Since I'm not native speaker, my writing is probably difficult to read. Hence please point out any mistakes.
I'm reading page 96 and 97 of Modal Logic written by Patrick Blackburn.
$\textbf{Proposition 2.59}$
Let $\tau$ be a modal similarity, and $\mathfrak{M}$ a $\tau$-model. Then, for any formula $\phi$ and any ultrafilter $u$ over $W$, $V(\phi)\in u$ iff $\mathfrak{ueM},u\Vdash\phi$. Hence for every state $w$ of \mathfrak{M} we have $w\leftrightsquigarrow \pi_w$.
$\textit{Proof.}$ The second claim of the proposition is immediate from the first one by the observation that $w\Vdash\phi\ \textrm{iff}\ w\in V(\phi)\ \textrm{iff}\ V(\phi)\in\pi_w$. The proof of the first claim is by induction on $\phi$. The basic case is immediate from the definition of $V^{ue}$. The proofs of the boolean cases are straightforward consequences of the defining properties of ultrafilters. As an example, we treat negation; suppose that $\phi$ is of the form $\neg\psi$, then
\begin{eqnarray} V(\neg\psi)\in u\ &\textrm{iff}&\ W-V\in u\\ &\textrm{iff}&\ V(\psi)\not\in u\ (\textrm{Definition of ultrafilter})\\ &\textrm{iff}&\ \mathfrak{ueM}, u\not\Vdash\psi\ (\textrm{induction hypothesis})\\ &\rm{iff}&\ \mathfrak{ueM}, u\Vdash\neg \psi \end{eqnarray}
Next, consider the case where $\phi$ is of the form $\Diamond \psi$ (we only treat the basic modal similarity type). Assume first that $\mathfrak{ueM},u\Vdash\Diamond \psi$. Then, there is an ultrafilter $u'$ such that $R^{ue}uu'$ and $\mathfrak{ueM},u'\Vdash\psi$. The induction hypothesis implies that $V(\psi)\in u'$, so by the definition of $R^{ue}$, $m_R(]V(\psi))\in u$. Now the result follows immediately from the observation that $m_R(V(\psi))=V(\Diamond \psi)$ The left-to-right implication requires a bit more work. Assume that $V(\Diamond\psi)\in u$. We have to find an ultrafilter $u'$ such that $V(\psi)\in u'$ and $R^{ue}uu'$. The latter constraint reduces to the condition that $m_R(X)\in u$ whenever $X\in u'$, or equivalently
$u'_0:=\{Y\ |\ l_R(Y)\in u\}\supset u'\cdots(*)$
We will first show that $u_0'$ is closed under intersection. Let $Y,Z$ be members of $u_0'$. By definition, $l_R(Y)$ and $l_R(Z)$ are in $u$. But then $l_R(Y \cap Z) \in u$, as $l_R(Y \cap Z) = l_R(Y ) \cap l_R(Z)$, as a straightforward proof shows. This proves that $Y\cap Z\in u_0'$. Next we make sure that for any $Y\in u_0', Y \cap V (\psi) \not=\varnothing$. Let $Y$ be an arbitrary element of $u_0'$, then by definition of $u_0', l_R(Y) \in u$. As u is closed under intersection and does not contain the empty set, there must be an element $x$ in $l_R(Y)\cap V (\Diamond)$. But then $x$ must have a successor $y$ in $V(\psi)$. Finally, $x \in l_R(Y)$ implies $y \in Y$. From the fact that $u_0'$ is closed under intersection, and the fact that for any $Y \in u_0', Y \cap V (\psi) \not= \varnothing$, it follows that the set $u_0'\cup \{V (\psi)\}$ has the finite intersection property. So the Ultrafilter Theorem provides us with an ultrafilter $u'$ such that $u_0'\cup \{V (\psi)\} \supset u'$. This ultrafilter $u'$ has the desired properties: it is clearly a successor of $u$, and the fact that $\mathfrak{ueM},u'\Vdash\psi$ follows from $V (\psi) \in u'$ and the induction hypothesis.$\dashv$
$\textbf{Proposition 2.61}$
Let $\tau$ be a modal similarity type, and let $\mathfrak{M}$ be a $\tau$-model. Then $\mathfrak{ueM}$ is m-saturated.
$\textit{Proof}.\ $ We only prove the proposition for the basic modal similarity type. Let $\mathfrak{M} = (W,R, V )$ be a model; we will show that its ultrafilter extension $\mathfrak{ueM}$ is m-saturated. Consider an ultrafilter $u$ over $W$, and a set $\Sigma$ of modal formulas which is finitely satisfiable in the set of successors of $u$. We have to find an ultrafilter $u'$ such that $R^ueuu'$ and $\mathfrak{ueM}, u'\Vdash \Sigma$. Define
$\Delta= \{V (\phi) | \phi \in \Sigma'\} \cup \{Y | l_R(Y) \in u\}\cdots(**)$,
where $\Sigma'$ is the set of (finite) conjunctions of formulas in $\Sigma$. We claim that the set $\Delta$ has the finite intersection property. Since both $\{V (\phi) | \phi \in \Sigma\}$ and $\{Y |l_R(Y) \in u\}$ are closed under taking intersections, it suffices to prove that for an arbitrary $\phi \in \Sigma'$ and an arbitrary set $Y \supset W$ for which $l_R(Y) \in u$, we have $V (\phi) \cap Y \not= \varnothing$. But if $\phi \in \Sigma'$, then by assumption, there is a successor $u''$ of $u$ such that $\mathfrak{ueM}, u'' \Vdash \phi$, or, in other words, $V (\phi) \in u'$. Then, $l_R(Y) \in u$ implies $Y \in u''$. Hence, $V (\phi)\cap Y$ is an element of the ultrafilter $u''$ and, therefore, cannot be identical to the empty set. It follows by the Ultrafilter Theorem that $\Delta$ can be extended to an ultrafilter $u'$. Clearly, $u'$ is the required successor of $u$ in which $\Sigma$ is satisfied. $\dashv$
Changing $[R^{ue}uu'\Leftrightarrow \{Y|\ l_R(Y)\in u\}\subset u'(*)]$ into $[R^{ue}uu_1\cdots u_n\Leftrightarrow Y_1\in u_1\ \textrm{or}\ \cdots\ \textrm{or}\ Y_n\in u_n,\ \textrm{whenever}\ l_R(Y_1,\ \ldots,\ Y_n)\in u]$, and $[\Delta= \{V (\phi) | \phi \in \Sigma'\} \cup \{Y | l_R(Y) \in u\}(**)]$ into $[\Delta_i=\{V(\phi)|\phi\in\Sigma_i\}\cup\{Y_i|l_R(Y_1,\ldots, Y_n)\in u\}$ (for each $1\leq i\leq n)$] in those proof, I tried to prove general cases of Proposition 2.59 and Proposition 2.61, but I couldn't.
Is this plan wrong, or are my efforts not enough ?
Please instruct me how to prove these proof.
