Changing measurable into small accessible cardinals

TL;DR

This paper provides a detailed proof of the properties of Prikry type forcing, demonstrating how it transforms a measurable cardinal into leph_omega, with implications for set theory and large cardinal axioms.

Contribution

It offers a comprehensive proof of Prikry forcing properties specifically for turning a measurable cardinal into leph_omega, clarifying its set-theoretic effects.

Findings

Confirmed the properties of Prikry forcing in this context

Established the transformation of measurable cardinals to leph_omega

Enhanced understanding of forcing techniques for large cardinals

Abstract

We give a detailed proof of the properties of the usual Prikry type forcing notion for turning a measurable cardinal into $\aleph_\omega$.

Equations

$$p=({\delta }_0,f_0\dots ,{\delta }_{n-1},f_{n-1},A,F)$$

$$p=({\delta }_0^p,f_0^p\dots ,{\delta }_{n^p-1}^p,f_{n^p-1}^p,A^p,F^p).$$

$$\{stem(p):p\in A\}\subseteq V_{\kappa }$$

$$A^{\ast }=\{\delta <\kappa :F(\delta )\le F^p(\delta ),F^q(\delta )\}\in \mathcal{U}.$$

$$stem(p{)}^{⌢}(A,F)\in \mathbb{P}$$

$$\mathbb{P}/p\simeq (\prod _{i<m}\mathrm{Col}(({\delta }_i^p{)}^{++},<{\delta }_{i+1}^p))\times \mathbb{P}/p^{\ge m},$$

$$\exists s^{⌢}(A,F)\le (A^1,F^1),s^{⌢}(A,F)\parallel \varphi ⟺s^{⌢}(A^1,F^1)\parallel \varphi .$$

$$B^s=\{\delta \in A^s:F^1(\delta )\le F^s(\delta )\}\in \mathcal{U}.$$

$$A^1={△}_s{B}^s=\{\delta <\kappa :s\in V_{\delta }\Rightarrow \delta \in B^s\}.$$

$$s^{⌢}(A^1,F^1)\le s^{⌢}(A^s,F^s)$$

$$s^{⌢}({\delta }_n,f_n{)}^{⌢}(A,F)\parallel \varphi ,$$

$$s^{⌢}({\delta }_n,F^2({\delta }_n){)}^{⌢}(A^2,F^2)\parallel \varphi .$$

$$s^{⌢}(\kappa ,g{)}^{⌢}(j(A^1),j(F^1))\parallel \varphi ,$$

$$s^{⌢}(\kappa ,f{)}^{⌢}(j(A^1),j(F^1))\parallel \varphi .$$

$$\exists f\le f_{\alpha },s_{\alpha }^{⌢}(\kappa ,f{)}^{⌢}(j(A^1),j(F^1))\parallel \varphi ⟹s_{\alpha }^{⌢}(\kappa ,f_{\alpha +1}{)}^{⌢}(j(A^1),j(F^1))\parallel \varphi .$$

$$s^{⌢}({\delta }_n,F^2({\delta }_n){)}^{⌢}(A^2,F^2)\parallel \varphi .$$

$$A^2={△}_s{A}^s\cap A^{\ast }.$$

$$s^{⌢}({\delta }_n,f_n{)}^{⌢}(A,F)\parallel \varphi ,$$

$$s^{⌢}(\delta ,F^3(\delta ){)}^{⌢}(A^3,F^3)\parallel \varphi .$$

$$s^{⌢}(\delta ,F^3(\delta ){)}^{⌢}(A^3,F^3)⊩\varphi .$$

$$p=s^{⌢}({\delta }_n,f_n{)}^{⌢}(A,F).$$

$$\forall \delta \in A^3,s^{⌢}(\delta ,F^3(\delta ){)}^{⌢}(A^3,F^3)⊩\varphi .$$

$$s^{⌢}(A^3,F^3)⊩\varphi .$$

Full text

The author’s research has been supported by a grant from IPM (No. 97030417).

1. introduction

In this short note, we present a proof of the following known result.

Theorem 1.1**.**

Assume $\operatorname{GCH}$ holds and $\kappa$ is a measurable cardinal. Then there exists a generic extension in which $\kappa=\aleph_{\omega}$.

We try to give the details as much as possible to make it accessible to general audience who has some familiarity with forcing and large cardinals (see [1] for preliminaries).

2. Proof of Theorem 1.1

Suppose that $\operatorname{GCH}$ holds and $\kappa$ is a measurable cardinal. Let $\mathcal{U}$ be a normal measure on $\kappa.$ Let also $j:V\to M\simeq\operatorname{Ult}(V,\mathcal{)}$ be the corresponding ultrapower embedding.

Lemma 2.1**.**

There exists $H\in V$ which a $\operatorname{Col}(\kappa^{++},<j(\kappa))_{M}$-generic filter over $M$.

Proof.

We have

1. (1)

$V\models$“$\operatorname{Col}(\kappa^{++},<j(\kappa))_{M}$ is $\kappa^{+}$-closed” 2. (2)

$V\models$“$|\{A\in M:A$ is a maximal antichain in $\operatorname{Col}(\kappa^{++},<j(\kappa))_{M}\}|\leq\kappa^{+}$”.

Thus we can easily find the required $H$. ∎

We are now ready to define our main forcing construction.

Definition 2.2**.**

A condition in ${\mathbb{P}}$ is of the form

[TABLE]

where

1. (1)

$n<\omega$. 2. (2)

$\delta_{0}<\dots<\delta_{n-1}$. 3. (3)

For $i<n-1,f_{i}\in\operatorname{Col}(\delta_{i}^{++},<\delta_{i+1})$. 4. (4)

$f_{n-1}\in\operatorname{Col}(\delta_{n-1}^{++},<\kappa)$. 5. (5)

$A\in\mathcal{U},$* and $\min(A)>\delta_{n-1}$.* 6. (6)

$F$* is a function with $\operatorname{dom}(F)=A.$* 7. (7)

For every $\delta\in A,F(\delta)\in\operatorname{Col}(\delta^{++},<\kappa)$. 8. (8)

$[F]_{\mathcal{U}}\in H.$**

Given a condition $p\in{\mathbb{P}},$ we denote it by

[TABLE]

We also set

• •

$stem(p)=(\delta^{p}_{0},f^{p}_{0}\dots,\delta^{p}_{n^{p}-1},f^{p}_{n^{p}-1})$, the stem of $p$.

• •

$u(p)=(A^{p},F^{p}),$ the upper part of $p$.

Definition 2.3**.**

Suppose $p,q\in{\mathbb{P}}.$

• (a)

$p$* is an extension of $q$, $p\leq q$, iff*

1. (1)

$n^{p}\geq n^{q}$. 2. (2)

For all $i<n^{q},\delta^{p}_{i}=\delta^{q}_{i}$. 3. (3)

For $n^{q}\leq i<n^{p},\delta^{p}_{i}\in A^{q}$. 4. (4)

For $i<n^{q},f^{p}_{i}\leq f^{q}_{i}$. 5. (5)

For $n^{q}\leq i<n^{p},f^{p}_{i}\leq F^{q}(\delta^{p}_{i})$. 6. (6)

$A^{p}\subseteq A^{q}.$** 7. (7)

For each $\delta\in A^{p},F^{p}(\delta)\leq F^{q}(\delta)$.

• (b)

$p$* is a direct extension of $q$, $p\leq^{*}q$, iff*

1. (1)

$p\leq q.$** 2. (2)

$n^{p}=n^{q}$.

We start by proving the basic properties of the forcing notion ${\mathbb{P}}$.

Lemma 2.4**.**

$({\mathbb{P}},\leq)$* satisfies the $\kappa^{+}$-c.c.*

Proof.

Let $A\subseteq{\mathbb{P}}$ be of size $\kappa^{+}$. Then, as

[TABLE]

has size $\kappa$, we can find $p,q\in A$ such that $stem(p)=stem(q).$ We claim that $p$ an $q$ are compatible. Since $[F^{p}]_{\mathcal{U}},[F^{q}]_{\mathcal{U}}\in H,$ we can find $[F]_{\mathcal{U}}\in H$ such that $[F]_{\mathcal{U}}\leq[F^{p}]_{\mathcal{U}},[F^{q}]_{\mathcal{U}}.$ Thus

[TABLE]

Set $A=A^{\ast}\cap A^{p}\cap A^{q}\in\mathcal{U}.$ Then

[TABLE]

and it extends both of $p$ and $q$. ∎

Lemma 2.5**.**

Suppose $p\in{\mathbb{P}}$ and $m<n^{p}$. Then

[TABLE]

where $p^{\geq m}=(\delta^{p}_{m},f^{p}_{m},\dots,\delta^{p}_{n^{p}-1},f^{p}_{n^{p}-1},A^{p},F^{p})$. Further, $({\mathbb{P}}/p^{\geq m},\leq^{*})$ is $\delta_{m}^{++}$-closed.

We now show that the forcing notion $({\mathbb{P}},\leq,\leq^{*})$ satisfies the Prikry property.

Lemma 2.6**.**

Suppose $q\in{\mathbb{P}}$ and $\phi$ is a statement of the forcing language of $({\mathbb{P}},\leq)$. Then there exists $p\leq^{*}q$ which decides $\phi$.

Proof.

We assume for simplicity that $n^{q}=0$ and $f^{q}_{0}=\emptyset.$ We write $q$ as $q=(A^{q},F^{q})$. The proof has four main steps.

Claim 2.7**.**

(Reduction to stems) There exists $q^{1}=(A^{1},F^{1})\leq^{*}q$ such that for any stem $s$,

[TABLE]

Proof.

For each stem $s$, if there exists $s^{\frown}(A,F)\leq(A^{q},F^{q})$ which decides $\phi,$ then let $(A^{s},F^{s})=(A,F)$ and otherwise set $(A^{s},F^{s})=(A^{q},F^{q})$.

Then $\{[F^{s}]_{\mathcal{U}}:s$ is a stem$\}\subseteq H$, and hence, we can find $[F^{1}]_{\mathcal{U}}\in H$ which extends all of them. For each stem $s$, set

[TABLE]

Let also

[TABLE]

We show that $q^{1}=(A^{1},F^{1})$ is as required. Thus suppose that $s$ is a stem and suppose there exists $s^{\frown}(A,F)\leq(A^{1},F^{1})$ which decides $\phi$.

It then follows that $s^{\frown}(A^{s},F^{s})$ decides $\phi$. But, by our construction,

[TABLE]

and hence $s^{\frown}(A^{1},F^{1})$ decides $\phi$ as well. ∎

Let $q^{1}=(A^{1},F^{1})$ be as in Claim 2.7.

Claim 2.8**.**

(Reduction to stem minus top element) There exists $q^{2}=(A^{2},F^{2})\leq^{*}q^{1}$ such that for any stem $s=(\delta_{0},f_{0},\dots,\delta_{n-1},f_{n-1})$, if $s^{\frown}(\delta_{n},f_{n})^{\frown}(A,F)\leq(A^{2},F^{2})$ and

[TABLE]

then

[TABLE]

Proof.

Let $D$ be the set of all conditions $f\in\operatorname{Col}(\kappa^{++},<j(\kappa))_{M}$ such that for any stem $s\in V_{\kappa},$ if there exists $g\in\operatorname{Col}(\kappa^{++},<j(\kappa))_{M}$ such that

[TABLE]

then

[TABLE]

We claim that $D\subseteq\operatorname{Col}(\kappa^{++},<j(\kappa))_{M}$ is dense. Thus suppose that $g\in\operatorname{Col}(\kappa^{++},<j(\kappa))_{M}.$ Let $(s_{\alpha}:\alpha<\kappa)$ enumerate all stems $s\in V_{\kappa},$ and define a decreasing sequence $(f_{\alpha}:\alpha\leq\kappa)$ of conditions in $\operatorname{Col}(\kappa^{++},<j(\kappa))_{M}$, such that $f_{0}=g$ and for any $\alpha<\kappa,$

[TABLE]

Then $f=f_{\kappa}\in D$ and it extends $g.$

Let $f=[F^{2}]_{\mathcal{U}}$. We may assume that $[F^{2}]_{\mathcal{U}}\leq[F^{1}]_{\mathcal{U}}$, and hence $A^{\ast}=\{\delta<\kappa:F^{2}(\delta)\leq F^{1}(\delta)\}\in\mathcal{U}.$

For any stem $s=(\delta_{0},f_{0},\dots,\delta_{n-1},f_{n-1})$, we have $A^{s}\in\mathcal{U},$ where $A^{s}$ consists of those $\delta_{n}\in A^{1}$ such that if there exists $s^{\frown}(\delta_{n},f_{n})^{\frown}(A,F)\leq s^{\frown}(A^{1},F^{1})$ which decides $\phi$, then

[TABLE]

Let

[TABLE]

Then $q^{2}=(A^{2},F^{2})$ is easily seen to be as requested. ∎

Claim 2.9**.**

(One point extension uniformization) There exists $q^{3}=(A^{3},F^{3})\leq^{*}q^{2}$ such that for any stem $s$, if $s^{\frown}(\delta_{n},f_{n})^{\frown}(A,F)\leq(A^{3},F^{3})$ and

[TABLE]

then for all $\delta\in A^{3},$

[TABLE]

Proof.

Let $s$ be a stem. Set

• •

$A^{s}_{0}=\{\delta\in A^{2}:s^{\frown}(\delta,F^{2}(\delta))^{\frown}(A^{2},F^{2})\Vdash~{}\phi\}$.

• •

$A^{s}_{1}=\{\delta\in A^{2}:s^{\frown}(\delta,F^{2}(\delta))^{\frown}(A^{2},F^{2})\Vdash~{}\neg\phi\}$.

• •

$A^{s}_{2}=\{\delta\in A^{2}:s^{\frown}(\delta,F^{2}(\delta))^{\frown}(A^{2},F^{2})\nparallel\phi\}$.

Let $i_{s}<3$ be such that $A^{s}=A^{s}_{i_{s}}\in\mathcal{U}.$ Let $A^{3}=\bigtriangleup_{s}A^{s}\in\mathcal{U},$ and set $F^{3}=F^{2}\upharpoonright A^{3}.$

We show that $q^{3}=(A^{3},F^{3})$ is as required. Thus suppose that $s^{\frown}(\delta_{n},f_{n})^{\frown}(A,F)\leq(A^{3},F^{3})$ and $s^{\frown}(\delta_{n},f_{n})^{\frown}(A,F)\parallel\phi$. Let us suppose that it forces $\phi.$ Then $\delta_{n}\in A^{s}_{0},$ and hence for all $\delta\in A_{3}$ such that $s^{\frown}(\delta,F^{3}(\delta))^{\frown}(A^{3},F^{3})$ is a condition, we have $\delta\in A^{s}_{0},$ and hence $s^{\frown}(\delta,F^{2}(\delta))^{\frown}(A^{2},F^{2})\Vdash~{}\phi.$ It follows that

[TABLE]

∎

Claim 2.10**.**

(Minimal extension counterexample) There exists $q^{4}=(A^{4},F^{4})\leq^{*}q^{3}$ which decides $\phi.$

Proof.

Suppose not. Let $p\leq q^{3}\leq$ decide $\phi,$ such that $n^{p}$ is minimal. By our assumption, $n^{p}>0$ and hence we can write $p$ as

[TABLE]

By Claim 2.9

[TABLE]

Note that if $p^{\ast}\leq s^{\frown}(A^{3},F^{3})$ and $n^{p^{\ast}}>n^{p}-1$, then for some $\delta\in A^{3},p^{\ast}\leq s^{\frown}(\delta,F^{3}(\delta))^{\frown}(A^{3},F^{3}),$ and hence

[TABLE]

This is in contradiction with the minimal choice of $n^{p}$. ∎

The lemma follows. ∎

Let $G$ be ${\mathbb{P}}$-generic over $V$. Let $C=(\delta_{n}:n<\omega)$ be the added Prikry sequence and for each $n<\omega$ set $G_{n}\subseteq\operatorname{Col}(\delta_{n}^{++},<\delta_{n+1})$ be the generic filter added by $G$.

Lemma 2.11**.**

Suppose $A\in V[G]$ and $A\subseteq\delta_{n}^{+}$. Then $A\in V[\prod_{i<n}G_{i}]$.

Proof.

Let $p\in G$ be such that $n^{p}>n$. Let also \smash{\underset{\raisebox{1.2pt}[0.0pt][0.0pt]{\sim}}{{A}}} be a ${\mathbb{P}}$-name for $A$ such that $\Vdash_{{\mathbb{P}}}$“\smash{\underset{\raisebox{1.2pt}[0.0pt][0.0pt]{\sim}}{{A}}}\subseteq\delta_{n}^{+}”. Let \smash{\underset{\raisebox{1.2pt}[0.0pt][0.0pt]{\sim}}{{B}}} be a ${\mathbb{P}}/p^{\geq n}$-name for a subset of $\prod_{i<n}\operatorname{Col}(\delta_{i}^{++},<\delta_{i+1})\times\delta_{n}^{+}$ such that

$\Vdash_{{\mathbb{P}}/p^{\geq n}}$“\forall\alpha<\delta_{n}^{+}[~{}(q,\alpha)\in\smash{\underset{\raisebox{1.2pt}[0.0pt][0.0pt]{\sim}}{{B}}})\iff q\Vdash_{\prod_{i<n}\operatorname{Col}(\delta_{i}^{++},<\delta_{i+1})}“\alpha\in\smash{\underset{\raisebox{1.2pt}[0.0pt][0.0pt]{\sim}}{{A}}}” ]”.

Let $(x_{\alpha}:\alpha\leq\delta_{n}^{+})$ be an enumeration of $\prod_{i<n}\operatorname{Col}(\delta_{i}^{++},<\delta_{i+1})\times\delta_{n}^{+}$. Define a $\leq^{*}_{{\mathbb{P}}}$-decreasing sequence $(p_{\alpha}:\alpha\leq\delta_{n}^{+})$ of conditions in ${\mathbb{P}}/p^{\geq n}$ such that for each $\alpha<\delta_{n}^{+},p_{\alpha+1}\|$“x_{\alpha}\in\smash{\underset{\raisebox{1.2pt}[0.0pt][0.0pt]{\sim}}{{B}}}”. This is possible as $({\mathbb{P}}/p^{\geq n},\leq^{*})$ is $\delta_{n}^{++}$-closed and satisfies the Prikry property. Then $p_{\delta_{n}^{+}}$ decides each “x_{\alpha}\in\smash{\underset{\raisebox{1.2pt}[0.0pt][0.0pt]{\sim}}{{B}}}”, and so, assuming $p_{\delta_{n}^{+}}\in G,$ we have

$a=\{\alpha<\kappa:\exists q\in\prod_{i<n}G_{i},p_{\delta_{n}^{+}}\Vdash$“(q,\alpha)\in\smash{\underset{\raisebox{1.2pt}[0.0pt][0.0pt]{\sim}}{{B}}}”$\}\in V[\prod_{i<n}G_{i}]$.

The result follows. ∎

The following is now immediate.

Lemma 2.12**.**

1. Suppose $\lambda<\kappa$ is a cardinal.
2. (a)

$V$* and $V[G]$ have the same bounded subsets of $\delta_{0}$.* 3. (b)

$Card^{V[G]}\cap[\delta_{0},\kappa)=\bigcup_{n<\omega}\{\delta_{n},\delta_{n}^{+},\delta_{n}^{++}\}$** 4. (c)

$V[G]\models$“ $\kappa=\delta_{0}^{+\omega}$”.

Now forcing over $V[G]$ by $\operatorname{Col}(\aleph_{0},<\delta_{0}),$ we get a model in which $\kappa$ becomes $\aleph_{\omega}.$ This completes the proof of Theorem 1.1.