%! TEX root = FC.tex
% vim: tw=50
% 18/02/2025 12PM

Cohen:

$\forall T \subseteq \zfc$ finite, $\exists T^*
\subseteq \zfc$ finite such that if $M$ is a
\gls{ctm} of $T^*$, then there is $N \supseteq M$
\gls{ctm} of $T + \neg \ch$. ($*$)

We have seen (\es{1}) that ($*$) implies
$\Con(\zfc) \implies \Con(\zfc + \neg \ch)$.

Simplified: If $M$ is a \gls{ctm} of $\zfc$, then
there is $N \supseteq M$ \gls{ctm} of $\zfc h=
\neg\ch$.

\textbf{Idea:} If $M$ is a \gls{ctm} of $\zfc$:
$\alpha \defeq \omega_1^M$; $\beta \defeq
\omega_2^M$; $f : \beta \to \mathcal{P}(\omega)$
injection. Force $N$ such that $f \in N$ and $M
\subseteq N$.

Observe that there is a \gls{ctm} $N$ such that
$f \in N$ and $M \subseteq N$. $M \cup
\tcl(\{f\})$ is transitive and
countable. Thus LSM % TODO
gives $N$ transitive countable with $M \cup
\tcl(\{f\}) \subseteq N$. Know $N \models \exists
g : \beta \to \mathcal{P}(\omega)$ is an
injection, but no clue what $\aleph_1$ and
$\aleph_2$ in $N$ are.

\textbf{How do we control what we add?}

\newpage
\section{Forcing}

\begin{fcdefnstar}[Forcing]
  \glsadjdefn{forcing}{forcing}{poset}%
  $(\Pbb, \le, \indicator{})$ is called a
  \emph{forcing poset} or \emph{forcing} if it is a
  partial order and $\indicator{} \in \Pbb$ and
  $\indicator{}$ is the largest element. Elements of
  $\Pbb$ are called \emph{condition}.
\end{fcdefnstar}

$p \le q$ is interpreted as
\begin{quote}
  ``$p$ is stronger than $q$''

  ``$q$ is weaker than $p$''
\end{quote}

\begin{note*}
  Unconventional.
\end{note*}

Alternative: ``Jerusalem convention''. Interpret
$p \le q$ as ``$q$ is stronger than $p$''.

We are \emph{not} following the Jerusalem
convention.

\begin{fcdefnstar}[Compatible]
  \glsadjdefn{comp}{compatible}{elements}%
  \glsadjdefn{incomp}{incompatible}{elements}%
  $p$ and $q$ are \emph{compatible} if there is $r
  \le p, q$. Otherwise, we say they are
  \emph{incompatible} (which we write as $p \perp
  q$).
\end{fcdefnstar}

\begin{fcdefnstar}[Antichain]
  \glsnoundefn{antic}{antichain}{antichains}%
  $A \subseteq P$ is an \emph{antichain} if any
  two distinct elements of $A$ are \gls{incomp}.
\end{fcdefnstar}

\begin{fcdefnstar}[Dense]
  \glsadjdefn{dense}{dense}{subset}%
  $D \subseteq \Pbb$ is \emph{dense} if $\forall p
  \in TODO$.
\end{fcdefnstar}

\begin{fcdefnstar}[Filter]
  \glsnoundefn{filt}{filter}{filters}%
  \glsnoundefn{filtb}{filter base}{filter bases}%
  $F \subseteq \Pbb$ is called a \emph{filter} if
  \begin{enumerate}[(a)]
    \item $\forall p, q \in F, \exists r \in F, r
      \le p, q$
    \item $\forall p \in F, \forall q \in \Pbb, q
      \ge p \implies q \in F$
  \end{enumerate}
  If $F$ only has property (a), we call it a
  \emph{filter base}, and then
  \[
    \{p \in \Pbb : \exists q \in F, q \le p\}
  \]
  is the \emph{filter generated from $F$}.
\end{fcdefnstar}

\begin{note*}
  \glsref[filt]{Filters} \emph{cannot} contain
  incompatible elements.
\end{note*}

\begin{fcdefnstar}[$D$-generic]
  \glsadjdefn{Dgen}{$D$-generic}{family}%
  If $\mathcal{D}$ is a family of \gls{dense}
  sets, then $F$ is called
  \emph{$\mathcal{D}$-generic} if $F$ is a
  \gls{filt} and $\forall D \in \mathcal{D}, D
  \cap F \neq \emptyset$.
\end{fcdefnstar}

\begin{fcthm}[]
  \label{lec8the:theorem}
  Assuming:
  - $\mathcal{D}$ is countable
  Then:
  there is a \dgen{} \gls{filt}.
\end{fcthm}

\begin{proof}
  Let $\mathcal{D} = \{D_n : n \in \Nbb\}$. Pick
  $p_0 \in D_0$ arbitrarily and define by
  recursion $p_{i + 1}$ by picking some $q \le
  p_i$ with $q \in D_{i + 1}$.

  Then $\{p_i : i \in \Nbb\}$ is a \gls{filtb}, so
  let $F$ be the \gls{filt} generated by it. Then
  it is \dgen{} by construction.
\end{proof}

\subsubsection*{Main example}

Fix any sets $X$ and $Y$
\[
  \Fn(X, Y)
  \defeq \{p : \text{$p$ is a finite fucntion with
  $\dom(p) \subseteq X$ and $\range(p) \subseteq Y$}\}
.\]
Define $p \le q \iff p \supseteq q$ and
$\indicator{} \defeq \emptyset$.

What does $p \perp q$ mean?
\[
  p \perp q
  \iff \exists x \in X, x \in \dom(p) \cap \dom(q)
  \wedge p(x) \neq q(x)
.\]

\begin{fclemma}[]
  % Lemma 1
  Assuming:
  - $F \subseteq \Pbb$ is a \gls{filt}
  Then:
  $\bigcup F$ is a function.
\end{fclemma}

\noproof

\begin{fclemma}[]
  % Lemma 2
  $D_x \defeq \{p \in \Pbb : x \in \dom(p)\}$ is
  \cloze{a \gls{dense} set. If $\mathcal{D} \defeq
  \{D_x : x \in X\}$, and $F$ is \dgen, then
  $\dom(\bigcup F) = X$.}
\end{fclemma}

\begin{proof}
  If $x \in X$, find $p \in F \cap D_x$, then $x
  \in \dom(p)$, TODO
\end{proof}

\subsection{Cohen Forcing}

\subsubsection*{Example 1}

$\Cbb \defeq \Fn(\omega, 2)$.

If $F$ is \dgen, then $\bigcup F : \omega \to 2$
(by above).

\begin{fclemma}[]
  % Lemma 3
  Assuming:
  - Fix $f : \omega \to 2$ and define
  \[
    N_f \defeq \{p \in \Pbb : \exists u, p(u) \neq
    f(u)\}
  .\]
  - $F$ is \glsref[Dgen]{$\mathcal{D} \cup \{N_f\}$}
  Then:
  $\bigcup F \neq f$.
\end{fclemma}

\begin{fccoro}[]
  There is no $F$ that is \cloze{\glsref[Dgen]{
  $\mathcal{D} \cup \{N_f \st f : \omega \to
  2\}$-generic}.}
\end{fccoro}

\emph{However:} if $M$ is a \gls{ctm} and you
consider
\[
  \mathcal{N}_f \defeq \{N_f : f \in M\}
,\]
then by \cref{lec8the:theorem}, there is a
\glsref[Dgen]{$\mathcal{D} \cup
\mathcal{N}_M$-generic}. And for this $F$, TODO

\subsubsection*{Example 2}

$\Fn(X, Y) \eqdef \Pbb$ as before.
\begin{align*}
  R_y
  &\defeq \{p \in \Pbb : y \in \range(p)\} \\
  \mathcal{R}
  &\defeq \{R_y : y \in Y\}
\end{align*}

\begin{fclemma}[]
  % Lemma 4
  Assuming:
  - $F$ is \dgen[\mathcal{D} \cup \mathcal{R}]{}
  Then:
  $\bigcup F : X \to Y$ is a surjection.
\end{fclemma}

\begin{fccoro}[]
  Assuming:
  - $|Y| > |X|$
  Then:
  there is no \dgen[\mathcal{D} \cup \mathcal{R}].
\end{fccoro}

\emph{However:} if $M$ is a \gls{ctm} and, e.g.
TODO

\subsubsection*{Example 3}

$\Fn(X \times Y, 2)$. Assume $X$ is infinite.
Consider
\[
  E_{y, y'}
  \defeq \{p \in \Pbb : \exists x \in X, p(x, y)
  \neq p(x, y')\}
.\]
This is \gls{dense} for $y \neq y'$.
\[
  \mathcal{E} \defeq \{E_{y, y'} : y \neq y' \in Y\}
.\]

\begin{fclemma}[]
  % Lemma 5
  Assuming:
  - $F$ is \dgen[\mathcal{D} \cup \mathcal{E}]
  Then:
  there is an injection from $Y$ into
  $\mathcal{P}(X)$.
\end{fclemma}

\begin{proof}
  Fix $y$ and define
  \[
    A_y \defeq \{x \in X : (\bigcup F) (x, y) = 1\}
  .\]
  $E_{y, y'}$ guarantees that $y \mapsto Ay$ is an
  injection.
\end{proof}

\begin{fccoro}[]
  Assuming:
  - $|Y| > |\mathcal{P}(\omega)|$
  Then:
  there is no \dgen[\mathcal{D} \cup \mathcal{E}].
\end{fccoro}

\emph{However:} if $M$ is a \gls{ctm} of $\zfc +
\ch$, $\alpha \def \omega_1^M$, $\beta \defeq
\omega_2^M$, $\Fn(\omega \times \omega_2^M, 2)$
Because $\beta < \omega_1$, $\mathcal{E}$ is
countable and so a \dgen[\mathcal{D} \cup
\mathcal{E}] exists.