%! TEX root = FC.tex % vim: tw=50 % 08/02/2025 12PM Want $T_{F, G, H} \vdash L1(F, G, H)$, $T_{F, G, H} \vdash L2(F, G, H)$ and $T_{F, G, H}$ proves existence of $R$. Convention: We say ``$T$ is sufficiently strong'' if $T \subseteq \zfc$ is finite and $T$ proves hte existence of all relevant operations such that they are absolute for transitive models of $T$. \begin{proof} Observe that by assumption, being an ``attempt'' is \gls{abs} for transitive models of $T$. Let $M \models T$ be transitive. \begin{enumerate}[(1)] \item To show: If $M \models R(\alpha, \ol{x}) = z$, then $R(\alpha, \ol{x}) = z$. If $M \models R(\alpha, \ol{x}) = z$, then $M \models \exists r, \text{ $r$ is an attempt and $r(\alpha, \ol{x}) = z$}$. Without the $\exists r$, this would be \gls{abs}. So when we include the existential quantifier, we get an \gls{uabs} sentence. Thus: there is $r$ such that $r$ is an attempt and $r(\alpha, \ol{x}) = z$. So $R(\alpha, \ol{x}) = z$. \item Other direction. Assume $r$ is an attempt with $r(\alpha, \ol{x}) = z$. Since $T_{F, G, H} \vdash L2(F, G, H)$, we have \[ M \models \exists r', \ub{\text{ $r'$ is an attempt and $(\alpha, \ol{x}) \in \dom r'$}}_{\text{\gls{abs}}} .\] Since it is \gls{abs}, $r'$ is a real attempt. By ???, $r'(\alpha, \ol{x}) = r(\alpha, \ol{x})$. Hence $M \models R(\alpha, \ol{x}) = z$. \qedhere \end{enumerate} \end{proof} \begin{note*} This uses the fact that ``$\Delta_1$'' concepts are \gls{abs}. \end{note*} \begin{fcdefnstar}[Delta1T property] \glssymboldefn{delot}% A property is called $\Delta_1^T$ if it's both $\sigo^T$ and $\pio^T$. \end{fcdefnstar} \textbf{Observe:} $\delot$ concepts are \gls{abs} (upwards from $\sigo$ and downwards from $\pio$). \subsubsection*{Typical Applications} Bounding a quantifier by operation. Let $F$ be an operation and $T$ strong enough to prove $F$ is an operation and \gls{abs}. \begin{align*} T &\vdash \forall x, \exists z, F(x) = z \\ T &\vdash \forall x, \forall z, \forall z', F(x) = z \wedge F(x) = z' \to z = z' \end{align*} Then the quantifiers $\exists y \in F(x)$ and $\forall y \in F(x)$ preserve \glsref[abs]{absoluteness}. \begin{align*} \exists y \in F(x) \psi &\iff \ub{\exists z \ub{(z = F(x) \wedge \exists y \in z\psi)}_{\text{\gls{abs}}}}_{\text{\gls{uabs}}} \\ &\iff \ub{\forall z \ub{(z = F(x) \to \exists y \in z\psi)}_{\text{\gls{abs}}} }_{\text{\gls{dabs}}} \end{align*} \subsubsection*{Applications} \begin{enumerate}[(1)] \item \glssymboldefn{Fml}% Encode formulas as elements of $\omega^{<\omega}$. \begin{center} \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline $\in$ & $=$ & $($ & $)$ & $\wedge$ & $\vee$ & $\neg$ & $\exists$ & $\forall$ & $v_0$ & $v_1$ & $v_2$ & $\cdots$ \\ \hline $0$ & $1$ & $2$ & $3$ & $4$ & $5$ & $6$ & $7$ & $8$ & $9$ & $10$ & $11$ & $\cdots$ \\ \hline \end{tabular} \end{center} $\forall v_0 \exists v_1 \neg v_0 \in v_1$ would be $(8, 9, 7, 10, 6, 8, 0, 10)$. $\mathrm{Fml} \subseteq \omega^{<\omega}$. So, $\mathrm{Fml}$ is \gls{abs} for some (sufficiently strong) finite fragment of $\zfc$ (see \es{1}). \item If $X$ is any set then \[ \text{``$X \models \varphi$''} \] (which means ``$(X, \in) \models \varphi$'') is defined by the usual (Tarski) recursion and thus also \gls{abs} (\es{1}). \end{enumerate} \subsection{The constructible hierarchy} Fix a set $X$. Define for each $\varphi \in \Fml$ and each $p \in X^{<\omega}$ (parameter) \[ D(\varphi, p, X) % \defeq \{a \in X : \forall w, w \in a % \leftrightarrow X \models \varphi(p, w)\} \defeq \{w \in X : X \models \varphi(p, w)\} \] the subset of $X$ defined by $\varphi$ with parameter $p$. For a sufficiently strong $T \subseteq \zfc$ finite, we have that $T$ proves that $D$ is an absolute operation (see \es{1}). \[ \mathcal{D}(X) \defeq \{D(\varphi, p, X) : \varphi \in \Fml, p \in X^{<\omega}\} .\] This is \gls{abs} for a sufficiently strong theory (use Replacement to get $\mathcal{D}(X)$). This $\mathcal{D}(X)$ is sometimes (misleadingly) called the ``definable power set of $X$'' (it is misleading because it is more like a ``definable (by $X$) power set of $X$''). ($\alpha \in \mathcal{D}(X) \iff \exists \varphi \in \Fml, \exists p \in X^{<\omega}, a = D(\varphi, p, X)$) Obvious: $\mathcal{D}(X) \subseteq \mathcal{P}(X)$. Also: If $X$ is transitive, then so is $\mathcal{D}(X)$. \begin{align*} L_0 &\defeq \emptyset \\ L_{\alpha + 1} &\defeq \mathcal{D}(L_\alpha) \\ L_{\lambda} &= \bigcup_{\alpha < \lambda} L_\alpha \end{align*} The constructible \gls{hier}. We usually write $L \defeq \bigcap_{\alpha \in \Ord} L_\alpha$. \textbf{Claim:} $L$ is a \gls{hier} (in the sense of Lecture). See \es{1}. By closure of \glsref[abs]{absoluteness} under transfinite recursion, the $L$-\gls{hier} is \gls{abs} for transitive models of $T \subseteq \zfc$ where $T$ is strong enough to prove that it exists. i.e. if $M \models T$ transitive and $\alpha \in \Ord \cap M$ and $M \models X = L_\alpha$, then $X = L_\alpha$. So \[ \bigcup_{\alpha \in \Ord \cap M} L_\alpha \subseteq M .\] The main theorem of next lecture will be: If $L \models \zf$ and $M \models \zf$ transitive, then \[ \bigcup_{\alpha \in \Ord \cap M} L_\alpha \models \zf .\] (Minimal $\zf$-model). \subsubsection*{Some first idea of what the $L$-hierarchy is like} Clearly, by induction, $L_\alpha \subseteq V_\alpha$, and clearly for $n \in \omega$, $L_n = V_n$. So $L_\omega = V_\omega$. \[ L_{\alpha + 1} \defeq \bigcup_{\varphi \in \Fml} \bigcup_{p \in L_\alpha^{< \omega}} \{D(\varphi, p, L_\alpha)\} .\] If $\alpha \ge \omega$, then \[ |L_{\alpha + 1}| \le \aleph_0 \cdot |L_\alpha^{<\omega}| = \aleph_0 \cdot |L_\alpha| .\] Thus $|L_\alpha| = |L_{\alpha + 1}|$. Therefore $\alpha < \omega_1$, $|L_\alpha| = \aleph_0$ and $|L_{\omega_1}| = \aleph_1$. This means: $V_{\omega + 1} \neq L_{\omega + 1}$ (since the first has size $2^{\aleph_0}$, while the second has size $\aleph_0$). Note: This does not mean $V \neq L$. ($V = L$ means $\forall x, \exists \alpha, x \in L_\alpha$). $V = L$ is called the ``axiom of constructibility''.