Wednesday, December 05, 2018

Section 2 of [Sh:410] Part V - The Final Piece

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

Clearly $N\cap\mu\subseteq N^+\cap\mu$ so assume by way of contradiction the containment is strict, and let

$\gamma(*)=\min(N^+\cap\mu\setminus N).$

If we set $A=N\cap (\omega_1,\mu]\cap\REG$, then our construction guarantees

$\ch_N\restr A= \ch_{N^+}\restr A$.  In particular, if $\mu$ is regular, then

$\sup(N^+\cap\mu)=\sup(N\cap\mu)$.  This also holds if $\mu$ is singular, as either $\cf(\mu)\leq\omega_1$ and $N\cap\mu$ is cofinal in $\mu$, or $\cf(\mu)\in A$ and we use the fact that $\sup(N^+\cap\cf(\mu))=\sup(N\cap\cf(\mu))$.

In either situation, it follows that  $N\cap\mu\setminus\gamma(*)\neq\emptyset$.

Let us define

$$\beta(*)=\min(N\cap\mu\setminus\gamma(*)).$$

Note that the interval $[\beta(*),\gamma(*))$ is (non-empty and) disjoint to $N$.  A simple argument by contradiction tells us that $\gamma(*)$ must be a regular cardinal, but this is impossible:

On the one hand, $\sup(N\cap\gamma(*))=\sup(N^+\cap\gamma(*))$ as $\gamma(*)\in A$, but on the other hand

$\sup(N\cap\gamma(*)\leq\beta(*)<\sup(N^+\cap\gamma(*))$

as $\beta(*)+1\in N^+\cap\gamma(*)$.

Thus, it must be the case that $N\cap\mu=N^+\cap\mu$, and so $X\subseteq N\cap\mu$ as required.

Section 2 of [Sh:410] Part IV - The Proof

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

Moving now to a proof of the theorem from our last post:

Let us assume are as in the same set-up as Part II of this series, that is, 

 $\mu\geq\aleph_\omega$ is a cardinal, and

$\lambda^*=\sup\{\pp(\epsilon):\omega_1<\epsilon\leq\mu\text{ and }\cf(\epsilon)=\aleph_0\}$

Let $\chi$ be a sufficiently large regular cardinal, and let $\langle M_\alpha:\alpha<\cf(\lambda^*)\rangle$ be a tower of elementary submodels of $\mathfrak{A}:=\langle H(\chi), \in, <^*\rangle$  built ``as usual'', and let 

$M^*:=\bigcup_{\alpha<\cf(\lambda^*)}M_\alpha.$

We claim that $\mathcal{P}:= M^*\cap [\mu]^{\aleph_0}$ enjoys the property demanded by the theorem.

To show this, let $X$ be a subset of $\mu$ of cardinality $\aleph_1$.  We are going to run what I like to call "THE argument", as it appears so often in these sorts of results.

We will build a sequence $\langle N_m:m<\omega\rangle$ of elementary submodels of $\mathfrak{A}$ so that the following conditions are satisfied:

  1. $X\subseteq \bigcup_{m<\omega}N_m$, 
  2. $|N_m|=\aleph_1$, and 
  3. every member of $[N_m]^{\aleph_0}$ is covered by a countable set from $M^*$, that is,
$$(\forall Y\in [N_m]^{\aleph_0})(\exists Z\in M^*\cap [M^*]^{\aleph_0})\bigl[ Y\subseteq Z\bigr].$$

This will establish the theorem, as we let $X_m = X\cap N_m$.  In this situation, a countable subset of $X_m$ is covered by a countable set from $M^*$. Without loss of generality this countable set is a subset of $\mu$, and therefore in $\mathcal{P}$.

Our construction will start by setting

$$N_0:=\Sk_{\mathfrak{A}}(\omega_1+1)$$.   Note that condition 3 is satisfied in this situation because $N_0$ is actually an element of $M^*$.


So suppose we are given $N_m$ satisfying conditions (2) and (3) above.  We define objects as follows:

  • $A_m:= N_m\cap (\omega_1,\mu]\cap\REG$
  • $N_m^+:=\Sk_{\mathfrak{A}}(N_m\cup X)$, and
  • $g_m = \ch_{N_m^+}\restr A_m$
We are now set up to apply Lemma 3 from Part I and Lemma 4 from Part II of this series.  Since $|A_m|=\aleph_1$, our assumptions tell us there is a set $\mathcal{B}$ such that 
  • $\mathcal{B}\subseteq M\cap [A_m]^{\aleph_0}$, and
  • $|\mathcal{B}|=\aleph_1$.
By Lemma 3, there is a function $f_m\in\prod A_m$ such that

  • $g_m<f_m$, and
  • $f_m\restr B\in M^*$ for every $B\in\mathcal{B}$.
We define  

$$N_{m+1} := \Sk_{\mathfrak{A}}(N_m\cup \{f_m(\theta):\theta\in A_m\}).$$

Lemma 4 tells us that Condition (3) is maintained for $N_{m+1}$, and the construction continues.

To finish, we define  

$$N=\bigcup_{m<\omega} N_m$$

and

$$N^+=\bigcup_{m<\omega} N_m^+$$.

Clearly $X\subseteq N^+\cap\mu$, so we finish if we know $N^+\cap \mu = N\cap\mu$.

This will follow by a standard argument, but I don't know that there is a crisp formulation of what we need in Shelah's writings.  We'll just show the equality in the next post.





Section 2 of [Sh:410] Part III - The Theorem

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

Our goal is to present a proof of the following theorem.

Theorem
Suppose $\mu\geq\aleph_\omega$ is a cardinal, and let
$$\lambda^*=\sup\{\pp(\epsilon):\omega_1<\epsilon\leq\mu\text{ and }\cf(\epsilon)=\aleph_0\}.$$

Then there is a set $\mathcal{P}\subseteq [\mu]^{\aleph_0}$ of cardinality $\lambda^*$ such that any $X\in[\mu]^{\aleph_1}$ can be written as a countable union of sets, each of whose countable subsets is covered by a set in $\mathcal{P}$, that is, given $X\subseteq \mu$ of cardinality $\aleph_1$, we can find sets $\langle X_n:n<\omega\rangle$ such that


  • $X=\bigcup_{n<\omega}X_n$, and
  • $(\forall Y\in [X_n]^{\aleph_0})(\exists Z\in \mathcal{P})\bigl[Y\subseteq Z\bigr]$.

The use of $\mu\geq\aleph_\omega$ is just to ensure that $\lambda^*$ is not defined as the supremum of the empty set.

This theorem is (a piece of) Claim 2.2 in Section 2 of [Sh:410].  The proof sketched by Shelah is dubious at a couple of points, and I don't think his full result goes through.  We are replacing his Observation 2.3 by the results in the previous two posts.


After we prove the theorem, we will look at some applications to combinatorial set theory.

Section 2 of [Sh:430] Part II - Even More Technical Lemmas

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

We deal with some more technical material regarding Skolem hulls.

Let $\mu\geq\aleph_\omega$ be a cardinal, and let 

$\lambda^*=\sup\{\pp(\epsilon):\omega_1<\epsilon\leq\mu\text{ and }\cf(\epsilon)=\aleph_0\}$

Let $\chi$ be a sufficiently large regular cardinal, and let $\langle M_\alpha:\alpha<\cf(\lambda^*)\rangle$ be a tower of elementary submodels of $\mathfrak{A}:=\langle H(\chi), \in, <^*\rangle$  built ``as usual'', and let 

$M^*:=\bigcup_{\alpha<\cf(\lambda^*)}M_\alpha.$

Lemma 4
Suppose we have a model $M\prec \mathfrak{A}$ of cardinality $\aleph_1$ such that 

$$(\forall Y\in [M]^{\aleph_0})(\exists Z\in M^*\cap [M^*]^{\aleph_0})\bigl[Y\subseteq Z\bigr].$$

Let $A= M\cap(\omega_1,\mu]\cap\REG$,  let $\mathcal{B}\subseteq M^*\cap[A]^{\aleph_0}$ be cofinal in $[A]^{\aleph_0}$ with $|\mathcal{B}|=\aleph_1$, and let $f$ be a function as in the conclusion of Lemma 3 of the previous post. Define a model $N$ by 

$$N:=\Sk_{\mathfrak{A}}(M\cup \{f(\theta):\theta\in A\}).$$

Then

$$(\forall Y\in [N]^{\aleph_0})( \exists Z\in M^*\cap [M^*]^{\aleph_0})\bigl[Y\subseteq Z\bigr].$$



Proof:

Given a countable $Y\subseteq N$, there are countable $p\subseteq M$ and $q\subseteq A$ such that

$$Y\subseteq\Sk_{\mathfrak{A}}(p\cup\{f(\theta):\theta\in q\}).$$

By our assumption, we may assume that $p$ and $q$ are both in $M^*$ (they are both covered by countable sets from $M^*$, and replacing them by larger sets just enlarges the Skolem hull.)  Moreover, we may assume $q\in \mathcal{B}$.

Given our choice of $f$ and $\lambda^*$, we know $\{f(\theta):\theta\in q\}$ is in $M^*$ by way of Lemma 3 of the preceding post.

If we choose $\alpha<\cf(\lambda^*)$ so that $p$, $q$, and $\{f(\theta):\theta\in q\}$ are all in $M_\alpha$, then the relevant Skolem hull can be computed in $M^*$ using $M_\alpha$, and so

$Y\subseteq Z:=\Sk_{\mathfrak{A}}(p\cup\{f(\theta):\theta\in q\})\in M^*\cap [M^*]^{\aleph_0}.$



Thursday, November 29, 2018

Section 2 of [Sh:410] Part I - Technical Lemmas

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

This post will lay down preliminary material for obtaining some results in [Sh:410] to be discussed in the next installment.

Let $\mu>\aleph_1$ be a cardinal, and let

$\lambda^*=\sup\{\pp(\epsilon):\omega_1<\epsilon\leq\mu\text{ and }\cf(\epsilon)=\aleph_0\}.$


Let $\chi$ be a sufficiently large regular cardinal, and let $M$ be an elementary submodel of $\langle H(\chi),\in,<^*\rangle$ of cardinality $\lambda^*$ containing everything relevant.

Note that $M$ will contain a family

$\bar{F}=\langle \bar{f}^A: A\subseteq (\omega_1,\mu]\cap\REG\text{ and }|A|=\aleph_0\rangle,$
where
$\bar{f}^A=\langle \bar{f}^{A,\lambda}:\lambda\in\pcf(A)\rangle,$
and
$\bar{f}^{A,\lambda}=\langle f^{A,\lambda}_\alpha:\alpha<\lambda\rangle$
is a minimally obedient (at cofinality $\omega_1$) universal sequence for $\lambda\in\pcf A$.  (See Section 5 of Abraham-Magidor.)  $_\square$

Lemma 1
Suppose $A$ satisfies
  • $A\subseteq (\aleph_1,\mu]\cap\REG$,
  • $A\in  M$, and
  • $|A|=\aleph_0$.
Then $f^{A,\lambda}_\alpha\in M$ for any $\lambda\in\pcf(A)$ and $\alpha<\lambda$.

Proof:  Given the preceding post, this follows easily from the definition of $\lambda^*$ as $\lambda^*+1\subseteq M$.  $_\square$

Lemma 2
Suppose $N$ is an $\omega_1$-presentable elementary submodel of $H(\chi)$, and let $A$ satisfy

  • $A\subseteq (\aleph_1,\mu]\cap\REG$,
  • $A\in N\cap M$, and
  • $|A|=\aleph_0$.
Then $\ch_N\restr A\in M$.

Proof.  This is an application of Corollary 5.9 in Abraham-Magidor.  Since $A\in N$, we know there are cardinals $\lambda_0>\lambda_1>\cdots>\lambda_n$ in $N\cap \pcf(A)$ such that
$$\ch_N\restr A = \sup\{f^{A,\lambda_0}_{\gamma_0},\dots,f^{A,\lambda_n}_{\gamma_n}\},$$
where $\gamma_n=\sup(N\cap\lambda_n)$.  Since $A\in M$, the previous lemma tells us $\ch_N\restr A\in M$.  $_\square$


Lemma 3
Suppose $A\subseteq (\aleph_1,\mu]\cap\REG$ is of cardinality $\aleph_1$, and $\mathcal{B}$ is a subset of  $M\cap[A]^{\aleph_0}$ of cardinality $\aleph_1$.  Give $g\in\prod A$ there is $f\in\prod A$ such that

  • $g<f$, and
  • $f\upharpoonright B\in M$ for every $B\in\mathcal{B}$.
(Note that we do not assume that $A$ is in $M$, only that the elements of $\mathcal{B}$ are.)
Proof.

Let $N$ be an $\omega_1$-presentable elementary submodel of $H(\chi)$ containing $A$, $\mathcal{B}$, and $g$.   Let $f=\ch_N\restr A$, so clearly $g<f$ as $A\subseteq N$.

Given $B\in\mathcal{B}$, we know $B\in N\cap M$ (as $\mathcal{B}\subseteq N$) and so the previous lemma applies giving us $f\restr B=\ch_N\restr B\in M$.  $_\square$







Wednesday, November 21, 2018

Section 2 of [Sh:410] Part 0: Two cardinals

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\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}
\)

This post pins down an easy observation, relevant to several results of Shelah:


Observation:

Suppose $\theta\leq\kappa<\mu$   Then following two cardinals are equal:

$\lambda_1:=\sup\{\pp_{<\theta}(\tau):\kappa<\tau\leq\mu\text{ and }\cf(\tau)<\theta\}$

$\lambda_2:=\sup\{\max\pcf(A):  A\subseteq (\kappa,\mu]\cap\REG \text{ and }|A|<\theta\}$.

Proof:

Suppose $\lambda<\lambda_1$, and choose a cardinal $\tau$ satisfying
  • $\kappa<\tau\leq\mu$,
  • $\cf(\tau)<\theta$, and
  • $\lambda<\pp_{<\theta}(\tau)$
This last statement means that we can find a set $A$ and an ultrafilter $U$ on $A$ such that

  • $A$ is cofinal in $\tau\cap\REG$,
  • $|A|<\theta$,
  • $U$ extends the cobounded filter on $A$,  and
  • $\lambda<\tcf(\prod A/ U)$
Replacing $A$ by $A\cap (\kappa,\mu]$ does not change the situation, as $A\cap\kappa+1$ is not in the ultrafilter $U$, and so $A$ and $U$ witness that $\lambda<\lambda_2$.


Now suppose $\lambda<\lambda_2$, and choose $A$ and $U$ such that
  • $A\subseteq (\kappa,\mu]\cap\REG$,
  • $|A|<\theta$,
  • $U$ an ultrafilter on $A$, and
  • $\lambda<\tcf(\prod A/U)$.
Let $\epsilon\leq\sup(A)$ be minimal with $A\cap\epsilon\in U$.  It is clear that $A$ is unbounded in $\epsilon$, and so $\epsilon$ is a singular cardinal of cofinality at most $|A|<\theta$.

The set $A\cap\epsilon$ and the ultrafilter $U$ restricted to $A\cap\epsilon$ witness that

$\lambda<\pp_{<\theta}(\epsilon)$

and this implies $\lambda<\lambda_1$.

Sunday, November 11, 2018

A note on rank and colorings

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


Given a function $f:\theta\rightarrow\lambda$ for cardinals $\theta$ and $\lambda$ with $\cf(\theta)$ uncountable, we let $||f||$ denote the Galvin-Hajnal rank of $f$ with respect to the unbounded filter on $\theta$.

Theorem
Suppose $\aleph_0<\theta<\lambda$ and there is a function $f:\theta\rightarrow\lambda$ such that $||f||\geq\lambda$.  Then there is a coloring $p:[\lambda]^{<\omega}\rightarrow\theta$ such that the range of $p\restr [A]^{<\omega}$ is unbounded in $\theta$ whenever $A\subseteq\lambda$ is of cardinality $\lambda$.

proof.

Let $\chi$ be a sufficiently large regular cardinal, and assume by way of contradiction that the conclusion fails.  Then we can an elementary submodel $M$ of $H(\chi)$ such that

  • $\theta$, $\lambda$, and $f$ are all in $M$,
  • $|M\cap\lambda|=\lambda$, but
  • $M\cap\theta$ is bounded in $\theta$.
Given such an $M$, we define $\delta=\sup(M\cap\theta)$.   Let us now agree to call a function $g:\theta\rightarrow\lambda$ bad if
  • $g\in M$, and
  • $g(\delta)<\otp(M\cap ||g||)$.
Notice that our function $f$ is bad, so there exists at least one bad function.

Claim: If $g$ is a bad function then there is another bad function $h$ such that $h\restr[\delta,\theta)<g\restr[\delta,\theta)$.

Notice that the claim gives us a contradiction: using $f$ as $g_0$, we can iterate the claim to produce a sequence of bad functions $\langle g_n:n<\omega\rangle$ so that $\langle g_n(\delta):n<\omega\rangle$ forms an infinite decreasing sequence of ordinals.

To prove the claim, we proceed as follows:

We know $g(\delta)<\otp(M\cap||g||)$, so choose $\beta\in M\cap ||g||$ such that
$$g(\delta)=\otp(M\cap\beta).$$

Since $\beta<||g||$, we can find $h:\theta\rightarrow\lambda$ such that $h<^* g$ and
$\beta\leq ||h||$.


Since $g$ and $\beta$ are in $M$, we may assume that $h$ is in $M$ as well, and therefore so is the ordinal
\begin{equation}
\gamma:=\sup\{i<\theta: g(i)\leq h(i)\}.
\end{equation}

Given our choice of $\delta$, it follows that $\gamma<\delta$, and so
\begin{equation}
h\restr [\delta,\theta)< g\restr [\delta,\theta).
\end{equation}

In particular,  $h(\theta)<g(\theta)$ and therefore
\begin{equation}
h(\theta)<g(\theta)=\otp(M\cap\beta)\leq\otp(M\cap ||h||),
\end{equation}
and so $h$ is indeed a bad function.




Thursday, June 13, 2013

Around Observation 5.5

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

I apologize for the lack of context for this post, but I'm just trying to get some ideas down.


Proposition: Assume \(\cf(\lambda)<\theta=\cf(\theta)<\lambda\).    Then

$(1)\quad\quad \pp(\lambda)=\pp_\theta(\lambda)\Longrightarrow\cov(\pp(\lambda),\lambda,\theta^+,\theta)=\pp(\lambda).$

Proof:   We know $\lambda<\pp(\lambda)$ so $\lambda<\cov(\pp(\lambda),\lambda,\theta^+,\theta)$. Thus, by the cov vs. pp theorem (Theorem 5.4 on page 87 of The Book) it suffices to prove the following:

If $\lambda\leq\mu\leq\pp(\lambda)$ and $\cf(\mu)=\theta$, then $\pp_{\Gamma(\theta)}(\mu)\leq\pp(\lambda)$.

Clearly for such a $\mu$ we have $\pp_{\Gamma(\theta)}(\mu)\leq\pp(\mu)$, so we must show

$(2)\quad\quad \lambda\leq\mu\leq\pp(\lambda)\wedge\cf(\mu)=\theta\Longrightarrow\pp(\mu)\leq\pp(\lambda)$

But  we know the following:

  • $\pp(\mu)=\pp_\theta(\mu)$ by definition.
  • $\pp(\lambda)=\pp_\theta(\lambda)$ by assumption, and
  • $\pp_\theta(\mu)\leq\pp_\theta(\lambda)$ by "inverse monotonicity" of $\pp_\theta$  (Conclusion 2.3(2) on page 57 of The Book)
Putting all these together gives us what we need.   Q.E.D.

As far as I know, it is still unknown if the hypothesis of the preceding proposition can fail.  We do know the following:

  • If $\lambda$ is a strong limit and $\cf(\lambda)\leq\theta=\cf(\theta)<\lambda$, then $\pp(\lambda)=\pp_\theta(\lambda)$.
  • Actually, instead of assuming $\lambda$ is a strong limit, it suffices to assume that $\pp_\theta(\mu)<\lambda$ for all sufficiently large $\mu<\lambda$ with $\cf(\mu)\leq\theta$.  This is part of Corollary 1.6 on page 321.
  • We could also assume $\pp(\lambda)<\lambda^{+(\cf(\lambda))^+}$ and obtain the same conclusion.
What is the point?  In the first place, the argument given above allows us to deduce Observation 5.5 on page 404.  In the second place...well, I'll write the next piece out tomorrow.   

Wednesday, June 12, 2013

Final Inequality Finale

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

Continuing our last post, we need to prove that $X$ is a subset of $M\cap\lambda$.  Let

$N:=N_{\sup(I)}.$

Note that $X\subseteq N\cap\lambda$, so it suffices to prove the following:

$N\cap\lambda\subseteq M.$


Suppose by way of contradiction that this fails,  and define

$(1)\quad\quad \gamma(*)=\min(N\cap\lambda\setminus M).$

Since $\mu+1\subseteq M$, we know

$(2)\quad\quad\mu<\gamma(*)<\lambda.$

Next, define
$(3)\quad\quad\beta(*)=\min(M\cap N\setminus\gamma(*)).$

Claim 1: $\beta(*)<\lambda$

Proof:  Clearly $\lambda$ and $\cf(\lambda)$ are elements of $M\cap N$.  Since $\cf(\lambda)<\mu$, we know $\cf(\lambda)+1\subseteq M$.  Since $\cf(\lambda)<\theta$  and $N\cap\theta$ is an initial segment of $\theta$, it follows that $\cf(\lambda)+1\subseteq N$ as well.  From this we deduce that $M\cap N$ contains a cofinal subset of $\lambda$, and hence $\beta(*)<\lambda$.   Q.E.D.

Putting things together, we have

$(4)\quad\quad\mu<\gamma(*)<\beta(*)<\lambda,$

and by definition of $\beta(*)$,

$(5)\quad\quad M\cap N\cap [\gamma(*),\beta(*))=\emptyset.$

We now analyze what sort of ordinal $\beta(*)$ might be.  In the first place, $\beta(*)$ must be a limit ordinal,  because otherwise its predecessor would violate (5).

Claim 2: $\beta(*)$ is a regular cardinal

Proof:  Suppose not, and let $\kappa(*)=\cf(\beta(*))$.  Since $\kappa(*)\in M\cap N$, it must be the case that

$(6)\quad\quad \kappa(*)<\gamma(*)<\beta(*).$

Let $f\in M\cap N$ be a strictly increasing map from $\kappa(*)$ onto a cofinal subset of $\beta(*)$, and let $\gamma<\kappa(*)$ be the least ordinal with  $\gamma(*)<f(\gamma)$.

We know $\gamma\in N$ (it is definable once we have $\gamma(*)$ available) but $\gamma$ cannot be an element of $M$ as otherwise we contradict (5).  This means that $\kappa(*)$ must be larger than $\mu$ as $N\cap\mu\subseteq M$.

Certainly $\kappa(*)$ is regular, so $\kappa(*)$ is a regular cardinal in $N\cap (\mu,\lambda)$.  Given our definition of $N$, we can find $i\in I$ such that $f$, $\gamma$, and $\kappa(*)$ are all in $N_i$.

Let $\mathfrak{a}=N_i\cap\lambda\cap\REG$.  We chose $\alpha_i$ and $\mu_i$ so that

$\kappa\in\mathfrak{a}\setminus\mu_i+1\Longrightarrow \sup(N_i\cap\kappa)<f_{\alpha_i}(\kappa).$

Since $i\in I$, it follows that $\mu_i\leq\mu$ and therefore

$(7)\quad\quad \gamma\leq\sup(N_i\cap\kappa(*))<f_{\alpha_i}(\kappa(*))<\kappa(*).$

Let $\delta(*)=f(f_{\alpha_i}(\kappa(*))$.  Since $f$ is strictly increasing, we know

$(8)\quad\quad \gamma(*)\leq f(\gamma)<\delta(*)<\beta(*).$

The ordinal $\delta(*)$ is in $N$ because everything needed to define it is available in $N_{i+1}$. However, since $i\in I$ it follows that the ordinal $\alpha_i$ is in $B$ and hence $f_{\alpha_i}\in M$.  Since $f$ and $\kappa(*)$ are also in $M$, we find $\delta(*)\in M$ as well.  But now we have contradicted (5).  Q.E.D.

Claim 3: $\beta(*)$ cannot be a regular cardinal

Proof:  This is almost the same argument as above.   Fix $i\in I$ with $\gamma(*)\in N_i$.  Then arguing as above we find

$(9)\quad\quad \gamma<\sup(N_i\cap\beta(*))<f_{\alpha_i}(\beta(*))<\beta(*)$

and   $f_{\alpha_i}(\beta(*))$ contradicts (5). Q.E.D.

Clearly Claim 2 and Claim 3 provide the needed contradiction and we have established what we need.

Final Inequality 2

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

Picking up right where we left off, suppose $X\in [\lambda]^{<\theta}$.  We will find an acceptable model $M$ with $X\subseteq M\cap\lambda$.

Start by fixing a continuous increasing $\in$-chain  $\langle N_i:i<\theta\rangle$  of elementary submodels of $\mathfrak{A}$ satisfying:

  • $\langle N_j:j\leq i\rangle\in N_{i+1}$,
  • $|N_i|<\theta$
  • $N_i\cap\theta$ is an initial segment of $\theta$
  • $\{X,p\}\in N_0$

This is certainly possible as $\theta$ is a regular cardinal.   Note that our requirement that $N_i\cap\theta$ is an initial segment of $\theta$ has a lot of power:  for example, it implies that $X\subseteq N_0$, and that $N_j\subseteq N_i$ whenever $j<i<\theta$.


Given $i<\theta$, let us consider $\mathfrak{a}:= N_i\cap\lambda\cap\REG$.  We know the following facts:
  • $|\mathfrak{a}|=|N_i|<\theta$, and
  • $\mathfrak{a}$ is cofinal in $\lambda$ (as $\cf(\lambda)<\theta$).
This means that $\prod\mathfrak{a}$ is "covered" by our family $\mathcal{F}$:  if $g\in\prod\mathfrak{a}$ then there is an $f\in\mathcal{F}$ and $\mu<\lambda$ such that $g(\kappa)<f(\kappa)$ whenever $\kappa\in\mathfrak{a}\cap (\mu,\lambda)$.

We apply the observation in the previous paragraph to the characteristic function of $N_i$ on $\mathfrak{a}$. This gives us an ordinals $\alpha_i<\lambda(1)$ and $\mu_i<\lambda$ such that
$$\kappa\in\mathfrak{a}\cap (\mu_i,\lambda)\Longrightarrow \sup(N_i\cap \kappa)<f_{\alpha_i}(\kappa).$$
Moreover, we can choose $\alpha_i$ and $\mu_i$ in $N_{i+1}$, as the above argument can be done in the model $N_{i+1}$.  This is important, because it guarantees that the function $i\mapsto\alpha_i$ is one-to-one.

Since $\cf(\lambda)<\theta=\cf(\theta)<\lambda$, there is a $\mu<\lambda$ such that 
$$|\{i<\theta:\mu_i\leq\mu\}|=\theta.$$
Fix such a $\mu$ (without loss of generality greater than $\cf(\lambda)$), and define 
$$I^*:=\{i<\theta:\mu_i\leq\mu\}$$.

Now look at the set $A^*=\{\alpha_i:i\in I^*\}$.   This set is in $[\lambda(1)]^\theta$ because we made sure that the function $i\mapsto\alpha_i$ is one-to-one.  By the definition of $\mathcal{P}$, there is a set $B^*\in\mathcal{P}$ such that $A^*\cap B^*$ is infinite.


Let $\alpha$ be the supremum of the first $\omega$ elements of $A^*\cap B^*$, and find $I\subseteq I^*$
of order-type $\omega$ so that

  • $\{\alpha_i:i\in I\}$ is strictly increasing,
  • each $\alpha_i$ is in $A^*\cap B^*$, and
  • $\sup\{\alpha_i:i\in I\}=\alpha$
Now let $B=B^*\cap \alpha$, and define

$M=\Sk_\mathfrak{A}(B\cup p\cup\mu+1)$

Clearly $M$ is an acceptable model, so we need to prove that $X\subseteq M$.

To be continued...




Tuesday, June 11, 2013

Final Inequality 1

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

Recall
  • $\cf(\lambda)<\theta=\cf(\theta)<\lambda$, 
  • $\lambda(1):=\cf_{<\theta}(\prod(\lambda\cap\REG), <_{J^\bd_\lambda})$, and
  • $\lambda(2)$ is the minimum cardinality of a family $\mathcal{P}\subseteq[\lambda(1)]^{<\lambda}$ such that for any $B\in[\lambda(1)]^\theta$ there is an $A\in\mathcal{P}$ with $A\cap B$ infinite.

We wish to prove $\cov(\lambda,\lambda,\theta,2)\leq\lambda(2)$.  This requires us to produce a family
$\mathcal{P}^*\subseteq[\lambda]^{<\lambda}$ such that
  • $|\mathcal{P}^*|\leq\lambda(2)$, and
  • for any $B\in [\lambda]^{<\theta}$ there is an $A\in\mathcal{P}^*$ with $B\subseteq A$.
We start by fixing a family $\mathcal{F}\subseteq \prod(\lambda\cap\REG)$ witnessing the definition of $\lambda(1)$, say $\mathcal{F}=\{f_\alpha:\alpha<\lambda(1)\},$ and let $\mathcal{P}\subseteq[\lambda]^{<\lambda}$ be as in the definition of $\lambda(2)$.





To find $\mathcal{P}^*$, let us assume $\chi$ is a regular cardinal much larger than any of the cardinals considered above, and let (as usual) $\mathfrak{A}$ denote the structure $\langle H(\chi),\in, <_\chi)$,where $<_\chi$ is some well-ordering of $H(\chi)$.  We will use $\Sk_\mathfrak{A}(B)$ to denote the Skolem hull of $B$ in the structure $\mathfrak{A}$.


Ad hoc definition:
Let us say that an elementary submodel $M$ of $\mathfrak{A}$ is acceptable if  
$M=\Sk_{\mathfrak{A}}(B\cup p\cup\mu+1)$
where
  • $B$ is an initial segment of some member of $\mathcal{P}$, 
  • $p=\{\theta,\lambda,\lambda(1),\lambda(2), \mathcal{F}, \mathcal{P},...\}$, and
  • $\mu<\lambda$
(We have been sloppy with $p$.  The intent is that $p$ encodes the parameters associated with the preceding discussion.)


If $M$ is an acceptable model, then $|M|<\lambda$.  Also note that there are at most $\lambda(2)=|\mathcal{P}|$ acceptable models.

We define $\mathcal{P}^*\subseteq[\lambda]^{<\lambda}$ to be all sets of the form $M\cap\lambda$ where $M$ is an acceptable model.

It remains to show that $\mathcal{P}^*$ has the required covering property.  We will do this in the next couple of posts.