Diamond, Continuum, and Forcing

Transcription

1 Diamond, Continuum, and Forcing Susan, Summer 2018 The Diamond Axiom Let s play a game. You choose a subset A of ω 1, and the game will be played in ω 1 rounds. On the α th round, I will attempt to guess the intersection of your set with the ordinal α. At the end of the game, we ll look at the set S of α for which I guessed correctly. A reasonable question to ask might be: how big can the set S be? I would have to be extremely lucky to guess correctly on a club. But maybe a stationary set isn t too much to ask for. The diamond axiom is the statement that without any information about the set A, I can always guess so that the set S is stationary. In fact, I can always do it with the same sequence of guesses! The Diamond Axiom: There exists a sequence of sets {g α : α ω 1 } satisfying the following properties: (i) For all α ω 1, g α α. (ii) For any A ω 1, the set S A = {α : g α = A α} is stationary. Such a sequence is called a diamond sequence. Diamond And CH Diamond is a smallness axiom. This may seem strange, since it gets used to build objects whose existence is independent of ZFC. In fact, it starts by asserting the existence of such an object! However, it makes sense if you think about it in terms of the number of subsets of ω 1. Diamond says that there are few enough subsets of ω 1 that all of them are guessable on a stationary set using a single sequence of guesses. One example of Diamond acting as a smallness axiom is the fact that in any universe in which Diamond is true, the Continuum Hypothesis is true as well: 1

2 Theorem: CH Proof. Let {g α : α ω 1 } be a diamond sequence, and let A be a subset of ω. Then the set S A = {α : g α = A α} is stationary. In particular, S A {α ω 1 : α > ω} is nonempty. Thus there exists some ordinal α > ω such that g α = A α = A. It follows that there can be at most ω 1 subsets of ω, and so the Continuum Hypothesis holds. Forcing Diamond The proof of the consistency of diamond using forcing is really cool. We force with the most obvious possible poset, and without any further cleverness on our part, the resulting object is a diamond sequence. The tricky parts of the proof lie entirely in verifying that it really does fall out just this easily. Theorem: is consistent with ZFC. Proof. Let M be a countable transitive model of ZFC. We wish to create an extension of M in which a diamond sequence exists. This means that the object we wish to produce is, at a minimum, a sequence {g α : α ω 1 } of sets satisfying g α α. We will force with the poset of all partial sequences of this form. That is, we take a poset P consisting of sequences of the form p = {g α : α β ω 1 }. If we have p = {g α : α β ω 1 } and q = {h α : α γ ω 1 } we say that p q if β γ, and for any α < γ we have g α = h α. Okay, here s the crazy part. When we forced the consistency of CH and CH, we had to carefully encode the object that we wanted to build into our poset. Here we haven t put anything remotely diamond-ey into the description of our poset. However, we are going to find that if we take a P-generic filter G over M, the union of G is a diamond sequence in M [G]. Let s prove it! 2

3 Step 1: Preservation of ω 1 First of all, since deals with ω 1 it will be important to verify that the ω 1 we re talking about in M matches the ω 1 in M [G]. To see that this is the case, let s take a countable descending sequence p 0 > p 1 > p 2 >... > p n >... in our poset P. For each n ω, we have p n = {g α : α β n }. If we take the union of all of these sequences, we get { q = g α : α } β n, n ω which is also an element of P, since the union of countably many countable ordinals is a countable ordinal. Furthermore, we know that q p n for all n ω. This tells us that every countable descending sequence in P has a lower bound in P. Let s think about what this means for the identity of ω 1. There is some object ω1 M M which M believes to be ω 1. The model M believes that ω1 M is the smallest uncountable ordinal. Being an ordinal is an absolute property, and countability translates to the existence of a surjection from ω to the countable set. Since these surjections can t disappear in going from M to M [G], all of the ordinals smaller than ω1 M will still be countable in M [G]. This means that the only way for ω1 M to be different from ω M [G] 1 is if it becomes countable in M [G]. Thus if ω1 M ω M [G] 1, then there exists a surjection f M [G] from ω to ω1 M. Since f M [G], there exists a P-name τ such that τ G = f. Since there exists p G such that M [G] = f is a surjection from ω to ω M 1, p τ is a surjection from ˇω to ˇω M 1. For each n, there exists α n ω1 M such that f(n) = α n. Thus there exists q n G such that q n τ(ň) = ˇα n. Define p 0 to be a common strengthening of p and q 0. for each n > 0, define p n to be a common strengthening of p n 1 and q n. The result is a descending chain p > p 0 > p 1 >... > p n >... 3

4 This chain must have a lower bound r. We can use this r to define a function h : ω ω M 1. For each n, there is exactly one α n such that r τ(ň) = ˇα n. Define h(n) = α n. It is clear that for all n, h(n) = f(n). However, we have just constructed h inside M using the forcing relation. This means that f must have existed in M all along, contradicting the fact that M believes ω M 1 is uncountable. It follows that ω M 1 must be the same ordinal as ω M 1 [G]. From here on out, we will just use the symbol ω 1 for both of these. Step 2: A Sequence in M [G] The set G is a filter in P, and for each ordinal α ω 1 the set is dense in P. This means that D α = {p P : g α is defined for q} p G p = {g α : α ω 1 }. We would like to prove that this set is a sequence in M [G]. Let s take two sets A, C M[G] such that A, C ω 1, and such that M [G] = C is a club. There exist P-names τ and σ such that A = τ G and C = σ G. There also exists p 0 G such that p 0 σ is a club. Now we re going to make an insanely ridiculous claim: because p 0 forces C to be a club, it automatically also forces A α to be equal to g α for one of the α s in C! That is, p 0 There exists α such that α σ and τ α = gˇ α. This is clearly what we want, but it seems too good to be true! So let s dig in and see why this would be the case. Suppose p 0 does not force this ridiculously awesome statement. Then there must be some filter G containing p 0 in which the statement is untrue in M [G]. Thus some strengthening of p 0 in G must force its negation. That is, there exists p 1 p 0 with p 1 G, and p 1 For all α, either α / σ or τ α gˇ α. 4

5 We re going to build a chain of poset elements q 0 q 1 q 2... as follows. First, we set q 0 = p 1. Then given q n, we will choose a q n+1 q n in G satisfying (i) There exists β n such that g βn (q n+1 \ q n ), and q n+1 ˇβ ˇσ. (ii) For each g α q n, either q n+1 ˇα τ or q n+1 ˇα τ. Let s verify that we actually can find a q n+1 satisfying both of these properties. Since p 0 G forces σ to be a club, we know that in M [G], C is an unbounded set. This means that we can find a β in C that is larger than any value α for which a g α is defined in q n. Since D β is dense in P for this β value, we can find an element r G that has a g β defined. This gives us an r that satisfies property (i). To find an element of G that satisfies property (ii), note that for any α such that g α is defined in q n, either α A, or α / A. Thus we can find some s G such that s forces all of the correct statements. That is, for each α either s α A or s α / A. Now we can take a common strengthening in G of r, s, and q n to give us our element q n1. Now that we have our sequence of q n s, we can define q = n ω q n = {g α : α γ ω 1 }. If each q n = {g α : α γ n ω 1 }, then we have γ = n ω γ n, and (i) gives us γ 0 < β 1 < γ 1 < β 2 < γ 2 < β 3 <..., which means that γ = n ω β n. Since q ( ˇβ n σ) for all n, and it must be the case that q ˇγ σ. q σ is a club, Now we ll use property (ii) to build a set B M. We ll define B γ to be the set of α such that q ˇα τ. Since, for all other elements, q ˇα / τ, it must be the case that q ((τ γ) = B). Now we define q = q {g γ = B}. But look what we ve done! We ve shown that q ˇγ σ and τ ˇγ = ǧ γ. 5

6 Which means that q also forces the weaker statement detailed below: q There exists α such that α σ and τ α = gˇ α. This contradicts the definition of p 1, proving that p 0 must have forced this statement to begin with! 6