma, we fix the following notation. A sunflower is a collection of sets, S, such that for some t, u ∩ v = t for all u = v in S. Moreover, the set t is called the centre of the sunflower. [2]). If a and b are positive integers, then any collection of b!ab + 1 sets of size ≤b contains a sunflower of size >a.

Lemma 6.3 (Erdős-Rado
Note that this result is provable without the Axiom of Choice. This can be seen by checking its proof, or alternatively by noting that all arithmetical statements are absolute to the constructible universe L, where the Axiom of Choice holds. It is easy to see that [A] <ω is Dedekind-finite if and only if there is a sequence A = A n | n < ω of disjoint nonempty finite subsets of A. We will call such a sequence a disjoint sequence. We fix some more notation: for any K ⊆ Add(A, 1) we define supp K = {supp(p) | p ∈ K}. Note that K is finite if and only if supp K is finite. We show that C k = {p ∈ C | | supp(p)| = k} is finite for all k < ω. It then follows that D = k<ω supp C k is an infinite union of finite sets and hence [A] <ω is Dedekindinfinite. Towards a contradiction, suppose that C k is infinite for some k < ω. Then supp C k contains arbitrarily large sunflowers by Lemma 6.3. Since their centres are all of size at most k, we can find two compatible conditions in C k , contradicting the assumption that C is an antichain.
(c) =⇒ (a): If [A] <ω is Dedekind-infinite, let A = A n | n < ω be a disjoint sequence and B n = i≤n A i . Define p n : B n → {0, 1} to take value 1 on A n and 0 on B n−1 . Then C = {p n | n < ω} is an infinite antichain. Definition 6.5. We say that Add(A, 1) has the finite decision property if for all formulas ϕ(ẋ), the set M ϕ(ẋ) of minimal elements of N ϕ(ẋ) = {p | p ϕ(ẋ)} with respect to restriction is finite. Lemma 6.6. The following are equivalent. 8 (a) Add(A, 1) has the finite decision property.
Proof. (a) =⇒ (b): Suppose that [A] <ω is Dedekind-infinite and let A = A n | n < ω be a disjoint sequence. Letẋ be the Add(A, 1)-name for the least n such that the canonical generic subset of A contains A n . Let ϕ(ẋ) denote a formula stating thatẋ is even. It is then easy to see that M ϕ(ẋ) is infinite.
(b) =⇒ (a): Suppose that ϕ(ẋ) is a formula such that M ϕ(ẋ) is infinite. We can assume that M ϕ(ẋ) k = {p ∈ M ϕ(ẋ) | |p| = k} is infinite for some k < ω, since one can otherwise construct a disjoint sequence. Let M = M ϕ(ẋ) k . Since M is infinite, supp M is infinite as well. Thus there are arbitrarily large sunflowers in supp M by Lemma 6.3. For any l < ω, let C ≥l M denote the set of finite subsets t of A such that there is some sunflower in supp M of size ≥l with centre t.

Subclaim. C ≥l
M is finite for some l < ω. Proof. Take any q ∈ Add(A, 1) with q ¬ϕ(ẋ) and let l = |q| + 1. We claim that C ≥l M is finite. To see this, suppose that C ≥l M is infinite. Then some t ∈ C ≥l M is disjoint from supp(q). Fix a sunflower N in supp M of size ≥l with root t. It follows that at most |q| elements of N can have nonempty intersection with supp(q). Since |N | > |q|, we can then find some r ∈ N with supp(q) ∩ supp(r) = ∅, so q and r are compatible. But q is incompatible with all the elements of M , which is a contradiction.
Using the previous claim, fix some l such that C ≥l M is finite. So there are only finitely many roots for sunflowers of size ≥l. Since there are arbitarily large sunflowers in supp M , there is some t ∈ C ≥l M such that there are arbitrarily large sunflowers in supp M with centre t.
There are only finitely many possible values that a condition can take on t, in fact at most 3 k many, where the third value stands for 'undefined'. Therefore, there is a condition q with the property: there are arbitrarily large subsets N of M such that supp N forms a sunflower with centre t and p ↾ t = q for all p ∈ N . Since q is a proper subset segment of p if |N | > 1, in particular q is not minimal in N and thus q / ∈ M .

Subclaim. q ϕ(ẋ).
Proof. Otherwise take some r ≤ q with r ¬ϕ(ẋ). Then r is incompatible with all elements of N . Moreover, take a subset N of M as above with |N | > |r|. Since supp N forms a sunflower with centre t and |N | > |r|, there is some s ∈ N with supp(r) ∩ supp(s) = t. We further have s ↾ t = q by the choice of N . Therefore r and s are compatible. But this contradicts the fact that r and s force opposite truth values of ϕ(ẋ).
By the previous claim, we have q ∈ M , contradicting the above facts.

Proof. (a) =⇒ (b): Assume that [A]
<ω is Dedekind-infinite, and let A n | n < ω be a disjoint sequence witnessing that. Suppose that G ⊆ Add(A, 1) is V -generic subset of A, and let B be {a | ∃p ∈ G, p(a) = 1}. Define f (n) = a whenever B ∩ A n = {a}, which by a density argument happens infinitely often. This defines an injection from an infinite subset of ω into A, and therefore A is Dedekind-infinite.
(b) =⇒ (c): Suppose that A is Dedekind-infinite in the generic extension, and let f : A → A be an injection which is not a bijection, then there is some a ∈ A such that f (x) = a for all x ∈ A. In V , by Dedekind-finiteness of A there, the set A \ {a} is strictly smaller in size, and thus witnessing that A was collapsed.
Let ϕ(ň,ǎ) denote the formulaḟ (ň) =ǎ. Let s n a be the set supp M ϕ(ň,ǎ) for n < ω and a ∈ A. By the finite decision property in Lemma 6.6, M ϕ(ň,ǎ) and thus s n a are finite.
Subclaim. For any n < ω, {a ∈ A | s n a = ∅} is finite. Proof. Fix n < ω and let B denote {a ∈ A | s n a = ∅}. Towards a contradiction, suppose that B is infinite. First assume that for all k < ω, there are only finitely a ∈ B with |s n a | = k. Then A k defined as |s n a |=k s n a is finite as well. Since the sets M ϕ(ň,ǎ) are disjoint as a ∈ A varies, A = A k | k < ω has an injective infinite subsequence and hence [A] <ω is Dedekind-infinite. Now assume that for some k < ω, there are infinitely many a ∈ B with |s n a | = k. Note that k = 0 by the definition of B. Since the sets M ϕ(ň,ǎ) are disjoint as a ∈ A varies, the set S = {s n a | a ∈ B} is infinite. This set contains arbitrarily large sunflowers by Lemma 6.3.
Let T be a sunflower in S of size 3 k + 1. Since the centre of T has size ≤k, there are at most 3 k possible values for restrictions to the centre of T . Hence we can find a = b in B and conditions p ∈ M ϕ(ň,ǎ) and q ∈ M ϕ(ň,ǎ) with p compatible with q. But this contradicts the fact that conditions in M ϕ(ň,ǎ) and M ϕ(ň,b) are pairwise incompatible.
Note that s n a = ∅ implies that M n a is either empty or contains only 1. Thus the subclaim implies that M n defined as a∈A M ϕ(ň,ǎ) is finite for all n < ω. In other words, there are only finitely many possible choices forḟ (n). However, 1 forces thatḟ has infinite range and thus n<ω M n is an infinite subset of A. This allows us to construct a disjoint sequence, in contradiction to the assumption that [A] <ω is Dedekind-finite.
Note that by the homogeneity of Add(A, 1) the above proof does not depend on the choice of a generic filter, and since collapsing A or making it Dedekind-infinite can be stated as a formula whose free variables are all canonical ground model names, there is no dependence on any specific condition either. Suppose that 1 forces thatẊ is a new subset of some ordinal η, and let ϕ(α) denote the formulaα ∈Ẋ. By the finite decision property in Lemma 6.6, M ϕ(α) is finite for all α < η. However, α<η M ϕ(α) is infinite, sinceẊ is a name for a new set of ordinal. Thus it is easy to construct a disjoint sequence.
We equip 2 A with the product topology. Moreover, let N p = {x ∈ 2 A | p ⊆ x} denote the basic open set associated to p ∈ Add(A, 1). Note that 2 A is a Hausdorff space. We will consider the following notion: a topological space is called extremally Note that f ∈ 2 A is in the closure of N I if and only if there is some p ∈ Add(A, 1) such that p ⊆ f and N p ∩ N I = ∅, which equivalently means that p ġ ∈Ṅ I , whereġ is the canonical generic function A → {0, 1} andṄ I is the re-interpretation of the union p∈I N p in the generic extension. But by the finite decision property, which holds by Lemma 6.6, there is a finite set, M , of minimal elements p such that p ġ ∈Ṅ I . In particular, C = p∈M N p is a finite union of clopen sets and thus clopen, and N I is dense in C, since for every p ∈ I, p ġ ∈Ṅ I . So the closure of N I is open, as wanted.