Saturday, November 30, 2019

The role of markers

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\def\REG{\sf {REG}}
\def\restr{\upharpoonright}
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\def\ran{\rm{ran}}
\DeclareMathOperator{\ch}{Ch}
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\DeclareMathOperator{\Sk}{Sk}
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\)

Definition of $P_\alpha$

Given $\alpha<\kappa$, we define the partial order $P_\alpha$ to be those conditions in $P_\kappa$ that satisfy

  • $w_p\subseteq\alpha$, and
  • $N\cap m_p(N)\subseteq\alpha$ for all $N\in\mathcal{N}_p$
The last condition says that the implicit coordinates arising from $N$ are all less than $\alpha$.  This looks different than what Aspero-Mota do because we're requiring that markers are actually elements of the model.  This is probably only a cosmetic difference; on the face of it, conditions with the property we demand should be dense in the Aspero-Mota poset.

Restrictions of conditions

Give a condition $p\in P_\kappa$ and $\alpha<\kappa$, we define $p\restr\alpha$ by setting
  • $\mathcal{N}_{p\restr\alpha}=\mathcal{N}_p$
  • for $N\in\mathcal{N}_{p\restr\alpha}$ we define $m_{p\restr\alpha}(N)=\min\{m_p(N),\min(N\setminus\alpha)\}$
  • $f_{p\restr\alpha}= f_p\restr \alpha$
Properties of Restrictions of Conditions

Claim 1:  If $p\in P_\kappa$ and $\alpha<\kappa$ then $p\restr\alpha$ is a condition in $P_\alpha$.

First check that $p\restr\alpha$ is actually a condition in $P_\kappa$, and the rest follows immediately from the definition of $p\restr\alpha$. $_\square$

Claim 2:   $p$ is an extension of  $p\restr\alpha$ in $P_\kappa$.

Claim 3:   If $q\leq p\restr\alpha$ in $P_\alpha$ then $q$ is compatible with $p$.

We define $r=\langle \mathcal{N}_r, m_r, f_r\rangle$ by setting

  • $f_r=f_p\cup f_q$
  • $\mathcal{N}_r = \mathcal{N}_q$
  • $m_r(N)=m_p(N)$ for $N\in\mathcal{N}_p$, and $m_r(N)=m_q(N)$ otherwise.
We must check that $r$ is a condition in $P_\kappa$ extending both $q$ and $p$.

Clearly $\mathcal{N}_r$, $m_r$, and $f_r$ all have the right form, so we need to check first that $f_r$ is consistent with the demands imposed by $\mathcal{N}_r$ and $m_r$.  This is immediate for coordinates in $w_r$ below $\alpha$, as $f_r$ and $f_q$ agree.    For coordinates at or above $\alpha$, we know that $f_r$ agrees with $f_p$ and so is consistent with the models in $\mathcal{N}_p$.  The new models that appear in $\mathcal{N}_q$ all have markers at most $\alpha$, and so impose no restrictions on these coordinates.

Note that $r$ extends $q$ (we've potentially added new coordinates, and increased the markers on models from $\mathcal{N}_p$ back to their original values), and also extends $p$ (as $\mathcal{N}_r$ and $f_r$ extend $\mathcal{N}_p$ and $f_p$, while the marker function has remained the same on models from $\mathcal{N}_p$.)

NB.  This is a place where allowing models of the same height to have unrelated markers might be relevant in the more general construction

Claim 4:  $P_\alpha$ is a regular suborder of $P_\kappa$.

If $\mathcal{A}$ is a maximal antichain in $P_\alpha$ and $p\in P_\kappa$, then $p$ is compatible with an element of $\mathcal{A}$ using the preceding argument, as we simply let $q$ be an extension of $p\restr\alpha$ in $\mathcal{A}$..

Friday, November 15, 2019

Baumgartner + Aspero-Mota III (defunct)

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\DeclareMathOperator{\ch}{Ch}
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\DeclareMathOperator{\supp}{supp}
\)

Let $\delta^*<\delta$ be greater than the height of any model in $\mathcal{N}_r$ and greater than anything appearing in the working part of $r$.

Claim: There is a $\beta\in M\cap\kappa$ such that $\sup(M\cap N\cap\kappa)<\beta$ whenever $N$ is a dangerous model.  (proved in email to Justin)

Note that $p$ is an extension of $r$ in $D$ with the property that

(a) $w_p\cap \beta = w_r$

(b) the working part of $p$ restricted to rows below $\delta^*$ and columns below $\beta$ is just the working part of $r$

(c) models in $\mathcal{N}_p$ of height less than $\delta^*$ are isomorphic to models already in $\mathcal{N}_r$.

So by elementarity, $M$ can see an extension $q$ of $r$ in $D$ with the above properties ($r$, $\delta^*$, and $\beta$ are in $M$!)

We claim that $q$ is compatible with $p$, by showing that the dangerous models and coordinates are rendered harmless.

(1) Suppose $N'$ is a dangerous model.  We know that the demands made by $N'$ are compatible with the working part of $r$ because they both came from the condition $p$.  Anything added by the working part of $q$ either lies above row $\delta^*$ or beyond column $\beta$, and so cannot possibly interfere with $N'$'s demands.  $M$ doesn't need to know what's going on with these coordinates at all, because we chose $q$ in such a way that our explicit commitments avoid them)

(2) When we copy out models from $\mathcal{N}_q$ because of the symmetry constraints, we attach the marker $0$ to these new models, so the copies have no explicit coordinates at all.  So the dangerous coordinates become irrelevant because of our use of markers.

Yes, it needs details.











Baumgartner + Aspero-Mota II (defunct)

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\)

Let's look at proving that $P_\kappa$ is proper.  Suppose $M^*$ is a countable elementary submodel of $H(\chi)$ for some sufficiently large regular $\chi$ with $\mathcal{P}_\kappa\in M^*$, and let $M$ be the trace of $M^*$ in $H(\kappa^+)$.

Suppose $p\in \mathcal{P}_\kappa$ with $M\in\mathcal{N}_p$.  We want to show that $p$ is $(M, P_\kappa$-generic.

Let $D\in M$ be dense and open in $P_\kappa$, and assume that $p\in D$.  We produce a $q\in M\cap D$ that is compatible with $p$

We reflect $p$ to a condition $r\in M\cap P_\kappa$ as follows:


  • $\mathcal{N}_r = \mathcal{N}_p\cap M$
  • $m_r$ is the restriction of $m_p$ to $\mathcal{N}_r$
  • $f_r$ is $f_p$ restricted to $M\cap \kappa$
Note that $r$ is in fact a condition in $M\cap P_\kappa$, and $p$ is an extension of $r$ in $D$.

The parts of $p$ that are exterior to $M$ are potentially of three types:
Type 1) There are the models in $\mathcal{N}_p$ that are of the height of $M$ or above 
Type 2) There are models in $\mathcal{N}_p$ that are of height below $M$ that are not in $M$ (in this case, they will be isomorphic to a model in $M$ via an isomorphism that extends to a copy of $M$ in $\mathcal{N}_p$)
Type 3) There are explicit coordinates of $p$ that lie outside of $M\cap\kappa$.

So suppose $q\leq r$ in $M\cap P_\kappa$.  Why might $q$ be incompatible with $p$?

The models of Type 1 should not matter here, as any requirements on implicit coordinates are imposed beyond anything available in $M$.

However:


  • A model of  of Type 2 is dangerous, because it could have implicit coordinates in $M\cap\kappa$ which $q$ explicitly violates. 
  • A type 3 explicit coordinate could be dangerous, as maybe $q$ added a model to $\mathcal{N}_r$ that looks fine inside of $M$, but when we copy it out for reasons of symmetry, the copy could cover the coordinate, and demand something incompatible with what we've already committed to
Let us call a model of Type 2 a dangerous model and a coordinate of Type 3 a dangerous coordinate.

We will find the $q$ we want using intricate combinatorics and insightful analysis to avoid danger.





Baumgartner+ Aspero-Mota I (edited)

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\)

We attempt to implement the Aspero-Mota technology in the context of adding Baumgartner clubs.

Let $\kappa$ be a regular cardinal.  A condition $p$ in $P_\kappa$ is a triple $(\mathcal{N}_p, m_p, f_p)$ where

  • $\mathcal{N}_p$ is a finite symmetric system of elementary submodels of $H(\omega_2)$
  • $m_p:\mathcal{N}_p\rightarrow\kappa+1$ such that $m_p(N)\in N\cap\kappa+1$
  • $f_p$ is a finite partial function with domain $w_p$ such that
    • $f_p(\alpha)$ is a condition $p_\alpha$ in the Baumgartner forcing, and
    • for $N\in \mathcal{N}_p$ and $\alpha\in w_p$, if $\alpha\in N\cap\kappa$ and $\alpha<w_p(N)$ then $p_\alpha(\delta)$ is defined and equal to $\delta$.  

$q\leq p$ means

  • $\mathcal{N}_q\leq \mathcal{N}_p$ as symmetric systems,
  • $m_p(N)\leq m_q(N)$ for $N\in\mathcal{N}_p$, and
  • $f_q\supseteq f_p$
We emphasize that just extending each component as above will not in general result in a condition, as the three pieces all interact.

The function $m_p$ is referred to as the marker function, and $m_p(N)$ is the marker of $N$ in $p$.

The function $f_p$ is called the working part of $p$.

The set $w_p$ is referred to as the the set of explicit coordinates of $p$

An ordinal $\alpha<\kappa$ is an implicit coordinate of $p$ if there is an $N\cap\mathcal{N}_p$ such that
  • $\alpha\in N\cap\kappa$, and
  • $\alpha< m_p(N)$.
We say that $N$ is a witness for the coordinate $\alpha$ in $p$.

Notice that in general the collection of implicit coordinates of $p$ may be infinite, but the collection of explicit coordinates is always finite.

If $\alpha$ is both an implicit and explicit coordinate of $p$ then $p_\alpha(\delta)=\delta$ whenever $N$ is a witness for $\alpha$ in $p$ and $\delta=N\cap\omega_1$.


The intent is that a model in $N$ in $\mathcal{N}_p$ imposes constraints on $p$ in all coordinates that $N$ can see below its marker, forcing that on those coordinates $N\cap\omega_1$ is a fixed point of the Baumgartner club being added.

For future reference:  We could shift the restrictions imposed by $N$ to the notion of extension, rather than putting it into the definition of condition.  In that case, $N$ would enforce the constraints on all further extensions.  That might make it easier in the end (as we can add to the side condition without worrying about whether we're consistent with the explicit part) but I'll run with what I've got above.










Sunday, October 06, 2019

Many Baumgartner Clubs V

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\def\ran{\rm{ran}}
\DeclareMathOperator{\ch}{Ch}
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\DeclareMathOperator{\Sk}{Sk}
\DeclareMathOperator{\supp}{supp}
\)


So suppose $r$ extends $q|_M$ in $M\cap\mathbb{P}$.  We show that $r$ and $q$ are compatible by explicitly building  a condition $s\in\mathbb{P}$.  The main point is showing that $r(\alpha)$ and $q(\alpha)$ are compatible for each $\alpha<\kappa$, and we'll use the usual argument for Baumgartner's forcing as $q(\alpha)(\delta)=\delta$ for all $\delta<\kappa$.

Let $S=\supp(r)\cup\supp(q)$ (a finite set).  For $\alpha\in\kappa\setminus S $ we know $r(\alpha)=C(r)$ and $q(\alpha)=C(q)$, and $C(r)$ extends $C(q)|_M$ in $M\cap\mathbb{B}$.  It follows that $r(\alpha)$ and $q(\alpha)$ are compatible by the usual argument for $\mathbb{B}$ -- in fact, $r(\alpha)\cup q(\alpha)= C(r)\cup C(q)$ can serve as such a witness uniformly for all such $\alpha$. (Thus $C(s)$ will be $C(r)\cup C(q)$.)

If $\alpha\in M\cap\kappa$ then $r(\alpha)$ extends $q(\alpha)|_M=q|_M(\alpha)$ in $M\cap \mathbb{B}$ we can find a suitable $s(\alpha)$ just as above.

So the only places where a different argument is required are those finitely many $\alpha\in\supp(q)$ that are not in $M$.

Given such an $\alpha$, we know that $q$ is $M$-suitable and so there is a $\beta\in M\cap\kappa$ such that $q(\alpha)=q(\beta)$.  Since $\alpha\notin M$ we know $r(\alpha)=C(r)$.

Thus, it suffices to prove that $C(r)$ and $q(\beta)$ are compatible in $\mathbb{B}$.  Since $\beta\in M$, we know $r(\beta)$ and $q(\beta)$ are compatible in $\mathbb{B}$ and $r(\beta)\leq C(r)$ by definition of $\mathbb{P}$. $_\square$.






Many Baumgartner Clubs IV

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\def\cf{\rm{cf}}
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\DeclareMathOperator{\ch}{Ch}
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\)

So let $q$ be an $M$-suitable extension of $p^*$.  What do we mean by $q|_M$?

Let's try defining $q|_M$ in the obvious way, by setting

$$q|_M(\alpha)=
\begin{cases}
q(\alpha)|_M &\text{if $\alpha\in \supp(q)\cap M\cap\kappa$, and }\\
C(q)|_M &\text{otherwise.}
\end{cases}
$$

So we reflect $q$ into $M$ componentwise, except in the places in $\supp(q)$ that lie outside of $M$, where we just default to the reflection of the core.  The use of $M$-suitability is to ensure that this little move doesn't come back to haunt us later.

Is $q|_M\in M$?   

Certainly each component is in $M\cap\mathbb{B}$ because we extend $p^*$.  Since $\supp(q)\cap M\cap\kappa$ is finite, we know it is in $M$ as well. Thus, we can define $q|_M$ inside of $M$.

Is $q|_M\in\mathbb{P}$?  

Each component is in $\mathbb{B}$, so we need to establish the existence of a core, and that each component extends that core.  This is easy, right?  The core will be $C(q)|_M$ and reflection into $M$ preserves extension in $\mathbb{B}$.


So the point now is to prove that any extension $r$ of $q|_M$ in $M\cap\mathbb{P}$ is compatible with the original condition $q$.

Many Baumgartner Clubs III

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Claim:  $p^*$ is $(M, P)$-generic.

Recall that given a condition $\sigma\in\mathbb{B}$, we can reflect $\sigma$ into the model $M$ by setting $\sigma|_M=\sigma\cap M$.   If $\sigma(\delta)=\delta$, then $\sigma|_M$ is a condition in $M\cap \mathbb{B}$, and any extension of $\sigma|_M$ in $M\cap\mathbb{B}$ is compatible with the original condition $\sigma$.  Our goal is to show that the set of conditions in in $\mathbb{P}$ that have ``useful reflections in $M$'' is dense below $p^*$, and then use this to prove the claim.


Definition
A condition $q\leq p^*$ in $\mathbb{P}$ is $M$-suitable if $(\forall\alpha<\kappa)(\exists\beta\in M\cap\kappa)[q(\alpha)|_M=q(\beta)|_M]$

Informally, $q$ is $M$-suitable if every ``column'' of $q$ (a condition $\mathbb{B}$) is identical to a column of $q$ that $M$ can see.

Lemma
The $M$-suitable conditions are dense below $p^*$ in $\mathbb{P}$.
proof.  Let $q$ be an extension of $p^*$.  The main point is that $M\cap\kappa$ is infinite while $\supp(q)$ is finite.

Because we extend $p^*$, we know that $q(\alpha)|_M$ is in $M\cap\mathbb{B}$ for each $\alpha<\kappa$.  Clearly ordinals $\alpha\in M\cap\kappa$ are irrelevant to our purpose.
Since $M\cap \kappa$ must contain an ordinal $\beta$ outside of $\supp(q)$, and so for any $\alpha<\kappa$ outside of $\supp(q)$ we know
$$q(\alpha)|_M = q(\beta)|_M$$,
and so we need only worry about ordinals $\alpha\in \supp(p)\setminus M$ for which no suitable $\beta\in M\cap\kappa$ can be found.

Given such an $\alpha$, we simply choose $\beta\in M\cap\kappa\setminus\supp(q)$, and replace $q(\beta)$ by $q(\alpha)$.

This is fine, as $q(\beta)=C(q)$ and $q(\alpha)\leq C(q)$.  Now we repeat for each problematic $\alpha$ to generate an $M$-suitable extension of $q$.  $_\square$

Really, we could have maybe defined $M$-suitable to be:

$$(\forall\alpha<\kappa)(\exists \beta\in M\cap\kappa)[q(\alpha)=q(\beta)]$$

and I think everything remains OK.









Many Baumgartner Clubs II

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\def\restr{\upharpoonright}
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\DeclareMathOperator{\ch}{Ch}
\DeclareMathOperator{\PP}{PP}
\DeclareMathOperator{\Sk}{Sk}
\DeclareMathOperator{\supp}{supp}
\)

Towards properness:

Let $M$ be a countable elementary submodel of $H(\chi)$ with $\mathbb{P}\in M$, and suppose $p\in M\cap \mathbb{P}$.

Note that $C(p)\in M$ and $\supp(p)\in M$, and so $p(\alpha)\in M\cap\mathbb{B}$ for all $\alpha<\kappa$ (outside of $M\cap\kappa$, these are all equal to $C(p)\in M$)

Let $\delta= M\cap\omega_1$, and define $p^*$ with domain $\kappa$ by setting $p^*(\alpha)=p(\alpha)\cup\{\langle\delta,\delta\rangle\}$.

By the usual Baumgartner argument, $p^*(\alpha)$ is an extension of $p(\alpha)$ in $\mathbb{B}$ for each $\alpha<\kappa$.

For $\alpha$ outside of $\supp(p)$, we have $p^*(\alpha)=C(p)\cup\{\langle\delta,\delta\rangle\}\in\mathbb{B}$.

For $\alpha\in \supp(p)$, we have

$p^*(\alpha)=p(\alpha)\cup\{\langle\delta,\delta\rangle\}\leq C(p)\cup\{\langle\delta,\delta\rangle\}$

So $p^*$ is in $\mathbb{P}$ by taking $\supp(p^*)=\supp(p)$ and $C(p^*)=C(p)\cup\{\langle\delta,\delta\rangle\}$.

Finally, $p^*$ is an extension of $p$ by construction.

Of course, we want to prove that $p^*$ is $(M, P)$-generic.







An attempt at many Baumgartner Clubs I

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\def\cf{\rm{cf}}
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\DeclareMathOperator{\ch}{Ch}
\DeclareMathOperator{\PP}{PP}
\DeclareMathOperator{\Sk}{Sk}
\DeclareMathOperator{\supp}{supp}
\)

I'm just taking a stab at the Aspero-Mota technology in a very simplified setting.  Aspero has a paper called "Adding Many Baumgartner Clubs", and I think what I do here is an even more simplified version of what he has there.  We'll see if it works.

Let $\mathbb{B}$ denote Baumgartner's forcing for adding a club of $\omega_1$ with finite conditions.

A condition $\sigma\in\mathbb{B}$ is a finite partial function from $\omega_1$ to $\omega_1$ that can be extended to a strictly increasing and continuous total function from $\omega_1$ to $\omega_1$.

Let $\kappa$ be a (regular?) cardinal.  We will consider a certain subset $\mathbb{P}$ of the product $\mathbb{B}^\kappa$.

We say that $p\in\mathbb{P}$ if

  1. $p(\alpha)\in\mathbb{B}$ for all $\alpha<\kappa$
  2. There is a condition $C(p)\in\mathbb{B}$ (the core of $p$) and a finite subset $\supp(p)$ (the support of $p$) of $\kappa$ such that
    • $p(\alpha)=C(p)$ for all $\alpha\in \kappa\setminus\supp(p)$, and
    • $p(\alpha)\leq C(p)$ for all $\alpha\leq\kappa$

We will attempt to prove that $\mathbb{P}$ is proper.







Tuesday, July 23, 2019

A Curiosity (Part IV)

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\DeclareMathOperator{\Sk}{Sk}
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We'll finish off the proof of our "toy" theorem by splitting into two (non-mutually exclusive) cases depending on the structure of the set $X$ from Part III.


Case 1:  $X$ contains unboundedly many $\lambda$ with $\cf(\lambda)>\cf(\mu)$.

For $\langle \lambda_\alpha:\alpha<\cf(\mu)\rangle$ be an increasing sequence of cardinals in $X$ such that
$$\cf(\mu)<\cf(\lambda)\leq\kappa<\lambda_\alpha.$$


For each $\alpha<\cf(\mu)$, we can find a set $A_\alpha$ of regular cardinals cofinal in $\lambda_\alpha$ such that
$$\tau=\tcf(\prod A/ J^{\bd}[A])$$
as $\cf(\lambda_\alpha)>\cf(\mu)\geq\aleph_0$ and $\lambda_\alpha$ eventually $\pp_\kappa$-closed.

If we let $A=\bigcup_{\alpha<\cf(\mu)}A_\alpha$ and $J$ the ideal on $A$ defined by

$$Y\in J\Longleftrightarrow Y\cap A_\alpha\text{ is bounded in }A_\alpha\text{ for all sufficiently large }\alpha$$

then $\tau=\tcf(\prod A/ J)$ (see Claim 1.10 on page 12 of Cardinal Arithmetic) and since $J$ is $\cf(\mu)$-complete, we have what we need.




Case 2:  $X$ contains unboundedly many $\lambda$ with $\cf(\lambda)\leq\cf(\mu)$.

We run a similar argument, except this time we can arrange $|A_\alpha|=\cf(\lambda_\alpha)\leq\cf(\mu)$, so in the end  $|A|=\cf(\mu)$.  This would put $\tau\in\PP(\mu)$, and we have a contradicted our assumption on $\tau$.


A Curiosity (Part III)

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\DeclareMathOperator{\ch}{Ch}
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\DeclareMathOperator{\Sk}{Sk}
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We're going to need a couple of basic facts about pseudopowers.  The first of these is from page 57 of Cardinal Arithmetic:

Inverse Monotonicity
If $\lambda<\mu$ are singulars of cofinality $\leq\kappa$ (with $\kappa<\lambda$) and $\mu\leq\pp_\kappa(\lambda)$ then
$$\PP_\kappa(\mu)\subseteq\PP_\kappa(\lambda).$$
In particular,
$$\pp_\kappa(\mu)\leq\pp_\kappa(\lambda).$$


Definition
Let $\mu$ be singular, and $\kappa<\mu$.  We say $\mu$ is eventually $\pp_\kappa$-closed if for all sufficiently large $\lambda<\mu$,
$$\cf(\lambda)\leq\kappa<\lambda\Longrightarrow \pp_\kappa(\lambda)<\mu.$$

The other basic fact about pseudopowers that we'll need is one of the main results of Chapter VIII in Cardinal Arithmetic:

Theorem (See Corollary 1.6 on page 321 of Cardinal Arithmetic)
Suppose $\cf(\mu)\leq\kappa<\mu$ and $\mu$ is eventually $\pp_\kappa$-closed.  Then
$$\PP_\kappa(\mu)=\PP(\mu).$$
Moreover, if $\mu$ is of uncountable cofinality, then and $\tau\in\PP_\kappa(\mu)$, then there is a set $A$ of regular cardinals cofinal in $\mu$ such that

  • $|A|=\cf(\mu)$, and
  • $\tau=\tcf(\prod A/ J^{\bd}[A]) $
where $J^{\bd}[A]$ is the ideal of bounded subsets of $A$. 

Let's look back now at theorem from the last post:

Suppose $\mu$ is singular and $\cf(\mu)\leq\kappa<\mu$ and $\tau\in\PP_\kappa(\mu)$.  If $\tau\notin \PP(\mu)$ then we know $\mu$ cannot be eventually $\pp_\kappa$-closed, that is, the set of singular cardinals $\lambda$ satisfying

  • $\cf(\lambda)\leq\kappa$, and
  • $\mu\leq\pp_\kappa(\lambda)$
is unbounded in $\mu$.

By inverse monotocity, we must have $\PP_\kappa(\mu)\subseteq\PP_\kappa(\lambda)$ for such $\lambda$.

Given $\alpha<\mu$, the least cardinal $\lambda<\mu$  with $\alpha<\lambda$ and $\pp_\kappa(\mu)\leq\pp_\kappa(\lambda)$ is necessarily eventually $\pp_\kappa$-closed, and so $\tau\in\PP(\lambda)$ for each such $\lambda$, witnessed by a $\cf(\lambda)$-complete ideal.

In summary:

If $\mu$ is singular, $\cf(\mu)\leq\kappa<\mu$, and $\tau\in\PP_\kappa(\mu)\setminus\PP(\mu)$, then the set of $\lambda<\mu$ such that 

  • $\cf(\lambda)\leq\kappa$, and
  • $\tau\in\PP(\lambda)$ via a $\cf(\lambda)$-complete ideal
is unbounded in $\mu$.   Let us agree to let $X$ denote the set of such $\lambda$.

to be continued...










Monday, July 22, 2019

A Curiosity (Part II)

\(
\DeclareMathOperator{\pp}{pp}
\DeclareMathOperator{\tcf}{tcf}
\DeclareMathOperator{\pcf}{pcf}
\DeclareMathOperator{\cov}{cov}
\def\cf{\rm{cf}}
\DeclareMathOperator{\otp}{otp}
\def\REG{\sf {REG}}
\def\restr{\upharpoonright}
\def\bd{\rm{bd}}
\def\subs{\subseteq}
\def\cof{\rm{cof}}
\def\ran{\rm{ran}}
\DeclareMathOperator{\ch}{Ch}
\DeclareMathOperator{\PP}{PP}
\DeclareMathOperator{\Sk}{Sk}
\)

In this post, we'll take a look at a simplified situation that doesn't have the same implications for covering numbers as the real theorem we're aiming for.  The proof is not difficult at all, but it does illustrate the main idea we'll be exploiting.  We'll state the proof here, and prove it next time.

Theorem
Suppose $\mu$ is singular, and $\kappa$ is a regular cardinal with $\cf(\mu)\leq\kappa<\mu$.  Then
$$\PP_\kappa(\mu)=\PP(\mu)\cup\PP_{\Gamma(\kappa^+,\cf(\mu)})(\mu).$$

Here,  a cardinal $\tau$ is in $\PP(\mu)$ if there is a cofinal $A\subseteq\mu\cap\REG$ of cardinality $\cf(\mu)$ and an ideal $J$ on $A$ extending the bounded ideal such that the reduced product $\prod A/J$ has true cofinality $\tau$, that is,
$$\tau=\tcf(\prod A/J).$$

We define $\PP_\kappa(\mu)$ similarly, except now we relax things and require $|A|\leq\kappa$.

Finally, membership in $\PP_{\Gamma(\kappa^+,\cf(\mu))}(\mu)$ demands that $|A|\leq\kappa$, but $J$ is required to be $\cf(\mu)$-complete as well.

The theorem says, roughly, that if a cardinal $\tau$ is representable in this fashion using a set $A$ of cardinality at most $\kappa$, then either it is representable using a set of size $\cf(\mu)$ (with no guarantees about the completeness of the ideal used), or through a $\cf(\mu)$-complete ideal on a set of cardinality at most $\kappa$.

In particular, the theorem is vacuously true if $\kappa=\cf(\mu)$, or if $\mu$ has countable cofinality.







Wednesday, July 10, 2019

A curiosity (Part I)

\(
\DeclareMathOperator{\pp}{pp}
\DeclareMathOperator{\tcf}{tcf}
\DeclareMathOperator{\pcf}{pcf}
\DeclareMathOperator{\cov}{cov}
\def\cf{\rm{cf}}
\DeclareMathOperator{\otp}{otp}
\def\REG{\sf {REG}}
\def\restr{\upharpoonright}
\def\bd{\rm{bd}}
\def\subs{\subseteq}
\def\cof{\rm{cof}}
\def\ran{\rm{ran}}
\DeclareMathOperator{\ch}{Ch}
\DeclareMathOperator{\PP}{pp}
\DeclareMathOperator{\Sk}{Sk}
\)

I want to take a few posts to prove the following (overly-dramatic) result:

Theorem (for example)
If $\mu$ has cofinality $\aleph_6$, then
$$\cov(\mu,\mu,\aleph_9,\aleph_3)=\cov(\mu,\mu,\aleph_7,\aleph_3)+\cov(\mu,\mu,\aleph_9,\aleph_6).$$

There is nothing special about the specific integers in the statement, other than that they are non-zero, and the relative ordering is important.   More generally, we have:

Theorem 1
If $\aleph_0<\sigma\leq\cf(\mu)<\theta<\mu$ and
$$(\forall\tau<\cf(\mu))\left[\cf([\tau]^{<\sigma},\subseteq)<\cf(\mu)\right]$$
THEN
$$\cov(\mu,\mu,\theta,\sigma)=\cov(\mu,\mu,(\cf\mu)^+,\sigma)+\cov(\mu,\mu,\theta,\cf\mu)).$$


Comments:

  1. $\cov(\mu,\kappa,\theta,\sigma)$ is defined to be the minimum cardinality of a set $\mathcal{P}\subseteq[\mu]^{<\kappa}$ such that that every element of $[\mu]^{<\theta}$ can be covered by a union of fewer than $\sigma$ elements of $\mathcal{P}$.
  2. So for example, $\cov(\mu,\kappa^+,\kappa^+, 2)$ is just the cofinality of the structure $([\mu]^{\kappa},\subseteq)$.
  3. In general $\cov(\mu,\kappa,\theta, 2)=\cov(\mu,\kappa,\theta,\aleph_0)$.
  4. In the theorem, it is important that $\sigma$ is uncountable, as we will prove this using Shelah's "cov vs. pp" theorem, which requires that hypothesis.

So in the right side of the statement of the theorem,  t the first covering number involves $<\sigma$-covering members of $[\mu]^{\cf(\mu)}$, while the second involves $<\cf(\mu)$-covering members of $[\mu]^{<\theta}$. 


Note as well the following corollary, which implies the "overly-dramatic" theorem we started with.

Corollary:
If $0<m<n<p<\omega$ then for $\mu$ of cofinality $\omega_n$ we have
$$\cov(\mu,\mu,\aleph_p,\aleph_m)=\cov(\mu,\mu,\aleph_{n+1},\aleph_m)+\cov(\mu,\mu,\aleph_p,\aleph_n).$$


This follows from Theorem 1 as the hypotheses needed are satisfied if $\cf(\mu)=\aleph_n$ for some $n<\omega$.

Friday, March 01, 2019

On a result of Gitik and Shelah

\(
\DeclareMathOperator{\pp}{pp}
\DeclareMathOperator{\tcf}{tcf}
\DeclareMathOperator{\pcf}{pcf}
\DeclareMathOperator{\cov}{cov}
\def\cf{\rm{cf}}
\DeclareMathOperator{\otp}{otp}
\def\REG{\sf {REG}}
\def\restr{\upharpoonright}
\def\bd{\rm{bd}}
\def\subs{\subseteq}
\def\cof{\rm{cof}}
\def\ran{\rm{ran}}
\DeclareMathOperator{\ch}{Ch}
\DeclareMathOperator{\PP}{pp}
\DeclareMathOperator{\Sk}{Sk}
\)

Granted, the title of this post is not very illuminating, as there are several theorems due to Gitik and Shelah.  What I have in mind is the following result:

Theorem 1.5 of [GiSh:412]
If $\mu>\kappa$, then the set $\{\cov(\mu,\kappa,\theta,\omega_1):\theta<\kappa\}$ is finite.

The result is in [GiSh:412], and is described as a ``non-GCH version'' of the Hajnal-Shelah theorem stating the set $\{\mu^\theta:2^\theta<\mu\}$ is finite.  It is a lovely illustration of the power of Shelah's cov vs. pp Theorem, and it is the use of this theorem that requires the fourth parameter in cov to be uncountable.

The proof in [GiSh:412] is very short, but it doesn't quite work as written.  We'll take a look at their proof and show it works except in the case where $\kappa$ is singular of uncountable cofinality, and then give an argument reducing this case to the one where $\kappa$ is regular. 

Lemma
If $\mu>\kappa$ and $\kappa$ is not singular of uncountable cofinality, then the set $\{\cov(\mu,\kappa,\theta,\omega_1):\theta<\kappa\}$ is finite.

Proof:

Assume this fails.  By monotonicity properties of cov, we can find an increasing sequence $\langle\theta_n:n<\omega\rangle$ of cardinals such that

$$\cov(\mu,\kappa,\theta_n,\omega_1)<\cov(\mu,\kappa,\theta_{n+1},\omega_1)$$

for each $n<\omega$.  Without loss of generality, we may assume each of these covering numbers is greater than $\mu$.

Since we have $\omega_1$ appearing as the fourth component, we can apply the Cov vs. pp Theorem (Theorem 5.4 of Chapter II of Cardinal Arithmetic), which tells us that if $\cov(\mu,\kappa,\theta,\omega_1)>\mu$, then

$$\cov(\mu,\kappa,\theta,\omega_1)=\sup\{\pp_{\Gamma(\theta,\omega_1)}(\chi): \kappa\leq\chi\leq\mu \text{ and }\cf(\chi)<\theta\}$$

In particular, for each $n<\omega$ we have

$$\mu<\cov(\mu,\kappa,\theta_n,\omega_1)=\sup\{\pp_{\Gamma(\theta_n,\omega_1)}(\chi): \kappa\leq\chi\leq\mu \text{ and }\cf(\chi)<\theta_n\}.$$

Given $n$, let $\chi_n$ be the least cardinal satisfying

  • $\kappa\leq\chi_n\leq\mu$,
  • $\omega_1\leq\cf(\chi_n)\leq\theta_n$, and
  • $\pp_{\Gamma(\theta_n,\omega_1)}(\chi_n)>\mu$.
Notice that $\chi_n$ must exist as $\cov(\mu,\kappa,\theta_n,\omega_1)>\mu$, and furthermore $\kappa<\chi_n$, because we have assumed $\kappa$ is not singular of uncountable cofinality -- this is the point where we use that assumption.

Since the sequence $\langle \chi_n:n<\omega\rangle$ is non-increasing, we can drop a few terms and assume that it is constant, say with value $\chi^*$.  Thus, there is a single cardinal $\chi^*$ such that for each $n<\omega$, $\chi^*$ is the least cardinal  satisfying

  • $\kappa<\chi^*\leq\mu$,
  • $\omega_1\leq \cf\chi^*<\theta_n$, and
  • $\pp_{\Gamma(\theta_n,\omega_1)}(\chi^*)>\mu$.

Now suppose $\kappa<\chi<\chi^*$ and $\omega_1\leq\cf(\chi)<\theta_n$.   Since $\chi^*$ is the least cardinal with the enumerated properties, we know 

$$\pp_{\Gamma(\theta_n,\omega_1)}(\chi)\leq\mu.$$

But we can then conclude

(*) $\pp_{\Gamma(\theta_n,\omega_1)}(\chi)<\chi^*$,

as otherwise Inverse Monotonicity (see part (4) of the No Hole Conclusion 2.3 of Chapter II of Cardinal Arithmetic) forces us to conclude

$$\pp_{\Gamma(\theta_n,\omega_1)}(\chi^*)\leq\pp_{\Gamma(\theta_n,\omega_1)}(\chi),$$

which would contradict our choice of $\chi^*$.


Since we have (*) holding for all relevant $\chi<\chi^*$, we can now apply a generalization of one of the main results of [Sh:371].  The actual theorem we need is Lemma 1.2 of the published version of [Sh:410], which tells us that since (*) holds for all sufficiently large $\chi<\chi^*$, we have

(**) $\pp_{\Gamma(\theta_n,\omega_1)}(\chi^*)=\pp_{\Gamma((\cf \chi^*)^+,\omega_1)}(\chi^*).$

(There are some issues with the proof of Lemma 1.2 in [Sh:410] that I will deal with in a future post, but the result still holds.)

Thus, the sequence $\langle\pp_{\Gamma(\theta_n,\omega_1)}(\chi^*):n<\omega\rangle$ is actually constant with value some $\lambda^*$.  The following claim immediately yields a contradiction:

Claim:  For all $n<\omega$, $\cov(\mu,\kappa,\theta_n,\omega_1)=\lambda^*$.

Proof of claim:
Let $n<\omega$ be given. Recall that by the cov vs. pp Theorem, we know 
$$\cov(\mu,\kappa,\theta_n,\omega_1)=\sup\{\pp_{\Gamma(\theta_n,\omega_1)}(\chi): \kappa\leq\chi\leq\mu \text{ and }\cf(\chi)<\theta_n\}$$
so in particular $$\lambda^*\leq\cov(\mu,\kappa,\theta_n,\omega_1).$$

Let $\chi$ be one of the cardinals used in computing the supremum.  By our assumptions on $\kappa$, we know $\kappa\neq\chi$ so $\kappa<\chi$.   If $\chi<\chi^*$, then by (*) we know

$$\pp_{\Gamma(\theta_n,\omega_1)}(\chi)<\lambda^*$$

On the other hand, if $\chi^*<\chi$ then because $\chi<\mu<\pp_{\Gamma(\theta_n, \omega_1)}(\chi^*)$, we know

$$\pp_{\Gamma(\theta_n,\omega_1)}(\chi)\leq \pp_{\Gamma(\theta_n, \omega_1)}(\chi^*)=\lambda^*$$

by inverse monotonicity.

Thus, $\cov(\mu,\kappa,\theta_n,\omega_1)=\lambda^*$. $_\square$



Now where did we use the assumption that $\kappa$ is not a singular cardinal of uncountable cofinality?   It was used when we needed to know $\kappa<\chi^*$, so that we can conclude (*) holds for all sufficiently large $\chi<\chi^*$ of the right cofinality.

Now how do we deal with this last case?  Let us assume that $\kappa$ is singular of uncountable cofinality, and find the sequence $\langle \theta_n:n<\omega\rangle\rangle$.   For each $n<\omega$ there is a cardinal $\kappa_n<\kappa$ such that

$$\cov(\mu,\kappa,\theta_n, \omega_1)=\cov(\mu,\kappa^*,\theta_n,\omega_1)$$

for all $\kappa^*\geq\kappa_n$ (as the covering numbers are non-increasing as we increase $\kappa^*$ to $\kappa$).

Since $\kappa$ has uncountable cofinality, there is a single $\kappa^*<\kappa$ (which can be taken to be regular) such that

$$\cov(\mu,\kappa,\theta_n,\omega_1)=\cov(\mu,\kappa^*,\theta_n,\omega_1)\text{ for all }n<\omega.$$

Now we generate a contradiction as in the preceding case.  $_\square$