lecture 10

inputs/lecture_10.tex

@@ -13,8 +13,24 @@ Applications of induction and recursion:
     Once we have such a function,
      $\{x\} \cup \bigcup \ran(F)$
      is a set as desired.
-
+     To get this $F$ using the recursion theorem,
-     \todo{insert formal application of recursion theorem}
+     pick $D$ such that
+     \[
+         (\emptyset, 0, \{x\}) \in D
+     \]
+     and
+     \[
+         (f, n+1, \bigcup\bigcup \ran(f)) \in D.
+     \]
+     The recursion theorem then gives a function
+     such that
+     \begin{IEEEeqnarray*}{rCl}
+         F(0) &=& \{x\},\\
+         F(n+1) &=& \bigcup\bigcup \ran(F\defon{n+1})\\
+                &=& \bigcup \bigcup \{\{x\}, x, \bigcup x, \ldots, \underbrace{\bigcup^{n-1} x}_{F(n)}\}
+                    = \bigcup F(n),
+     \end{IEEEeqnarray*}
+     i.e.~$F(n+1) = \bigcup F(n)$.
 \end{proof}
 
 \begin{notation}
@@ -26,7 +42,27 @@ Applications of induction and recursion:
     such that $F(\alpha) = \bigcup \{\cP(F(\beta)): \beta < \alpha\}$.
 \end{lemma}
 \begin{proof}
-    \todo{TODO}
+    Use the recursion theorem with $R = \in $
+    and $(w,x,y) \in D$ iff
+    \[
+    y = \bigcup \{\cP(\overline{y}) : \overline{y} \in \ran(w)\}.
+    \]
+    This function has the following properties:
+    \begin{IEEEeqnarray*}{rCl}
+        F(0) &=& \bigcup \emptyset = \emptyset,\\
+        F(1) &=& \bigcup \{\cP(\emptyset)\} = \bigcup \{\{\emptyset\} \} = \{\emptyset\},\\
+        F(2) &=& \bigcup \{\cP(\emptyset), \cP(\{\emptyset\})\} = \bigcup \{\{\emptyset\}, \{\emptyset, \{\emptyset\} \} \} = \{\emptyset, \{\emptyset\} \},\\
+        \ldots
+    \end{IEEEeqnarray*}
+
+    It is easy to prove by induction:
+    \begin{enumerate}[(a)]
+        \item Every $F(\alpha)$ is transitive.
+        \item $F(\alpha) \subseteq F(\beta)$ for all $\alpha \le \beta$.
+        \item $F(\alpha+1) = \cP(F(\alpha))$ for all $\alpha \in \OR$.
+        \item $F(\lambda) = \bigcup \{F(\beta) :\beta < \lambda\}$
+            for $\lambda \in \OR$ a limit.
+    \end{enumerate}
 \end{proof}
 \begin{notation}
     Usually, one write $V_\alpha$ for $F(\alpha)$.
@@ -37,4 +73,97 @@ Applications of induction and recursion:
     such that $x \in V_\alpha$,
     i.e.~$V = \bigcup \{V_{\alpha} : \alpha \in \OR\}$.
 \end{lemma}
+\begin{proof}
+    We use induction on the well-founded $\in$-relation.
+    Let $A = \bigcup \{V_\alpha : \alpha \in \OR\}$.
+    We need to show that $A = V$.
+    By induction it suffices to prove that for every $x \in V$,
+    if $\{y : y \in x\} \subseteq  A$, then $x \in A$.
+    The hypothesis says that for all $y \in x$,
+    there is some $\alpha$ with $y \in V_\alpha$.
+    Write $\alpha_y$ for the least such $\alpha$.
+    By \AxRep, $\{\alpha_y : y \in x\}$
+    is a set and we may let
+    $\alpha = \sup \{\alpha_y : y \in x\} \ge \alpha_y$
+    for all $y \in x$.
+    Then $y \in V_{\alpha_y} \subseteq  V_\alpha$
+    for all $y \in x$.
+
+    In other words $x \subseteq V_\alpha$,
+    hence $x \in V_{\alpha+1}$.
+\end{proof}
+
+\begin{lemma}[\vocab{Transitive collapse}/\vocab{Mostowski collapse}]
+    \yalabel{Mostowski Collapse}{Mostowski}{lem:mostowski}
+    Let $R$ be a binary set-like relation on a class $A$.
+    Then $R$ is well-founded iff there is a transitive class $B$
+    such that
+    \[
+         (B, \in\defon{B}) \cong (A, R),
+    \]
+    i.e.~there is an isomorphism $F$,
+    that is a function $F\colon  B \to A$
+    with $x \in y \iff (F(x),F(y)) \in R$ for $x,y \in B$.
+\end{lemma}
+\begin{proof}
+    ``$\impliedby$'' Suppose that $R$ is ill-founded
+    (i.e.~not well-founded).
+    Then there is some $(y_n : n < \omega)$ such that $y_n  \in A$
+    and $(y_{n+1}, y_n) \in R$ for all $n < \omega$.
+    But then if $F$ is an isomorphism as above,
+    \[
+    F^{-1}(Y_{n+1}) \in F^{-1}(Y_n)
+    \]
+    for all $n < \omega$ $\lightning$
+
+    ``$\implies$ '' Suppose that $R$ is well-founded.
+    We want a transitive class $B$ and a function $F\colon  B \leftrightarrow A$
+    such that
+    \[
+    x \in y \iff (F(x), F(y)) \in R.
+    \]
+    Equivalently $G\colon  A \leftrightarrow B$
+    with $(x,y) \in R$ iff $G(x) \in G(y)$ for all $x,y \in A$.
+
+    In other words, $G(y) = \{G(x) : (x,y) \in R\}$.
+    Such a function $G$ and class $B$ exist by the recursion theorem.
+\end{proof}
+
+\begin{lemma}[\vocab{Rank function}]
+    Let $R$ be a well-founded and set-like binary relation
+    on a class $A$.
+    Then there is a function $F\colon  A \to \OR$,
+    such that for all $x,y \in A$
+    \[(x,y) \in R \implies F(x) < F(y) F(x) < F(y).\]
+\end{lemma}
+\begin{proof}
+    By the recursion theorem,
+    there is $F$ such that
+    \[
+    F(y) = \sup \{F(x) + 1 : (x,y) \in R\}.
+    \]
+    This function is as desired.
+\end{proof}
+
+This does not skip any ordinals,
+as $F(y)$ is the least ordinal $> F(x)$
+for all $(x,y) \in R$.
+Thus $\ran(F)$ is transitive.
+So either $\ran(F) = \OR$
+or $\ran(F) \in \OR$.
+This $F$ is called the \vocab{rank function} for $(A, R)$.
+\begin{notation}
+    \[
+    \rk_R(x) = \|x\|_R  \coloneqq  F(x),
+    \]
+    and
+    \[
+    \rk_R \rank(R) \coloneqq \ran(F).
+    \]
+
+    In the special case that $R$ is a linear order on $A$,
+    hence a well-order,
+    $\rank(R)$ is called the \vocab{order type} of $R$
+    (or of $(A,R)$), written $\otp(R)$.
+\end{notation}
 

inputs/lecture_11.tex

@@ -158,21 +158,21 @@ so $|a| = \aleph_\beta$ for some $\beta \le |\alpha|$.
         \[
         \ran(\Gamma\defon{\aleph_\alpha \times  \aleph_\alpha}) \supseteq \aleph_\alpha,
         \] 
-        as otherwise $\Gamma\defon{\aleph_\alpha \times  \alepha_\alpha}: \alepha_{\alpha} \times  \alepha_\alpha \to  \eta$
+        as otherwise $\Gamma\defon{\aleph_\alpha \times  \aleph_\alpha}: \aleph_{\alpha} \times  \aleph_\alpha \to  \eta$
-        would be a bijection for some $\eta < \alepha_\alpha$,
+        would be a bijection for some $\eta < \aleph_\alpha$,
-        but $\alepha_\alpha$ is a cardinal.
+        but $\aleph_\alpha$ is a cardinal.
 
-        Suppose that $\ran(\Gamma\defon{\aleph_\alpha \times  \alepha_\alpha}) \supsetneq \aleph_\alpha$.
+        Suppose that $\ran(\Gamma\defon{\aleph_\alpha \times  \aleph_\alpha}) \supsetneq \aleph_\alpha$.
         Then there exist $\eta, \eta' < \aleph_\alpha$ with
         \[
-        \Gamma((\eta, \eta')) = \alepha_\alpha.
+        \Gamma((\eta, \eta')) = \aleph_\alpha.
         \] 
 
         So $\Gamma\defon{\{(\gamma,\delta) : (\gamma,\delta) <^\ast (\eta, \eta'\}}$
-        is bijective onto $\alepha_\alpha$.
+        is bijective onto $\aleph_\alpha$.
         If $(\gamma,\delta) <^\ast (\eta, \eta')$,
         then $\max \{\gamma,\delta\} \le \max \{\eta, \eta'\}$.
-        Say $\eta \le  \eta' < \alepha_\alpha$
+        Say $\eta \le  \eta' < \aleph_\alpha$
         and let $\aleph_\alpha = |\eta'|$.
         There is a surjection
         \[f\colon  \underbrace{(\eta +1)}_{ \le \aleph_\beta} \times \underbrace{(\eta' + 1)}_{\sim  \aleph_\beta} \twoheadrightarrow \aleph_\alpha.\]

logic.sty

@@ -135,5 +135,9 @@
 \newcommand{\concat}{{}^\frown}
 \DeclareMathOperator{\hght}{height}
 
+\DeclareSimpleMathOperator{rank}
+\DeclareSimpleMathOperator{rk}
+\DeclareSimpleMathOperator{otp}
+
 \newcommand\lecture[3]{\hrule{\color{darkgray}\hfill{\tiny[Lecture #1, #2]}}}