Tuesday, September 1, 2015
UCI Summer School, part 5 (Monroe Eskew)
This is just a placeholder, for now. My notes for this part are quite rough, so it will be a while before I will try to record it here. The next installment of these notes will cover Brent Cody's lectures on some results about the number of normal measures.
UCI Summer School, part 4: Measure algebras (Monroe Eskew)
Here are some more applications of the ideas we have been considering.
Definition: \mathcal{B} is a measure algebra if it is a complete Boolean algebra equipped with some function \mu:\mathcal{B}\rightarrow [0,1] with \mu(0)=0, \mu(1)=1, \mu(b)>0 for b\neq 0, and \mu countably additive (i.e., if \langle b_i:i<\omega\rangle is an antichain, then \mu(\sum b_i)=\sum \mu(b_i).
Exercise: All measure algebras are c.c.c.
Example: Let \kappa be a cardinal. We will describe a topology on {}^\kappa 2. Fix some x\in [\kappa]^{<\omega} and s:x\rightarrow 2 (i.e., s is a finite domain partial function from \kappa to 2). Then basic open sets are of the form \mathcal{O}_s=\{r\in {}^\kappa 2: \forall \alpha\in x(r(\alpha)=s(\alpha))\}. Let \mathcal{B}_\kappa be the \sigma-algebra generated by these basic open sets, and define \mu on \mathcal{B}_\kappa by setting \mu(\mathcal{O}_s=\frac{1}{2^{|s|}} (standard theorems from real analysis give that \mu extends uniquely to a countably additive probability measure on \mathcal{B}_\kappa.
Define \mathrm{Null}=\{A\in\mathcal{B}_\kappa: \mu(A)=0\}. Then \mathcal{R}_\kappa:=\mathcal{B}_\kappa/\mathrm{Null} is a measure algebra.
Exercise: Prove that \mathcal{R}_\kappa forces 2^\omega\ge \kappa.
Exercise: If \mathcal{A} is a measure algebra and \Vdash_A \dot{\mathcal{B}} is a measure algebra, then \mathrm{r.o.}(\mathcal{A}\ast \dot{\mathcal{B}}) is a measure algebra. (Note: this is not as easy as it may seem at first since for example \mathcal{A} might even add new reals which can be measures of elements of \mathcal{B}! We use r.o. for the Boolean completion here since the letter \mathcal{B} is overloaded).
Exercise: If \mathcal{A} is a complete sublagebra of a measure algebra \mathcal{B}, then \mathcal{A} is a measure algebra.
Continuing along this line,
Theorem: If \mathcal{B} is a measure algebra, \mathcal{A} a complete subalgebra of \mathcal{B}, and G\subseteq \mathcal{A} is generic over V, then in V[G] we have that \mathcal{B}/G is a measure algebra.
Note that in V[G], G is a filter on \mathcal{B}, so \mathcal{B}/G=\{[b]_G:b\in \mathcal{B}\}. We use G^* for the dual ideal. It's important to distinguish between the orderings of the two Boolean algebras here, and will be good to see how to translate between them using the forcing relation.
Lemma: If \mathcal{B} is complete and \mathcal{A} is a complete subalgebra and G\subseteq \mathcal{A} is generic, then \mathcal{B}/G is complete in V[G].
Proof of Lemma: Suppose \langle [b_\alpha]_G:\alpha<\kappa\rangle \in P(\mathcal{B}/G)\cap V[G]. For each \alpha<\kappa, let X_\alpha:=\{b:1\Vdash_{\mathcal{A}} [b]_{\dot{G}}\le [\dot{b}_\alpha]_{\dot{G}}\}. Let c_\alpha=\sum X_\alpha\in V (meet taken in \mathcal{B}).
We claim that 1\Vdash_{\mathcal{A}} [c_\alpha]_{\dot{G}}=[\dot{b}_\alpha]_{\dot{G}} for each \alpha--this suffices to prove the lemma. The proof of the claim is straightforward but a little tedious. First we show 1\Vdash_{\mathcal{A}} [c_\alpha]_{\dot{G}}\le[\dot{b}_\alpha]_{\dot{G}}. If this doesn't hold, then there are d,p so that p\in \mathcal{A}, d\wedge p\neq 0, and
p\Vdash [\check{d}]\le [\check{c}_\alpha] \textrm{ and }[\check{d}]\wedge [\dot{b}_\alpha]=0.
The first conjunct implies that p\wedge d\le c_\alpha in \mathcal{B}, and since c_\alpha is a lub for X_\alpha, there is some b\in X_\alpha so that p\wedge d\wedge b\neq 0. So 1\Vdash [p\wedge d\wedge b]\le [b_\alpha] by the definition of b\in X_\alpha. But the second conjunct gives p\wedge d\wedge b_\alpha=0, contradiction.
Now to show 1\Vdash_{\mathcal{A}} [c_\alpha]_{\dot{G}}\ge[\dot{b}_\alpha]_{\dot{G}}, assume for a contradiction that there are p,a\in \mathcal{A} so that p\Vdash [\check{a}]\le [\dot{b}_\alpha] and p\Vdash [a\wedge \neg c_\alpha]\neq 0. Now p forces [p\wedge a \wedge \neg c_\alpha]\le [\dot{b}_\alpha]. Trivially, \neg p forces [p\wedge a \wedge \neg c_\alpha]=0. So it's just outright forced that [p\wedge a \wedge \neg c_\alpha]\le [b_\alpha] and thus p\wedge a\wedge \neg c_\alpha \in X_\alpha, which contradicts c_\alpha is an upper bound for X_\alpha, completing the proof of the claim and the lemma. \Box
Proof of Theorem: Let \mu be a measure on \mathcal{B}, \mathcal{A} a complete subalgebra of \mathcal{B}. Define
\mu(b\mid a)=\frac{\mu(a \wedge b)}{\mu(a)}.
(We say the measure of b conditioned on a).
Definition: For a\in A,b\in B, \epsilon>0, say a is \epsilon-stable for b if for all x\le a in \mathcal{A}, |\mu(b\mid x)-\mu(b\mid a)|<\epsilon.
Lemma: For all b\in B and for all \epsilon>0 the set \{a\in A:a \textrm{ is }\epsilon-\textrm{stable for} b\} is dense in A.
Proof: Exercise. An interesting one.
In V[G], let \nu:\mathcal{B}/G\rightarrow [0,1] be given by \nu([b])=r if for every \epsilon>0 there is some a\in G so that a is \epsilon-stable for b and |\mu(b\mid a)-r|<\epsilon. The idea is that G could add new reals, so we can only have approximations to the measure of [b] using ground model reals attached to the members of \mathcal{A}.
We can check that this is well-defined: if [b]_G=[c]_G, then some a\in G forces b\Delta c\in G^*. This means that a\perp (b\Delta c), so \mu(a\wedge (b\Delta c)=0. Therefore \mu(b\mid x)=\mu(c\mid x) for all x\le a. Suppose r_0\neq r_1 both satisfy \nu(b)=r_i. Take \epsilon<|r_1-r_0|. Let a_0,a_1\in G be \epsilon/4-stable for b with |\mu(b\mid a_i)-r_i|<\epsilon/4. Now take a\le a_0, a_1 in \mathcal{A}. By a triangle inequality argument, we have |r_1-r_0|<\epsilon, a contradiction.
Exercise: Check that \nu(b)>0 for all b\neq_G 0.
Exercise: Check that \nu is countably additive. First prove that it is finitely additive.
\Box.
Now suppose P(Z)/I is a measure algebra. The duality theorem (ccc case) says that for any \theta, \mathcal{R}_\theta\ast P(Z)/\bar{I}\cong P(Z)/I\ast j(\mathcal{R}_\theta),
where j:V\rightarrow M is the generic embedding in V[G], G generic for P(Z)/I.
The right hand side of this isomorphism is a measure algebra, since P(Z)/I is a measure algebra by assumption, and j(\mathcal{R}_\theta) is a measure algebra of M, a model which is closed under countable sequences (and so has all the countable sequences to witness countable additivity and completeness of the Boolean algebra).
We have a map e:\mathcal{R}_\theta\rightarrow \mathrm{r.o.}(P(Z)/I\ast j(\mathcal{R}_\theta), so \mathcal{R}_\theta is isomorphic to a complete subalgebra of the right hand side. Now if H\subseteq \mathcal{R}_\theta is generic, then B/e''H is a measure algebra. Therefore P(Z)/\bar{I} is also a measure algebra.
A real-valued measurable cardinal is a cardinal \kappa which carries a \kappa-additive probability measure on all subsets of \kappa which gives measure 0 to singletons. It is atomless if every set of positive measure has a subset of strictly smaller positive measure.
Corollary: If \langle \kappa_i:i<\theta\rangle is a sequence of measurable cardinals, then if \gamma=\sup \kappa_i, \mathcal{R}_\gamma forces all \kappa_i to be atomless real-valued measurable cardinals (RVMs).
We note the fact that if \kappa is atomlessly RVM, then 2^\omega\ge \kappa, so we can't get class many RVMs.
However, if \kappa is strongly compact, then \mathcal{R}_\kappa forces that for all regular \lambda\ge \kappa, there is a countably additive real-valued probably measure \mu_\lambda on \lambda giving measure 0 to all subsets of size <\lambda.
Friday, August 7, 2015
UCI Summer School, part 3: Applications of Duality Theorem (Monroe Eskew)
We now turn towards applications of the duality theorem. It is recommended that the reader recalls the notation (I,j,K,J,\hat{H},e,\iota, etc.) from the previous lecture before proceeding.
The basic idea is that one uses the isomorphism there:
\mathcal{B}(\dot{\mathbb{P}\ast P(Z)/J})\equiv \mathcal{B}(P(Z)/I\ast j(\mathbb{P})/\dot{K})
to calculate the quotient algebra P(Z)/J as \mathcal{B}(P(Z)/I\ast j(\mathbb{P})/\dot{K})/e''H.
As discussed near the end of the last lecture, under certain assumptions, the statement of the duality theorem becomes somewhat simpler. The first examples will fall into this case.
Special Case: If I is \kappa-complete and \mathbb{P} is \kappa-c.c., then the hypothesis of the duality theorem holds, K=\{0\} and J=\bar{I}, the ideal generated by I in V^\mathbb{P}.
Exercise: Show that in the above case, if p\in \mathbb{P} and A\in (P(Z)/I)\cap V, then
\iota(p,\check{A})=(A,j(\dot{p})).
A further simplification will be that we will usually start with a measurable cardinal \kappa and take I to be the dual to the measure on \kappa, so P(\kappa)/I is the trivial Boolean algebra.
A measurable cardinal \kappa has a 2-saturated, \kappa-complete ideal, namely the dual to the measure on \kappa, and under GCH every cardinal \kappa carries a \kappa^{++}-saturated, \kappa-complete ideal, namely the ideal of bounded subsets. This motivates the following natural questions, which are the main focus of this lecture:
Question: Suppose \mu\le \kappa^+ is a regular cardinal. Is it consistent that there is a cardinal \kappa which is not measurable, but still \kappa carries a \mu-saturated, \kappa-complete ideal? (Here we want the amount of saturation to be exactly \mu).
Digression: does the answer change if we require \kappa to be a successor cardinal?
For the case where \kappa is a successor cardinal, \kappa^+-saturation is the strongest we can hope to achieve.
Exercise: Prove using the method of generic ultrapowers that if \kappa is a successor cardinal then there is no \kappa-complete, \kappa-saturated ideal on \kappa.
Kunen showed that if \kappa is a successor cardinal, then getting a \kappa^+-saturated ideal on \kappa requires large cardinals much stronger than a measurable, although we won't do this argument here (you can find it in this previous Specinar post. We now return to the original question.
Answer to question 1, if \mu<\kappa (\mu regular): We will use the basic technique of computing the quotient algebra P(\kappa)/J in V[H] using the duality theorem. Start with \kappa measurable in the ground model. Let \theta\ge \kappa, and consider \mathrm{Add}(\mu,\theta) which adds \theta Cohen subsets of \mu. \mathrm{Add}(\mu,\theta) is \kappa-c.c., and under the GCH it is even \mu^+-c.c. Let I=U^*, where U is a \kappa-complete normal ultrafilter on \kappa (here the star means taking the dual ideal). Let j:V\rightarrow M be the ultrapower embedding.
The duality theorem gives the isomorphism:
\mathrm{Add}(\mu,\theta)\ast P(\kappa)/\bar{I}\cong P(\kappa)/I\ast \mathrm{Add}(\mu,j(\theta))\cong\mathrm{Add}(\mu,j(\theta)),
since P(\kappa)/I is trivial.
If H is generic for \mathrm{Add}(\mu,\theta) over V, then
e''H=\{(1,j(p)):p\in H\}.
So
P(\kappa)/\bar{I}\cong \mathrm{Add}(\mu,j(\theta))/e''H\cong \mathrm{Add}(\mu,j(\theta)).
In V[H], P(\kappa)/\bar{I}\cong \mathcal{B}(\mathrm{Add}(\mu,j(\theta)), so \bar{I} is (\mu^{<\mu})^+ saturated. Furthermore, it is easy to check that \bar{I} is \kappa-complete, and 2^\mu\ge \kappa in V[H], so \kappa is not measurable. This answers question 1 for the case where \mu<\kappa. \Box
We might ask what large cardinal properties of \kappa are implied by this ideal hypothesis.
Proposition: If \kappa carries a \kappa-complete \mu-saturated ideal for some \mu<\kappa, then:
The basic idea is that one uses the isomorphism there:
\mathcal{B}(\dot{\mathbb{P}\ast P(Z)/J})\equiv \mathcal{B}(P(Z)/I\ast j(\mathbb{P})/\dot{K})
to calculate the quotient algebra P(Z)/J as \mathcal{B}(P(Z)/I\ast j(\mathbb{P})/\dot{K})/e''H.
As discussed near the end of the last lecture, under certain assumptions, the statement of the duality theorem becomes somewhat simpler. The first examples will fall into this case.
Special Case: If I is \kappa-complete and \mathbb{P} is \kappa-c.c., then the hypothesis of the duality theorem holds, K=\{0\} and J=\bar{I}, the ideal generated by I in V^\mathbb{P}.
Exercise: Show that in the above case, if p\in \mathbb{P} and A\in (P(Z)/I)\cap V, then
\iota(p,\check{A})=(A,j(\dot{p})).
A further simplification will be that we will usually start with a measurable cardinal \kappa and take I to be the dual to the measure on \kappa, so P(\kappa)/I is the trivial Boolean algebra.
A measurable cardinal \kappa has a 2-saturated, \kappa-complete ideal, namely the dual to the measure on \kappa, and under GCH every cardinal \kappa carries a \kappa^{++}-saturated, \kappa-complete ideal, namely the ideal of bounded subsets. This motivates the following natural questions, which are the main focus of this lecture:
Question: Suppose \mu\le \kappa^+ is a regular cardinal. Is it consistent that there is a cardinal \kappa which is not measurable, but still \kappa carries a \mu-saturated, \kappa-complete ideal? (Here we want the amount of saturation to be exactly \mu).
Digression: does the answer change if we require \kappa to be a successor cardinal?
For the case where \kappa is a successor cardinal, \kappa^+-saturation is the strongest we can hope to achieve.
Exercise: Prove using the method of generic ultrapowers that if \kappa is a successor cardinal then there is no \kappa-complete, \kappa-saturated ideal on \kappa.
Kunen showed that if \kappa is a successor cardinal, then getting a \kappa^+-saturated ideal on \kappa requires large cardinals much stronger than a measurable, although we won't do this argument here (you can find it in this previous Specinar post. We now return to the original question.
Answer to question 1, if \mu<\kappa (\mu regular): We will use the basic technique of computing the quotient algebra P(\kappa)/J in V[H] using the duality theorem. Start with \kappa measurable in the ground model. Let \theta\ge \kappa, and consider \mathrm{Add}(\mu,\theta) which adds \theta Cohen subsets of \mu. \mathrm{Add}(\mu,\theta) is \kappa-c.c., and under the GCH it is even \mu^+-c.c. Let I=U^*, where U is a \kappa-complete normal ultrafilter on \kappa (here the star means taking the dual ideal). Let j:V\rightarrow M be the ultrapower embedding.
The duality theorem gives the isomorphism:
\mathrm{Add}(\mu,\theta)\ast P(\kappa)/\bar{I}\cong P(\kappa)/I\ast \mathrm{Add}(\mu,j(\theta))\cong\mathrm{Add}(\mu,j(\theta)),
since P(\kappa)/I is trivial.
If H is generic for \mathrm{Add}(\mu,\theta) over V, then
e''H=\{(1,j(p)):p\in H\}.
So
P(\kappa)/\bar{I}\cong \mathrm{Add}(\mu,j(\theta))/e''H\cong \mathrm{Add}(\mu,j(\theta)).
In V[H], P(\kappa)/\bar{I}\cong \mathcal{B}(\mathrm{Add}(\mu,j(\theta)), so \bar{I} is (\mu^{<\mu})^+ saturated. Furthermore, it is easy to check that \bar{I} is \kappa-complete, and 2^\mu\ge \kappa in V[H], so \kappa is not measurable. This answers question 1 for the case where \mu<\kappa. \Box
We might ask what large cardinal properties of \kappa are implied by this ideal hypothesis.
Proposition: If \kappa carries a \kappa-complete \mu-saturated ideal for some \mu<\kappa, then:
- \kappa is weakly Mahlo
- \kappa has the tree property.
Exercise: Prove (1) of the proposition using generic ultrapowers.
Proof of Proposition (2): Suppose T is a \kappa-tree. If G\subseteq P(\kappa)/I is generic, then in V[G], T has a branch b given by taking any member of level \kappa of j(T), where j is the generic ultrapower embedding. Now for each \alpha<\kappa, S_\alpha=\{x\in T_\alpha: \exists p(p\Vdash \check{x}\in \dot{b})\}<\mu by the saturation. Now \bigcap_{\alpha<\kappa} S_\alpha is a \kappa-tree all of whose levels have size <\mu<\kappa. It is well-known (or a good exercise) that such trees have cofinal branches. \Box
Definition: An ideal I is nowhere prime if there is no I-positive set A so that I\upharpoonright A is prime (i.e., dual to an ultrafilter).
Exercise: Show that if there is a nowhere prime, \kappa-complete, \mu^+-saturated ideal, where \mu<\kappa, then 2^\mu\ge \kappa.
We continue with Question 1 with other arrangements of \mu and \kappa.
Answer to question 1, if \mu=\kappa^+: Start with \kappa measurable with 2^\kappa=\kappa^+, U a normal ultrafilter and j:V\rightarrow M the ultrapower embedding. Let \langle \mathbb{P}_\alpha,\dot{\mathbb{Q}}_\alpha:\alpha<\kappa\rangle be the Easton support iteration where for regular \alpha, \Vdash_{\mathbb{P}_\alpha} \dot{Q}=\mathrm{Add}(\alpha,1). It is straightforward to verify that \mathbb{P}_\kappa has the \kappa-c.c., so we are in the special case again.
Note that j(\mathbb{P}_\kappa)=\mathbb{P}_\kappa\ast \mathbb{Add}(\alpha,1)\ast (\mathbb{P}_{\kappa+1,j(\kappa)})^M (We use the notation \mathbb{P}_{\xi,j(\kappa)} for j(\mathbb{P}_\kappa)_{\xi,j(\kappa)}). The tail part is computed differently in M than in V, e.g., the support is on M-regular cardinals.
If G_\kappa\subseteq \mathbb{P}_\kappa is generic over V, then the special case of the duality theorem says that in V[G_\kappa], P(\kappa)/\bar{I}\cong (\mathbb{P}_{\kappa,j(\kappa)})^M. However, P(\kappa)/\bar{I}\cong(\mathbb{P}_{\kappa,j(\kappa)})^M does not have the \kappa^+-c.c. since there are M-regular cardinals between \kappa^+ and j(\kappa). So \bar{I} is not \kappa^+-saturated.
Now we could also satisfy the hypothesis of the duality theorem of adding a j(\mathbb{P}_\kappa)=P(\kappa)/J\cong \mathrm{Add}(\alpha,1)\ast (\mathbb{P}_{\kappa+1,j(\kappa)})^M generic filter \hat{H} over M in a different way. By a standard technique, a j(\mathbb{P})-generic over M exists in V[G_{\kappa+1}] (where G_{\kappa+1} is \mathbb{P}_\kappa\ast \mathrm{Add}(\alpha,1)-generic). This is because we clearly get a generic for the initial part \mathbb{P}_\kappa\ast \mathrm{Add}(\alpha,1), just G_{\kappa+1} itself. For the tail, (\mathbb{P}_{\kappa+1,j(\kappa)})^{M[G_{\kappa+1}]} is is j(\kappa)-c.c. of size j(\kappa) in M[G_{\kappa+1}], so M[G_{\kappa+1}] thinks the poset has at most j(\kappa) maximal antichains. In V[G_{\kappa+1}], |j(\kappa)|=\kappa^+, and the poset is \kappa^+-closed (in M[G_{\kappa+1}], but also in V[G_{\kappa+1}] by the agreement between these models). So we can construct a generic by hand in V[G_{\kappa+1}] by enumerating all of the maximal antichains in M[G_{\kappa+1}]. This completes the construction of \hat{H} in the extension by \mathrm{Add}(\kappa,1).
In this construction, we have that j(\mathbb{P})/K\cong \mathbb{P}_\kappa\ast \mathrm{Add}(\kappa,1), since the Boolean algebra homomorphism j(\mathbb{P}_\kappa)\rightarrow \mathbb{P}_\kappa\ast \mathrm{Add}(\kappa,1) given by p\mapsto \|p\in \hat{H}\|. has kernel exactly K as defined in the last lecture, and the map is surjective since the codomain completely embeds into the domain. So in the duality theorem calculation, we obtain an ideal J so that P(\kappa)/J\cong \mathrm{Add}(\kappa,1). So J is a \kappa^+-saturated ideal on \kappa. \Box
Note that \kappa is inaccessible.
Exercise: Prove that \kappa is weakly compact in V[G_\kappa]. (Hint: use the tree property characterization.)
Exercise: Prove that \kappa is not measurable in V[G_\kappa], but it is measurable in V[G_{\kappa+1}].
Remark: By forcing with (\mathbb{P}_{\kappa,j(\kappa)})^{M[G_{\kappa+1}]} instead of just \mathrm{Add}(\kappa,1) to add the j(\mathbb{P})-generic, we can get a nowhere prime \kappa complete \kappa^+-saturated ideal on \kappa in V[G_{\kappa+1}].
Answer to question 1, if \mu=\kappa: We will find an example so that \kappa is not weakly compact (compare to earlier results for saturation below \kappa), and in fact the quotient algebra is isomorphic to a \kappa-Suslin tree.
In the exercises, we will describe how to construct, for \alpha regular with \alpha^{<\alpha}=\alpha, a forcing \mathbb{Q}_\alpha which adds an \alpha-Suslin tree \dot{T}_\alpha so that \mathcal{B}(\mathbb{Q}_\alpha\ast \dot{T}_\alpha)\cong\mathrm{Add}(\alpha,1). This is due to Kunen.
In the construction for \mu=\kappa^+ we got a model (there called V[G_\kappa]) where there was an inaccessible \kappa and an ideal J on \kappa so that
P(\kappa)/J\cong\mathrm{Add}(\kappa,1)\cong \mathbb{Q}_\kappa\ast \dot{T},
where \dot{T} is the \kappa-Suslin tree added by \mathbb{Q}_\kappa.
Now start with this to be our ground model V. Let H\subseteq \mathbb{Q}_\kappa be generic. We want to show that in V[H], there is an ideal J_1 on \kappa so that P(\kappa)/J_1\cong T.
If G\subseteq \mathrm{Add}(\kappa,1) is generic over V, then take in V[G] an embedding
j:V\rightarrow M
which was constructed before. We want to extend the embedding to V[H].
Now G\in M since M is closed under \kappa-sequences in V[G]. We can extend j to V[G] by constructing a generic \hat{G} for \mathrm{Add}(j(\kappa))^M over M with \hat{G}\upharpoonright \kappa=G (using the standard method; cf the second exercise following previous construction).
In V, by duality theorem there are J_1 and K so that
\mathbb{Q}_\kappa\ast P(\kappa)/J_1\cong P(\kappa)/J\ast j(\mathbb{Q}_\kappa)/K.
In this case, K is a maximal ideal since the j(\mathbb{Q}_\kappa)-generic over M is already just added by P(\kappa)/J. So in V[H], P(\kappa)/J_1\cong T.
Now we turn to Kunen's forcing construction. Conditions in Kunen's forcing \mathbb{Q} are normal trees of successor ordinal height <\kappa which are homogeneous: for all t\in T not on the top level, T_t\cong T, where T_t is the tree \{s\in T: t\le_T s\} with the order inherited from T.
Exercise:
\Box
We will do one last application to construct a precipitous ideal on a cardinal \kappa which is not measurable so that its quotient algebra is \kappa^+ closed.
Start with \kappa measurable and 2^\kappa>\kappa^+. We will use the Easton support iteration \langle \mathbb{P}_\alpha,\dot{\mathbb{Q}}_\alpha:\alpha<\kappa\rangle, where \dot{\mathbb{Q}}_\alpha=\dot{\mathrm{Add}(\alpha^+,1)} for inaccessible \alpha<\kappa (and is trivial forcing otherwise).
Then \mathbb{P}_\kappa is \kappa-c.c., and forces that for all inaccessible \alpha<\kappa, 2^\alpha=\alpha^+ (this is a standard coding trick that was assigned as an exercise in one of Spencer Unger's lectures here). By duality,
\mathbb{P}_\kappa\ast P(\kappa)/\bar{I}\equiv j(\mathbb{P}_\kappa).
If G_\kappa\subseteq \mathbb{P}_\kappa is generic, then j(\mathbb{P}_\kappa)/e''G_\kappa\equiv \mathbb{P}_{\kappa,j(\kappa)}. Since M[G_\kappa] is closed under \le \kappa sequences in V[G_\kappa], this tail is \kappa^+-closed forcing over V[G_\kappa].
However, GCH holds at every inaccessible \alpha<\kappa and fails at \kappa in V[G_\kappa]. By a reflection argument, \kappa cannot be measurable in V[G_\kappa]. \Box
Exercise: Show that if H\subseteq P(\kappa)/\bar{I} is generic, then \kappa is measurable in V[G_\kappa\ast H].
Note that \kappa is inaccessible.
Exercise: Prove that \kappa is weakly compact in V[G_\kappa]. (Hint: use the tree property characterization.)
Exercise: Prove that \kappa is not measurable in V[G_\kappa], but it is measurable in V[G_{\kappa+1}].
Remark: By forcing with (\mathbb{P}_{\kappa,j(\kappa)})^{M[G_{\kappa+1}]} instead of just \mathrm{Add}(\kappa,1) to add the j(\mathbb{P})-generic, we can get a nowhere prime \kappa complete \kappa^+-saturated ideal on \kappa in V[G_{\kappa+1}].
Answer to question 1, if \mu=\kappa: We will find an example so that \kappa is not weakly compact (compare to earlier results for saturation below \kappa), and in fact the quotient algebra is isomorphic to a \kappa-Suslin tree.
In the exercises, we will describe how to construct, for \alpha regular with \alpha^{<\alpha}=\alpha, a forcing \mathbb{Q}_\alpha which adds an \alpha-Suslin tree \dot{T}_\alpha so that \mathcal{B}(\mathbb{Q}_\alpha\ast \dot{T}_\alpha)\cong\mathrm{Add}(\alpha,1). This is due to Kunen.
In the construction for \mu=\kappa^+ we got a model (there called V[G_\kappa]) where there was an inaccessible \kappa and an ideal J on \kappa so that
P(\kappa)/J\cong\mathrm{Add}(\kappa,1)\cong \mathbb{Q}_\kappa\ast \dot{T},
where \dot{T} is the \kappa-Suslin tree added by \mathbb{Q}_\kappa.
Now start with this to be our ground model V. Let H\subseteq \mathbb{Q}_\kappa be generic. We want to show that in V[H], there is an ideal J_1 on \kappa so that P(\kappa)/J_1\cong T.
If G\subseteq \mathrm{Add}(\kappa,1) is generic over V, then take in V[G] an embedding
j:V\rightarrow M
which was constructed before. We want to extend the embedding to V[H].
Now G\in M since M is closed under \kappa-sequences in V[G]. We can extend j to V[G] by constructing a generic \hat{G} for \mathrm{Add}(j(\kappa))^M over M with \hat{G}\upharpoonright \kappa=G (using the standard method; cf the second exercise following previous construction).
In V, by duality theorem there are J_1 and K so that
\mathbb{Q}_\kappa\ast P(\kappa)/J_1\cong P(\kappa)/J\ast j(\mathbb{Q}_\kappa)/K.
In this case, K is a maximal ideal since the j(\mathbb{Q}_\kappa)-generic over M is already just added by P(\kappa)/J. So in V[H], P(\kappa)/J_1\cong T.
Now we turn to Kunen's forcing construction. Conditions in Kunen's forcing \mathbb{Q} are normal trees of successor ordinal height <\kappa which are homogeneous: for all t\in T not on the top level, T_t\cong T, where T_t is the tree \{s\in T: t\le_T s\} with the order inherited from T.
Exercise:
- Show that Kunen's forcing is \kappa-strategically closed. Hint: the strategy will go by choosing a particular branch through each of the small trees chosen in a play of the game so far.
- Show that \mathbb{Q}\ast \dot{T} has a \kappa-closed dense subset, and deduce that \mathbb{Q}\ast \dot{T}\cong \mathrm{Add}(\kappa).
- Show that \dot{T} is a Suslin tree.
\Box
We will do one last application to construct a precipitous ideal on a cardinal \kappa which is not measurable so that its quotient algebra is \kappa^+ closed.
Start with \kappa measurable and 2^\kappa>\kappa^+. We will use the Easton support iteration \langle \mathbb{P}_\alpha,\dot{\mathbb{Q}}_\alpha:\alpha<\kappa\rangle, where \dot{\mathbb{Q}}_\alpha=\dot{\mathrm{Add}(\alpha^+,1)} for inaccessible \alpha<\kappa (and is trivial forcing otherwise).
Then \mathbb{P}_\kappa is \kappa-c.c., and forces that for all inaccessible \alpha<\kappa, 2^\alpha=\alpha^+ (this is a standard coding trick that was assigned as an exercise in one of Spencer Unger's lectures here). By duality,
\mathbb{P}_\kappa\ast P(\kappa)/\bar{I}\equiv j(\mathbb{P}_\kappa).
If G_\kappa\subseteq \mathbb{P}_\kappa is generic, then j(\mathbb{P}_\kappa)/e''G_\kappa\equiv \mathbb{P}_{\kappa,j(\kappa)}. Since M[G_\kappa] is closed under \le \kappa sequences in V[G_\kappa], this tail is \kappa^+-closed forcing over V[G_\kappa].
However, GCH holds at every inaccessible \alpha<\kappa and fails at \kappa in V[G_\kappa]. By a reflection argument, \kappa cannot be measurable in V[G_\kappa]. \Box
Exercise: Show that if H\subseteq P(\kappa)/\bar{I} is generic, then \kappa is measurable in V[G_\kappa\ast H].
Monday, July 27, 2015
UCI Summer School part 2: Duality Theorem (Monroe Eskew)
The Duality Theorem gives a general technique for forcing to make an ideal whose quotient algebra has various properties. It appears in Matthew Foreman's "Calculating quotient algebras of generic embeddings." My version of these notes omits a lot of the dots which indicate that certain objects are just names in a forcing extension. This is for aesthetic reasons, and hopefully does not lead to confusion.
Duality Theorem: Suppose I is a precipitous ideal on Z and \mathbb{P} is any partial order. If: there is a further generic extension of the extension by P(Z)/I so that if j:V\rightarrow M\subseteq M\subseteq V[G] is the ultrapower embedding from G\subseteq P(Z)/I, there is H\subseteq \mathbb{P} generic over V and \hat{H}\subseteq j(\mathbb{P}) generic over M and some extension of j to \hat{j}:V[H]\rightarrow M[\hat{H}].
Then: there is a \mathbb{P}-name for an ideal J on Z and a P(Z)/I-name for an ideal K on j(\mathbb{P}) and a canonical isomorphism \iota:\mathcal{B}(\dot{\mathbb{P}\ast P(Z)/J})\equiv \mathcal{B}(P(Z)/I\ast j(\mathbb{P})/\dot{K}).
So a very general statement of lifting a generic ultrapower map to a forcing extension gives a useful isomorphism for computing P(Z)/J in the generic extension by \mathbb{P}. We remark that in some cases, this will be an equivalence.
Note: In what follows, we tacitly identify all of the posets involved with their Boolean completions. Occasionally for emphasis, this identification will be explicit.
Proof: Assume (1). There is some A\in I^+ and some P(Z)/I-name for a forcing \dot{\mathbb{Q}} so that
A\Vdash_{P(Z)/I}(\Vdash_\dot{\mathbb{Q}} \dot{H_0}\subseteq j(\mathbb{P}) \textrm{ is generic over }M \textrm{ and }H:=j^{-1}[H_0]\subseteq \mathbb{P}\textrm{ is generic over }V).
Note that the set \{p\in\mathbb{P}:\quad \Vdash_{P(A)/I\ast \mathbb{Q}} j(p)\not\in H_0\} cannot be dense, since it is the complement of the generic j^{-1}[H_0]. So there is p_0\in \mathbb{P} so that for all p\le p_0, \| j(p)\in H_0 \|_{P(A)/I\ast \mathbb{Q}}\neq 0. We will constrain ourselves to work below this p_0 in \mathbb{P} and below A in P(Z)/I. For simplicity, assume that A=Z and p_0=1_\mathbb{P}.
In V^{P(Z)/I}, define
K=\{p\in j(\mathbb{P}):\quad \Vdash_{\mathbb{\dot{Q}^G}} p\not\in H_0\}.
Let G\ast h be generic for P(Z)/I\ast j(\mathbb{P})/K. From the j(\mathbb{P})/K-generic h, we can define a j(\mathbb{P})-generic \hat{H}=\{p:[p]_K\in h\}.
Claim: The following properties of H_0 are also true of \hat{H}:
Duality Theorem: Suppose I is a precipitous ideal on Z and \mathbb{P} is any partial order. If: there is a further generic extension of the extension by P(Z)/I so that if j:V\rightarrow M\subseteq M\subseteq V[G] is the ultrapower embedding from G\subseteq P(Z)/I, there is H\subseteq \mathbb{P} generic over V and \hat{H}\subseteq j(\mathbb{P}) generic over M and some extension of j to \hat{j}:V[H]\rightarrow M[\hat{H}].
Then: there is a \mathbb{P}-name for an ideal J on Z and a P(Z)/I-name for an ideal K on j(\mathbb{P}) and a canonical isomorphism \iota:\mathcal{B}(\dot{\mathbb{P}\ast P(Z)/J})\equiv \mathcal{B}(P(Z)/I\ast j(\mathbb{P})/\dot{K}).
So a very general statement of lifting a generic ultrapower map to a forcing extension gives a useful isomorphism for computing P(Z)/J in the generic extension by \mathbb{P}. We remark that in some cases, this will be an equivalence.
Note: In what follows, we tacitly identify all of the posets involved with their Boolean completions. Occasionally for emphasis, this identification will be explicit.
Proof: Assume (1). There is some A\in I^+ and some P(Z)/I-name for a forcing \dot{\mathbb{Q}} so that
A\Vdash_{P(Z)/I}(\Vdash_\dot{\mathbb{Q}} \dot{H_0}\subseteq j(\mathbb{P}) \textrm{ is generic over }M \textrm{ and }H:=j^{-1}[H_0]\subseteq \mathbb{P}\textrm{ is generic over }V).
Note that the set \{p\in\mathbb{P}:\quad \Vdash_{P(A)/I\ast \mathbb{Q}} j(p)\not\in H_0\} cannot be dense, since it is the complement of the generic j^{-1}[H_0]. So there is p_0\in \mathbb{P} so that for all p\le p_0, \| j(p)\in H_0 \|_{P(A)/I\ast \mathbb{Q}}\neq 0. We will constrain ourselves to work below this p_0 in \mathbb{P} and below A in P(Z)/I. For simplicity, assume that A=Z and p_0=1_\mathbb{P}.
In V^{P(Z)/I}, define
K=\{p\in j(\mathbb{P}):\quad \Vdash_{\mathbb{\dot{Q}^G}} p\not\in H_0\}.
Let G\ast h be generic for P(Z)/I\ast j(\mathbb{P})/K. From the j(\mathbb{P})/K-generic h, we can define a j(\mathbb{P})-generic \hat{H}=\{p:[p]_K\in h\}.
Claim: The following properties of H_0 are also true of \hat{H}:
- \Vdash_{P(Z)/I\ast \mathbb{Q}} \hat{H} is j(\mathbb{P})-generic over M.
- \Vdash_{P(Z)/I\ast \mathbb{Q}} j^{-1}[\hat{H}] is \mathbb{P}-generic over V.
- For all p\in \mathbb{P}, \not\Vdash_{P(Z)/I\ast \mathbb{Q}} j(p)\not\in \hat{H}.
Proof of Claim: For (1), suppose D\in M is open dense in j(\mathbb{P}). Then \{[d]_K:d\in D\textrm{ and }d\not\in K\} is dense in j(\mathbb{P})/K. Otherwise, there would exist p\in j(\mathbb{P})/K so that p\wedge d\in K for all d\in D. But this is impossible because we could then force with \mathbb{Q} over V[G] to get a generic H_0 containing p (as p\not\in K), and then H_0\cap D=\emptyset, contradicting genericity of H_0 over M.
The remaining parts of the claim can be checked similarly, and are left as an exercise. \Box.
Now let e:\mathbb{P}\rightarrow \mathcal{B}(P(Z)/I\ast \dot{j(\mathbb{P})/K}) be defined by e(p)=\|j(p)\in \hat{H}\|.
By (3) of the claim above, \mathrm{ker}(e)=0. Also e preserves Boolean operations simply by the elementarity of j. By (2) of the claim, e is a regular embedding (maps maximal antichains pointwise to maximal antichains).
Exercise: e:\mathbb{P}\rightarrow \mathbb{Q} is a regular embedding iff for every q\in\mathbb{Q} there is p\in \mathbb{P} so that for every p'\le p, e(p') is compatible with q.
Thus, if H\subseteq \mathbb{P} is generic over V, we can force with the quotient \mathcal{B}(P(Z)/I\ast \dot{j(\mathbb{P})/K})/e"H over V[H] to obtain a generic G\ast h for P(Z)/I\ast j(\mathbb{P})/K. By the definition of e, we have j_G"H\subseteq \hat{H}, where \hat{H} is defined from h as before. So we can extend the embedding j_G to \hat{j}:V[H]\rightarrow M[\hat{H}].
In V[H] we can finally define J=\{A\subseteq Z:1\Vdash [\mathrm{id}]\not\in \hat{j}(A)\}, where the forcing is with the quotient \mathcal{B}(P(Z)/I\ast \dot{j(\mathbb{P})/K})/e"H. In V, let
\iota(p,\dot{A})=e(p)\wedge \|[id]\in\hat{j}(\dot{A})\|.
Exercise: \iota is order and incompatibility preserving.
It remains to show that the range of \iota is dense. So take an arbitrary (B,\dot{q})\in P(Z)/I\ast j(P)/K. By strengthening this condition, we may assume without loss of generality that there is f:Z\rightarrow \mathbb{P} in V so that B\Vdash [[f]_M]_K=\dot{q}.
By regularity of e (using the characterization in the exercise), there is a p so that for all p'\le p, e(p')\wedge (B,\dot{q})\neq 0. Let \dot{A} be a \mathbb{P}-name for a subset of Z such that p \Vdash \dot{A}=\{z\in B: f(z)\in H\} and \neg p \Vdash \dot{A}\in J^+.
We check that (p,\dot{A}) is actually a condition, which involves checking that p\Vdash \dot{A}\in J^+. So take a generic G\ast \dot{h} containing e(p)\wedge (B,\dot{q}) (which is nonzero by choice of p). Then clearly B\in G and since [[f]_M]_K=q\in h, we have [f]_M=j(f)([\mathrm{id}])\in\hat{H}. Therefore [\mathrm{id}]\in \hat{j}(A), so this generic G\ast \dot{h} shows that it is not forced by \mathcal{B}(P(Z)/I\ast \dot{j(\mathbb{P})/K}/e"H that [\mathrm{id}]\not\in \hat{j}(A)\}.
By definition \iota:=\iota(p,\dot{A}) forces j(p)\in \hat{H} and [\mathrm{id}]\in\hat{j}(\dot{A}). Since \hat{j} extends j and it's forced that \dot{A}\subseteq B, \iota must force B\in G. And since j(p)\Vdash_{j(\mathbb{P})} j(\dot{A})=j(\{z:j(f)(z)\in \hat{H}\}), \iota must force \dot{q}=[j(f)(\mathrm{id})]_K\in h. \Box
Remark: Suppose K as in the Duality Theorem is forced to be principal, i.e., there is m so that \Vdash K=\{p\in j(\mathbb{P}:p\le \neg m\}.
Then the Duality Theorem is easily seen to be an equivalence.
We can compute some nice properties of the ideal J arising from the previous theorem.
Proposition: Using the notation of the previous theorem, J is forced to be precipitous, with the same completeness as I. If I is normal, then J is also normal. Also, if \bar{G}\subseteq P(Z)/J is generic over V[H] and G\ast h=\iota[H\ast \bar{G}] and \hat{j}:V[H]\rightarrow M[\hat{H}] are as before, then V[H]^Z/{\bar{G}}=M[\hat{H}] and \hat{j} is the ultrapower embedding.
Finally, we relate J to the ideal in V[H] generated by I.
Proposition: Suppose K as in the Duality Theorem is forced to be principal, with m so that \Vdash K=\{p\in j(\mathbb{P}):p\le \neg m\}. Suppose further that there exist f and A so that A\Vdash \dot{m}=[f]_G and \dot{B} is a \mathbb{P}-name for \{z\in A: f(z)\in \dot{H}\}. Then \bar{I}\upharpoonright B=J\upharpoonright B, and A\setminus B\in J, where \bar{I} is the ideal in V[H] generated by I.
Friday, July 17, 2015
UCI Summer School part 1: Basics (Monroe Eskew)
The UCI summer school in set theory just finished, and I'll be posting my notes on this blog. They will not be added in chronological order, and the final product will be very different from the actual presentation, assume that all errors were added by me.
There are some exercises which we solved during problem sessions during the day; I will preserve these in the text and can post solutions as well if there is interest.
We start off with Monroe Eskew's lectures on the duality theorem.
Basic facts: This section is meant to be supplemented by my earlier postings about ideal properties. The basic situation is that I is an ideal on Z\subseteq P(X), G is a generic for the forcing P(Z)/I, and j:V\rightarrow M=V^Z/G is the generic ultrapower by G.
Definition: An ideal I has the disjointing property if every antichain A in P(Z)/I has a pairwise disjoint system of representatives.
Exercise: If I is a \kappa-complete, \kappa^+-saturated ideal, then I has the disjointing property. If I is a normal ideal on Z\subseteq P(X), and I is |X|^+-saturated, then I has the disjointing property.
Lemma: Suppose A\in I^+ and I has the disjointing property, and [A]\Vdash \dot{\tau} \in V^Z/\dot{G}. Then there is f:Z\rightarrow V such that A\Vdash \tau = [f]_G.
Proof: Exercise.
Lemma: Suppose I is a countably closed ideal with the disjointing property. Let \kappa=\mathrm{crit}(j). Then I is precipitous, and the generic ultrapower M is closed under \kappa-sequences from V[G].
Proof: Precipitousness follows from the combinatorial characterization of precipitousness. Suppose that \Vdash \langle \tau_\alpha:\alpha<\kappa\rangle \subseteq V^Z/G. By the previous lemma, for all \alpha<\kappa, there is f_\alpha:Z\rightarrow V in V such that \Vdash \tau_\alpha=[f_\alpha]_G and k:Z\rightarrow \mathrm{ON} in V such that \Vdash \kappa=[k]_G. Finally, take f:Z\rightarrow V given by f(z)=\langle f_\alpha(z):\alpha<k(z)\rangle. In M, [f]_G=\langle \tau_\alpha:\alpha<\kappa\rangle. \Box
Definition: I is (\lambda,\kappa)-presaturated if for any sequence of antichains \langle A_\alpha:\alpha<\gamma<\lambda\rangle, the set \{X:\forall\alpha<\gamma\, |\{a\in A_\alpha:a\cap x\in I^+\}|<\kappa\} is dense.
Exercise: (\lambda^+, \kappa^+)-presaturation and \kappa-completeness imply that the generic ultrapower M is closed under \lambda-sequences from V[G]. Also show this for normal and (\lambda^+,|X|^+)-presaturated ideals.
We will use the notation \mathcal{B}(\mathbb{P}) to denote Boolean completion of a poset. As we have seen in previous posts, this is very handy when dealing with generic ultrapowers.
Exercise: \mathbb{P} and \mathbb{Q} are separative posets. The following are equivalent:
- \mathcal{B}(\mathbb{P})=\mathcal{B}(\mathbb{Q}).
- There is a \mathbb{P}-name \dot{h} and a \mathbb{Q}-name \dot{g} so that \Vdash_\mathbb{P} \dot{h} \textrm{ is }\mathbb{Q}-\textrm{generic over }V, \Vdash_\mathbb{Q} \dot{g} \textrm{ is }\mathbb{P}-\textrm{generic over }V, and\Vdash_\mathbb{P} \dot{g}^{\dot{h}^\dot{G}}=\dot{G} \textrm{ and } \Vdash_\mathbb{Q} \dot{h}^{\dot{g}^\dot{H}}=\dot{H}.
Tuesday, June 16, 2015
Tight stationarity and tree-like scales
Friday, April 17, 2015
Spencinar: Morley's categoricity theorem, part 1
Unfortunately, I missed posting about some interesting lectures given at the Very Informal Gathering, and the Southern California logic meeting at Caltech. We'll return with this series of posts following Spencer Unger's talks at the first two seminars of the quarter. There may be a larger amount of errors than usual in my transcription because of my limited knowledge of model theory. Basic model theory will be helpful to follow these notes, but I'll do my best to fill in the definitions. The notation follows Marker's text closely.
A general question is:
Given a first order theory T, what can we say about the number of non-isomorphic models of T of size \lambda, as \lambda varies through the cardinals?
This is addressed by Shelah's classification theory. Historically, the first step was Morley's categoricity theorem (1965) which states:
A general question is:
Given a first order theory T, what can we say about the number of non-isomorphic models of T of size \lambda, as \lambda varies through the cardinals?
This is addressed by Shelah's classification theory. Historically, the first step was Morley's categoricity theorem (1965) which states:
- If T is a first-order theory in a countable language and T is categorical in some uncountable cardinal, then it is categorical in all uncountable cardinals.
The proof of this theorem, which was the content of the lectures, involves the ideas of stability, indiscernibles, saturated models, and splitting types. It's somewhat different from the one I recall from the basic model theory class I took many years ago.
To begin with, we can define a rank R on sets of formulas (i.e., types) with parameters in some large saturated model \mathcal{C} (the "monster model"). Let p be a set of formulas. Then, working in \mathcal{C},
- R(p)=-1 if p is not satisfiable.
- R(p)\ge 0 if p is satisfiable.
- R(p)\ge \alpha+1 if for any finite p_0\subseteq p, there is a formula \psi such that R(p_0\cup\{\psi\})\ge \alpha and R(p\cup\{\neg \psi\})\ge \alpha.
The rank R(p) is then defined inductively to be the minimum value so that all of the above conditions are satisfied. Note: this isn't what is known as Morley rank. Andrew Marks noticed that this rank is the Cantor-Bendixson rank on the Stone space of complete types.
Some properties of the rank:
- (Monotonicity) If p\vdash q then R(p)\le R(q).
- (Finite character) For every p, there is a finite p_0\subseteq p so that R(p)=R(p_0).
- Proof: If R(p)=\alpha, there is a finite p_0 witnessing R(p)\not\ge \alpha+1, and monotonicity gives the reverse inequality.
- (Invariance) If f is an automorphism of \mathcal{C}, then R(p)=R(f(p)).
- Proof is by induction on the rank.
Definition: T is \lambda-stable if for any A\subseteq C of size \lambda, |S^n(A)|\le \lambda. Here S^n(A) is the set of complete n-types over A, that is, maximal satisfiable sets of formulas with parameters over A and free variables x_1,\ldots, x_n.
Theorem: Let T be a theory in a countable language. Then T is \aleph_0-stable iff R(\{\bar{x}=\bar{x}\})<\infty, i.e., has some ordinal value. In fact, if these equivalent conditions hold, then R(\{\bar{x}=\bar{x}\})<\omega_1.
Proof: Suppose R(\{\bar{x}=\bar{x}\})\ge \omega_1. We will inductively construct a complete binary tree whose nodes are finite sets of formulas p_\eta for \eta\in 2^{<\omega} and whose branches will give 2^{\aleph_0} many different complete types, contradicting stability. For the construction, we will maintain that R(p_\eta)\ge\omega_1.
Let p_\emptyset=\{\bar{x}=\bar{x}\}. Given p_\eta, there are formulas \psi_{\eta,i}(\bar{x},\bar{a}_{\eta,i}) for each i<\omega_1 witnessing that R(p_\eta)\ge i+1. Using the fact that the language of T is countable together with \aleph_0-stability, there are \psi and t so that for an unbounded set of i<\omega_1, \psi=\psi_{\eta,i} and t=\mathrm{tp}(\bar{a}_{\eta,i}) (the set of all formulas satisfied by \bar{a}_{\eta,i}, over the finitely many parameters in p_\eta). Pick a_\eta realizing t. Then define
- p_{\eta\circ 0}=p_\eta\cup\{\psi(\bar{x},\bar{a}_\eta)\}
- p_{\eta\circ 1}=p_\eta\cup\{\neg\psi(\bar{x},\bar{a}_\eta)\}
Using invariance of rank under automorphisms and the fact that in \mathcal{C} for any two tuples realizing the same type there exists an automorphism sending one to the other, the ranks of these sets of formulas are at least i for an unbounded set of i<\omega_1, so they must be \ge\omega_1, allowing the construction to continue. The branches of the tree we construct can be extended to complete types, and the construction ensures that each of these complete types is distinct (for two different branches, look at the first place of disagreement).
For the converse, assume R(\bar{x}=\bar{x})<\infty and T is not stable. Then there are A\subseteq \mathcal{C}, |A|\le \lambda such that there are distinct \{p_i:i<\lambda^+\}\subseteq S^n(A). By assumption, for any i<\lambda^+, there is \alpha_i so that R(p_i)=\alpha_i. By finite character, there is q_i\subseteq p_i finite with R(q_i)=R(p_i)=\alpha_i. By counting, |\{q_i:i<\lambda^+\}|\le \omega. Let i\neq j be such that q_i=q_j, so p_i,p_j are different completions of q_i=q_j. This means there is a formula \psi with \neg\psi\in p_i and \psi\in p_j. This gives R(q_j)\ge R(p_j)+1, a contradiction to the choice of q_j. \Box.
Monday, January 26, 2015
Spencinar: Forcing clubs in stationary subsets of P_\kappa(\lambda)
The first part of this talk follows closely the paper "Forcing closed unbounded sets" by Uri Abraham and Saharon Shelah ([AS]). The second part will follow the paper "Nonsplitting subset of P_\kappa(\kappa^+) by Moti Gitik ([G]).
This generalizes to higher regular cardinals \kappa assuming some cardinal arithmetic (GCH with \kappa the successor of a regular cardinal suffices) and fatness of S, which says that for any club E\subseteq \kappa, S\cap E contains closed subsets of arbitrarily large order type <\kappa (this is a result of J. Stavi).
The focus of this talk will be to obtain analogues for these theorems for subsets of P_\kappa(\lambda)=\{x\subseteq \lambda: |x|<\kappa\}.
This situation is not as clear as for subsets of ordinals, and in fact there are some sets whose stationarity is quite absolute (By stationary and club I mean here in the sense of Jech, see this previous post). For example, Theorem 6 in [AS] gives a stationary subset of P_{\aleph_1}(\omega_2) whose stationarity is preserved by any forcing which preserves \aleph_2:
Theorem: Let W\subseteq V be a transitive inner model such that \omega_2^W=\omega_2^V, and let S=(P_{\aleph_1}(\omega_2))^W. Then S is stationary in V.
Proof: The idea of the proof is to fix an arbitrary club C\subset P_{\aleph_1}(\omega_2) and find submodels which are ordinals, or somehow coded by ordinals in W. Then these submodels will automatically be in S.
By Kueker's theorem, there is a function F:[\omega_2]^{<\omega}\rightarrow \omega_2 so that any x\in P_{\aleph_1}(\omega_2) closed under F is in C (this uses the fact that we're working on \aleph_1 and \aleph_2, but this is a minor technical point). Let \alpha be an ordinal closed under F. If \alpha is countable, then we're done; otherwise \alpha has cardinality \aleph_1, hence also W-cardinality \aleph_1, so fix a bijection h:\omega_1\rightarrow \alpha in W. Then there is \xi<\omega_1 with h[\xi] closed under F, and h[\xi] is in W and hence in S, so we're done again. \Box
In [G], it's shown that any forcing which adds reals to W destroys the club-ness of S.
This paper gives some examples of sets whose stationary can be destroyed. We won't go through those results here. Instead, we now turn towards a result in [G] which also gives examples of this kind.
Let \kappa be supercompact in W, and let V be obtained from W by Radin forcing. We won't need to know much about Radin forcing, just:
- the continuum function and all cardinalities are preserved,
- there is a club C\subseteq \kappa of W-inaccessible cardinals,
- \kappa remains (sufficiently) supercompact in V.
From now on, work in V. Let A be any subset of the set of all t\in P_\kappa(\kappa^+)\cap W such that V\vDash ``|t| \textrm{ is a successor cardinal}".
We will find a forcing that adds a club in S:=P_\kappa(\kappa^+)\setminus A which is <\kappa-distributive (this is crucial, so that P_\kappa(\kappa^+) itself does not change), and has the \kappa^+-c.c.
Define the poset \mathbb{P} to be the collection of subsets of S of size <\kappa which have a maximum element and are closed under increasing unions, ordered by end-extension.
Claim: \mathbb{P} is <\kappa-distributive.
Let \langle D_\beta: \beta<\alpha\rangle be a sequence of dense open subsets of \mathbb{P} for some \alpha<\kappa. Then we will prove that \bigcap_{\beta<\alpha} D_\beta is also dense. So let x\in \mathbb{P}.
Let M\prec (H(\theta),\in,\triangleleft,A,\langle D_\beta\rangle,x,\ldots) be an elementary substructure of size \kappa so that M\cap \kappa^+\in \kappa^+ and M is closed under <\kappa-sequences. Let g\in W be a bijection from \kappa onto M\cap \kappa^+. Using <\kappa-closure of M, construct \langle M_i:i<\kappa\rangle a continuous IA chain of elementary submodels of M of size <\kappa so that \alpha+1\subseteq M_0 and g``i\subseteq M_i. There is a club E\subseteq \kappa such that for every i\in E, i\in C and M_i\cap \kappa^+=g``i. Let \langle i_\beta:\beta<\kappa\rangle be the increasing enumeration of E, and let N_\beta=M_{i_\beta}.
We will inductively construct a decreasing sequence of conditions \langle x_\beta:\beta\le \alpha\rangle such that:
- x_0\le x,
- x_{\beta+1}\in D_\beta,
- N_\beta\cap \kappa^+=\max(x_\beta),
- x_\beta\in N_{\beta+1}.
Take x'_0\in N_0, x'_0\le x, and let x_0=x'_0\cup\{N_0\cap \kappa^+\}.
Now we construct x_\beta, assuming that x_\gamma has been constructed for every \gamma<\beta. If \beta=\gamma+1 for some \gamma, then pick x'_\beta to be the \vartriangleleft-least in N_\beta\cap D_\beta extending x_\gamma, and define x_\beta=x'_\beta\cup\{N_\beta\cap \kappa^+\}. This is a valid condition since N_\beta\cap \kappa^+ has W-cardinality in C, so it can't be a member of A.
If \beta is limit, then let x'_\beta be the closure of \bigcup \{x_{\gamma}:\gamma<\beta\} under increasing unions, and x_\beta=x'_\beta\cup\{N_\beta\cap \kappa^+\}. By internal approachability, x_\beta\in N_{\beta+1}.
It is easy to see that:
Claim: For every \gamma<\beta, x_\beta is an end-extension of x_{\gamma}.
We now check that x_\beta\in \mathbb{P}. If not, then there is t\in x_\beta\cap A. Since t\subseteq N_\beta\cap \kappa^+, we must have |t|^W\le i_\beta. Furthermore, |t|^W is a successor cardinal of W and i_\beta is W-inaccessible, so |t|^W<i_\beta.
Since i_\beta is W-regular, there is some \gamma<\beta with t\subseteq g``i_\gamma=N_\gamma\cap \kappa^+. But this is impossible by the claim.
The \kappa^+-c.c. follows by a standard \Delta-system argument.
Gitik used this forcing as the building block of an iteration to produce a stationary subset Z of P_\kappa(\kappa^+) so that the non-stationary ideal restricted to that stationary set is \kappa^+-saturated, which is very interesting in light of results that show that the whole nonstationary ideal is not saturated. Another way to look at this is that it shows the consistency of the failure of a natural analogue of Solovay's splitting theorem for stationary subsets of P_\kappa(\lambda).
Assuming that \kappa is supercompact, the set Z of all t\in P_\kappa(\kappa^+)\cap W such that W\vDash ``|t| \textrm{ is a successor cardinal}" is stationary. The idea of the iteration is that we will destroy the stationary of the "bad sets" to while maintaining the stationary of Z. Maintaining the stationarity of Z is achieved through extending the supercompactness embedding to the final model, but this is not an easy task in this case.
An earlier version of this post had several occurrences of V which should have been W.
An earlier version of this post had several occurrences of V which should have been W.
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