w23-logic-2/inputs/lecture_06.tex

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\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}