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