diff --git a/inputs/lecture_06.tex b/inputs/lecture_06.tex index 1a52274..8bfebf0 100644 --- a/inputs/lecture_06.tex +++ b/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} + + + + + + + + diff --git a/logic2.tex b/logic2.tex index c06909b..bf32004 100644 --- a/logic2.tex +++ b/logic2.tex @@ -29,6 +29,7 @@ \input{inputs/lecture_03} \input{inputs/lecture_04} \input{inputs/lecture_05} +\input{inputs/lecture_06} \cleardoublepage