%! TEX root = FC.tex % vim: tw=50 % 28/01/2025 12PM \begin{fcdefnstar}[Absolute formula] \glsadjdefn{absm}{absolute}{formula}% We say \emph{$\varphi$ is absolute for $M$} if for all $x_1, \ldots, x_n \in M$, we have \[ M \models \varphi(x_1, \ldots, x_n) \iff \varphi(x_1, \ldots, x_n) \text{ is true} .\] \end{fcdefnstar} Clearly, if $\varphi$ is \gls{abs} for $M$ and \gls{abs} for $N$, then it's \gls{abs} between $M$ and $N$. Cohen's proof becomes: If $M$ is a countable transitive set such that $M \models \ZFC$, then there is a a countable transitive set $N \supseteq M$ such that $N \models \ZFC + \neg \CH$. \newpage \section{Transitive Models} Observation: If $M$ is transitive and $M \models \text{``$e$ is empty''}$, then $e = \emptyset$. This is because if $w \in e$, then $w \in e$ and $e \in M$ gives us that $w \in M$ by transitivity, so $M \models w \in e$, so $M \models \text{``$e$ is not empty''}$. \begin{fclemmastar}[] Assuming: - $M$ is transitive Then: \[ M \models \text{Extensionality} + \text{Foundation} .\] \end{fclemmastar} \begin{proof} Extensionality: $\forall x \forall y (\forall w (w \in x \leftrightarrow w \in y) \to x = y)$. Take $x \neq y$, $x, y \in M$. Without loss of generality take $z \in x \setminus y$. Then $z \in x$ and $x \in M$ so $z \in M$. So $M \models z \in x \wedge z \notin y$. So $M \models \neg\forall w (w \in x \leftrightarrow w \in y)$. Foundation: $\forall x(x \neq \emptyset \to \exists m(m \in x \wedge \forall w \neg (w \in m \wedge w \in x)))$. Take $x \in M$. $M \models x \neq \emptyset$ so $x$ is not empty. So find $m \in x$ which is $\in$-minimal. Then since $x \in M$ as well, we have $m \in M$. Therefore $x$ has an $\in$-minimal element in $M$. \end{proof} \subsection{Absoluteness for transitive models} \begin{fcdefnstar}[Bounded quantifier] We call a quantifier of the form $\exists x \in y, \varphi$ or $\forall x \in y, \varphi$ a \emph{bounded quantifier}. (Defined by $\exists x \in y, \varphi \defeq \exists x (x \in y \wedge \varphi)$ and $\forall x \in y \varphi \defeq \forall x (x \in y \to \varphi)$). \end{fcdefnstar} \begin{fcdefnstar}[Closed under bounded quantification] A class of formulas $\Gamma$ is \emph{closed under bounded quantification} if whenever $\varphi$ is in $\Gamma$, then so are $\exists x \in y, \varphi$ and $\forall x \in y, \varphi$. \end{fcdefnstar} \begin{fcdefnstar}[Delta0] \glssymboldefn{delz}% $\Delta_0$ is the smallest class of formulas containing the atomic formulas that is closed under propositional connectives and bounded quantifiers. Let $T$ be any theory. Then $\Delta_0^T$ is the class of formulas equivalent to a $\Delta_0$ formula in $T$. \end{fcdefnstar} \begin{fcthmstar}[] $\delz$ formulas are \cloze{\gls{absm} for transitive models.} \end{fcthmstar} \begin{proof} By induction: \begin{enumerate}[(1)] \item All atomic formulas are \gls{absm} by the substructure lemma. \item Propositional connectives: exactly the same proof as in the substructure lemma. \item Assume that $\varphi$ is \gls{absm} and show that $\exists x \in y, \varphi$ and $\forall x \in y, \varphi$ are absolute. \begin{itemize} \item $\exists x \in y, \varphi$: If $\exists x \in y, \varphi$ is true for some $y \in M$, then pick a witness $x \in y$. Since $y \in M$, we have $x \in M$. By the induction hypothesis, we have that $M \models \varphi(x, y)$. Thus $M \models \exists x (x \in y \wedge \varphi(x, y))$. If $M \models \exists x(x \in y \wedge \varphi(x, y))$, then $x \in y \wedge \varphi(x, y)$ is true. \item $\forall x \in y, \varphi$: Similar. \qedhere \end{itemize} \end{enumerate} \end{proof} \begin{fccorostar}[] Assuming: - $T$ is any theory - $M \models T$ is transitive Then: $\delzt$-formulas are \gls{absm} for $M$. \end{fccorostar} \begin{fcdefnstar}[$Sigma1$, $Pi1$] \glssymboldefn{Pi0Sigma1}% A formula is called $\Sigma_1$ if it is of the form $\exists x_1, \ldots \exists x_n, \varphi$ where $\varphi$ is $\delz$. It is called $\Pi_1$ if it is of the form $\forall x_1, \ldots, \forall x_n, \varphi$ where $\varphi$ is $\delz$. \end{fcdefnstar} (same for $\Sigma_1^T$, $\Pi_1^T$). % TODO % \begin{fccorostar}[] % $\Sigma_1^T$-formulas are % \cloze[false]{\gls{uabs}} for this model and $\Pi_1^T$ are % \cloze{\gls{dabs}}. % \end{fccorostar} \begin{proof} Just definition of the semantics of $\exists, \forall$. \end{proof} \begin{example*} What is $\delz$? \begin{enumerate} \item $x \in y$ \item $x = y$ \item $x \subseteq y$: $\iff \forall w \in x (w \in y)$ \item $z = \{x\}$: $\iff x \in z \wedge \forall w \in z (w = x)$ \item $z = \{x, y\}$ \item $z = (x, y) = \{\{x\}, \{x, y\}\}$ \item $z = \emptyset$: $\iff \forall w \in z(w \neq w)$ \item $z = x \cup y$: $\iff x \subseteq z \wedge y \subseteq z \wedge \forall w \in z (w \in x \vee w \in y)$ \item $z = x \cap y$ \item $z = x \setminus y$ \item $z = x \cup \{x\}$ \item $z \text{ is transitive}$ \item $z = \bigcup x$ \end{enumerate} \end{example*} \begin{fcdefnstar}[Absolute function] \glsadjdefn{absf}{absolute}{function}% Let $M$ a transitive set, and $F : M^n \to M$ (note that this means that $M$ is closed under $F$). We say $F$ is \emph{absolute for $M$} if there is a formula $\Phi$ which is \gls{absm} for $M$ such that \[ F(x_1, \ldots, x_n) = z \iff \Phi(x_1, \ldots, x_n, z) .\] \end{fcdefnstar} Observation: So, if $M$ is closed under pairing ($\forall x, y \in M, \{x, y\} \in M$), then the pairing operation $x, y \mapsto \{x, y\}$ is \gls{absf}, and therefore $M \models \text{Pairing}$. Similarly for union. \begin{fclemmastar}[] Assuming: - $\varphi$ is \gls{absm} for $M$ - $F, G_1, \ldots, G_n$ are \gls{absf} operations on $M$ Then: \begin{align*} \psi(x_1, \ldots, x_m) &\defeq \varphi(G_1(x_1, \ldots, x_m), \ldots, G_n(x_1, \ldots, x_m)) \\ H(x_1, \ldots, x_m) &\defeq F(G_1(x_1, \ldots, x_m), \ldots, G_n(x_1, \ldots, x_m)) \end{align*} are \gls{absm} for $M$. \end{fclemmastar} \begin{proof} Check the definitions! \end{proof} \begin{example*} More examples: \begin{enumerate}[(1)] \setcounter{enumi}{13} \item $z$ is an ordered pair: \[ \exists s \in z, \exists d \in z, \exists x \in s, \exists y \in d, (\forall w \in s, (w = x) \wedge \forall v \in d, (v = x \wedge v = y) \wedge \forall w \in z, (w = s \vee w = d)) .\] \item $z = a \times b$ \item $z \text{ is a relation}$ \item $z = \dom x$ \item $z = \range x$ \item $z \text{ is a function}$ \item $z \text{ is injective}$ \item $z \text{ is surjective}$ \item $z \text{ is bijective}$ \end{enumerate} \end{example*}