Wednesday, June 12, 2013

Final Inequality Finale

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

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.

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