lecture 6

inputs/lecture_06.tex

@@ -1 +1,235 @@
-\lecture{06}{}{AC and the well ordering theorem}
+\lecture{06}{2023-11-06}{}
+
+\begin{theorem}[Zorn]
+    \yalabel{Zorn's Lemma}{Zorn}{thm:zorn}
+    Let $(a, \le )$ be a partial order with $a \neq \emptyset$.
+    Assume that $b \le  a$ with $b \neq \emptyset$
+    and $ b$ linearly ordered, $b$ has an upper bound,
+    Then $a$ has a maximal element.
+\end{theorem}
+\begin{proof}
+    Fix $(a, \le )$ as in the hypothesis.
+    Let $A \coloneqq  \{ \{(b,x) : x in b\}  : b \le  a, b \neq \emptyset\}$.
+    Note that $A$ is a set (use separation on  $\cP(\cP(a) \times  \bigcup \cP(a))$).
+    Note further that if $b_1 \neq  b_2$,
+    then $\{(b_1, x) : x \in  b_1\} $
+    and $\{(b_2, x) : x \in b_2\}$ are disjoint.
+    Hence the axiom of choice
+    gives us a choice function $f$ on $A$,
+    i.e.~$\forall b \in  \cP(a) \setminus \{\emptyset\} .~(f(b) \in b)$.
+
+    Now define a binary relation $\le^\ast$:
+    We let $W$ denote the set of all well-orderings $\le'$
+    of subsets $b \subseteq a$,
+    such that for all $u,v \in b$
+    if $u \le' v$ then $u \le v$
+    and for all $u \in  b$
+    and
+    \[
+    B_u^{\le'} \coloneqq  \{ w \in  a : w \text{ is an $\le$-upper bound of $\{v \in  b : v \le' u\}$}\}
+    \]
+    then $B_u^{\le'} \neq \emptyset$ and $f(B^{\le'}_u) = u$.
+
+    \begin{claim}
+        If $\le', \le'' \in W$,
+        then $\le' \subseteq \le''$ or $\le'' \subseteq \le'$.
+    \end{claim}
+    \begin{subproof}
+        Let $\le' \in W$ be a well-ordering of $b \subseteq a$
+        and let $\le'' \in W$ be a well-ordering on $c \subseteq a$.
+        We know that wlog.~$(b, \le') \cong (c, \le'')$
+        or $\exists v \in c .~(b, \le') \cong (c, \le'')\defon{v}$.
+        Let $g\colon b \to c$ or $g\colon b \to c\defon{v}$ be a witness.
+        We want to show that $g = \id$.
+        Suppose that $g \neq \id$.
+        Let $u_0 \in  b$ be $\le'$-minimal such that $g(u_0) \neq u_0$.
+        Writing $\overline{g} \coloneqq  g\defon{\{w \in b: w <' u_0\}}$,
+        then $(b, \le ')\defon{u_0} \cong (c, \le'') \defon{g(u_0)}$
+        and $\overline{g}$ is in fact the identity on $\{w \in b | w \le' u_0\}$
+        but this means $\{w \in b | w <' u_0\} = \{w \in c | w <'' g(u_0)\}$
+        and $B_{u_0}^{\le'} = B_{g(u_0)}^{\le''} \neq \emptyset$.
+        Then $u_0 = f(B_{u_0}^{\le'}) = f(B_{g(u_0)}^{\le''}) = g(u_0)$.
+        Thus $g$ is the identity.
+    \end{subproof}
+    Given the claim, we can now see that $\bigcup W$ is a well order $\le^{\ast\ast}$
+    of $a$.
+    Let $B = \{w \in a | w \text{ is a $\le$-upper bound of $b$}\}$
+    (this is not empty by the hypothesis).
+    Suppose that $b$ does not have a maximum.
+    Then $B \cap b = \emptyset$.
+    Now $f(B) = u_0$
+    and let
+    \[
+    \le^{\ast\ast} = \le^{\ast} \cup  \{(u,u_0) | u \in b\} \cup  \{(u_0,u_0)\}.
+    \]
+    Then $B = B_{u_0}^{\le^{\ast\ast}}$.
+    So $\le^{\ast\ast} \in W$, but now $n_0 \in b$.
+    So $b$ must have a maximum.
+\end{proof}
+
+\begin{remark}
+    Over $\ZF$ the axiom of choice and \yaref{thm:zorn}
+    are equivalent.
+\end{remark}
+
+\begin{corollary}[Hausdorff's maximality principle]
+    Let $a \neq  \emptyset$.
+    Let $A \subseteq  \cP(a)$ be such that $\forall  B \subseteq  A$,
+    if $x \subseteq y \lor y \subseteq x$
+    for all $x,y \in B$,
+    then there is some $z \in  A$
+    such that $x \subseteq z$ for all $x \in B$.
+    Then $A$ contains a $\subseteq$-maximal element.
+\end{corollary}
+
+\begin{remark}[Cultural enrichment]
+    Other assertion which are equivalent
+    to the axiom of choice:
+    \begin{itemize}
+        \item Every infinite family of non-empty sets
+            $\langle  a_i : i \in I \rangle$
+            has non-empty product,
+            i.e.
+            \[
+            \prod_{i \in I} a_i \neq  \emptyset.%\footnote{This is clearly true.}
+            \]
+        \item Every set can be well-ordered.%\footnote{This is clearly false.}
+    \end{itemize}
+\end{remark}
+% \begin{remark}
+%     The axiom of choice is true.
+% \end{remark}
+
+\pagebreak
+\subsection{The Ordinals}
+\begin{goal}
+    We want to define nice representatives of the equivalence classes
+    of well-orders.
+    % TODO theorem
+\end{goal}
+Recall that (AoI) states the existence of an inductive set $x$.
+We can hence form the smallest inductive set
+\[
+\omega \coloneqq  \bigcap \{ x : x \text{ is inductive}\}
+\]
+Note that $\omega$ exists, as it is a subset of the inductive
+set given by AoI.
+We call $\omega$ the set of \vocab{natural numbers}.
+
+\begin{notation}
+    We write $0$ for $\emptyset$,
+    and $y + 1$ for $y \cup  \{y\}$.
+\end{notation}
+With this notation the AoI is equivalent to
+\[
+\exists  x_0.~(0 \in  x_0 \land \forall  n. ~(n \in x_0 \implies n+1 \in x_0)).
+\]
+
+We have the following principle of induction:
+\begin{lemma}
+    \yalabel{Induction}{Induction}{lem:induction}
+    Let $A \subseteq \omega$ such that $0 \in A$
+    and for each $y \in A$, we have that $y + 1 \in A$.
+    Then $A = \omega$.
+\end{lemma}
+\begin{proof}
+    Clearly $A$ is an inductive set,
+    hence $\omega \subseteq A$.
+\end{proof}
+
+\begin{definition}
+    A set $x$ is \vocab{transitive},
+    if $\forall y \in x.~y \subseteq x$.
+\end{definition}
+\begin{definition}
+    A set $x$ is called an \vocab{ordinal} (or \vocab{ordinal number})
+    iff $x$ is transitive
+    and for all $y, z \in x$,
+    we have that $y = z$, $y \in z$ or $y \ni z$.
+\end{definition}
+Clearly, the $\in$-relation is a well-order on an ordinal $x$.
+\begin{remark}
+    This definition is due to \textsc{John von Neumann}.
+\end{remark}
+
+\begin{lemma}
+    Each natural number (i.e.~element of $\omega$)
+    is an ordinal.
+\end{lemma}
+\begin{proof}
+    We use \yaref{lem:induction}.
+    Clearly $\emptyset$ is an ordinal.
+    Now let $\alpha$ be an ordinal.
+    We need to show that $\alpha + 1$ is an ordinal.
+    It is transitive, since $\alpha$ is transitive
+    and $\alpha \subseteq (\alpha + 1)$.
+
+    Let $x, y \in (\alpha+1)$.
+    If $x, y \in \alpha$, we know that $x = y \lor x \in y \lor x \ni y$
+    since $\alpha$ is an ordinal.
+    Suppose $x = \alpha$.
+    Then either $y = x$ or $y \in \alpha = x$.
+\end{proof}
+
+\begin{lemma}
+    $\omega$ is an ordinal.
+\end{lemma}
+\begin{proof}
+    $\omega$ is transitive:
+
+    Let $y \in \omega$. Let us show by \yaref{lem:induction},
+    that $y \subseteq \omega$.
+    For $y = \emptyset$ this is clear.
+
+    Suppose that $y \in \omega$ with $y \subseteq \omega$.
+    But now $\{y\} \subseteq \omega$,
+    so $y + 1 = y \cup \{y\} \subseteq \omega$.
+
+
+    $\omega$ is well-ordered by $\in$:
+
+    We do a nested induction. First let
+    \[
+    \phi(y,z) \coloneqq  y \in z \lor y \ni z \lor y = z.
+    \] 
+    We want to show:
+    \begin{enumerate}[(a)]
+        \item $\phi(0,0)$
+        \item $\forall z \in \omega. ~\phi(0, z) \implies \phi(0,z+1)$.
+        \item $\forall  y \in \omega.((\forall z' \in \omega.~\phi(y,z')) \implies (\forall z \in  \omega.~\phi(y+1, z')))$.
+    \end{enumerate}
+    (a) and (b) are trivial.
+    Fix $y \in \omega$ and
+    suppose that $\forall z' \in \omega .~\phi(y, z')$.
+    We want to show that $\forall z \in  \omega .~\phi(y+1, z)$.
+
+    We already know that $\forall  z \in \omega.~\phi(0,z)$ holds
+    by (b).
+    In particular, $\phi(0,y+1)$ holds,
+    so $\phi(y+1, 0)$ is true, since $\phi$ is symmetric.
+    Now if $\phi(y+1,z)$ is true,
+    we want to show $\phi(y+1,z+1)$ is true as well.
+    We have $y + 1 \in z \lor y + 1 = z \lor y + 1 \ni z$
+    by assumption.
+    \begin{itemize}
+        \item If $y + 1 \in z \lor y+1 = z$, then clearly $y + 1 \in z + 1$.
+        \item If $y +1 \ni z$, then either $z = y$ or $z \in y$.
+        \begin{itemize}
+            \item In the first case, $z+1 = y+1$.
+            \item Suppose that $z \in y$.
+                Then by the induction hypothesis $\phi(y, z+1)$ holds.
+                If $y \in z+1$, then $\{y,z\}$ would violate AoF.
+                If $y = z+1$, then $z + 1 \in y + 1$.
+                If $z+1 \in y$, then $z+1 \in y+1$ as well.
+        \end{itemize}
+    \end{itemize}
+        
+\end{proof}
+
+
+
+
+
+
+
+

logic2.tex

@@ -29,6 +29,7 @@
 \input{inputs/lecture_03}
 \input{inputs/lecture_04}
 \input{inputs/lecture_05}
+\input{inputs/lecture_06}
 
 
 \cleardoublepage