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Merge branch 'master' of https://git.abstractnonsen.se/josia-notes/w23-logic-2

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Josia Pietsch 2024-02-16 19:28:29 +01:00
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2
.gitea/workflows/build.yaml
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@ -12,7 +12,7 @@ jobs:
- name: Prepare pages
run: |
mkdir public
mv build/logic2.pdf build/logic2.log README.md public
mv build/*.pdf build/*.log README.md public
- name: Deploy to pages
uses: actions/pages@v1
with:

2
inputs/lecture_01.tex
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@ -109,6 +109,7 @@
If $A \sim B$, there is a bijection $h\colon A \to B$.
\end{theorem}
\begin{proof}
\gist{%
Let $f\colon A \hookrightarrow B$ and $g\colon B \hookrightarrow A$
be injective.
We need to define a bijection $h\colon A \to B$.
@ -146,6 +147,7 @@
It is clear that this is bijective.
\todo{missing picture $f(A^{\text{odd}}) \subseteq B^{\text{even}}$, $f(A^\infty) = B^\infty$}.
}{Preimage sequence}
\end{proof}
\begin{definition}

10
inputs/lecture_02.tex
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@ -1,4 +1,5 @@
\lecture{02}{2023-10-19}{Topology on $\R$}
\gist{%
\begin{definition}
A set $O \subseteq \R$ is called \vocab{open} in $\R$ iff it is the union
of a set of open intervals.
@ -20,6 +21,7 @@
\begin{remark}+
$\{O \subseteq \R\} \sim 2^{\aleph_0} < \cP(\R)$.
\end{remark}
}{}
\begin{definition}
We call $x \in \R$
@ -28,9 +30,11 @@
there is some $y \in A$, $y \in (a,b)$, $y \neq x$.
We write \vocab{$A'$} for the set of all accumulation points of $A$.
\end{definition}
\gist{%
\begin{example}
$\{\frac{1}{n+1} | n \in \N\}' = \{0\}$.
\end{example}
}{}
\begin{lemma}
\label{lem:closedaccumulation}
@ -38,9 +42,11 @@
\end{lemma}
\begin{refproof}{lem:closedaccumulation}
``$\implies$''
\gist{%
Let $A$ be closed. Suppose that $x \in A' \setminus A$.
Then there exists $(a,b) \ni x$
disjoint from $A$. Hence $x \not\in A' \lightning$
}{trivial.}
``$\impliedby$''
Suppose $A' \subseteq A$.
@ -48,6 +54,7 @@
$A \subseteq \R$ is closed iff all Cauchy sequences
in $A$ converge in $A$.
\end{claim}
\gist{%
\begin{subproof}
Let $A$ be closed and $\langle x_n : n \in \omega \rangle$
a Cauchy sequence in $A$.
@ -62,6 +69,7 @@
we may pick $x_n \in (x - \frac{1}{n+1}, x + \frac{1}{n+1}) \cap A$
for all $n < \omega$.
\end{subproof}
}{}
Now if $A' \subseteq A$ and $A$ were not closed,
there would be some Cauchy-sequence $(x_n)$
@ -92,6 +100,7 @@ We want to prove two things:
Then $P \sim \R$.
\end{lemma}
\begin{proof}
\gist{%
It suffices to find an injection $f\colon \R \hookrightarrow P$.
We have $\underbrace{\{0,1\}^{\omega}}_{\text{infinite 0-1-sequences}} \sim \R$,
hence it suffices to construct $f\colon \{0,1\}^\omega\hookrightarrow P$.
@ -132,4 +141,5 @@ We want to prove two things:
and $f(t') \in [a_{t'\defon{n}}, b_{t'\defon{n}}]$
which are disjoint.
Thus $f(t) \neq f(t')$, i.e.~$f$ is injective.
}{Cantor scheme.}
\end{proof}

6
inputs/lecture_03.tex
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@ -77,7 +77,7 @@ all condensation points are accumulation points.
\begin{subproof}
$P \neq \emptyset$: $\checkmark$
$P \subseteq P'$ (i.e. $P$ is closed):
$P \subseteq P'$:
% \begin{IEEEeqnarray*}{rCl}
% P &=& \{x \in A | \text{every open neighbourhood of $x$ is uncountable}\}\\
% &\subseteq & \{x \in A | \text{every open neighbourhood of $x$ is at least countable}\} = P'.
@ -97,7 +97,7 @@ all condensation points are accumulation points.
But then $(a,b) \cap A$ is at most countable
contradicting $ x \in P$.
$P' \subseteq P$ :
$P' \subseteq P$ (i.e.~$P$ is closed):
Let $x \in P'$.
Then for $a < x < b$ the set
$(a,b) \cap P$
@ -111,7 +111,7 @@ all condensation points are accumulation points.
\]
\end{refproof}
\todo{Alternative proof of Cantor-Bendixson}
\gist{\todo{Alternative proof of Cantor-Bendixson}}{}
% \begin{remark}
% There is an alternative proof of Cantor-Bendixson, going as follows:
% Fix $A \subseteq \R$ closed.

14
inputs/lecture_04.tex
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@ -5,13 +5,14 @@
\section{\texorpdfstring{$\ZFC$}{ZFC}}
% 1900, Russel's paradox
\todo{Russel's Paradox}
% \todo{Russel's Paradox}
$\ZFC$ stands for
\begin{itemize}
\item \textsc{Zermelo}’s axioms (1905), % crises around 19000
\item \textsc{Fraenkel}'s axioms,
\item the \yaref{ax:c}.
\end{itemize}
\gist{
\begin{notation}
We write $x \subseteq y$ as a shorthand
for $\forall z.~(z \in x \implies z \in y)$.
@ -45,6 +46,7 @@ $\ZFC$ stands for
\forall u.~((u \in z) \iff (u \in x \land u \not\in y)).
\]
\end{notation}
}{Trivial, boring notation.}
$\ZFC$ consists of the following axioms:
\begin{axiom}[\vocab{Extensionality}]
\yalabel{Axiom of Extensionality}{(Ext)}{ax:ext} % (AoE)
@ -148,13 +150,15 @@ $\ZFC$ consists of the following axioms:
\begin{axiomschema}[\vocab{Replacement} (Fraenkel)]
\yalabel{Axiom of Replacement}{(Rep)}{ax:rep}
Let $\phi$ be some $\cL_{\in }$ formula
with free variables $x, y$.\todo{Allow more variables}
with free variables $x, y$.
Then
\[
\forall x \in a.~\exists !y \phi(x,y) \implies \exists b.~\forall x \in a.~\exists y \in b.~\phi(x,y).
\begin{IEEEeqnarray*}{l}
\forall v_1 \ldots \forall v_p.~\\
\left[\left( \forall x \exists! y.~\phi(x,y,\overline{v})\right) \to \forall a .~\exists b .~\forall y.~(y \in b \leftrightarrow \exists x (x \in a \land \phi(x,y, \overline{v}))\right]
\end{IEEEeqnarray*}
%\forall x \in a.~\exists !y \phi(x,y) \implies \exists b.~\forall x \in a.~\exists y \in b.~\phi(x,y).
% \forall v_1 \ldots \forall v_p .~\forall x.~ \exists y'.~\forall y.~(y = y' \iff \phi(x))
% \implies \forall a.~\exists b.~\forall y.~(y \in b \iff \exists x.~(x \in a \land \phi(x))
\]
\end{axiomschema}
\begin{axiom}[\vocab{Choice}]

8
inputs/lecture_05.tex
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@ -23,6 +23,7 @@
\]
\end{definition}
\gist{%
\begin{definition}
For sets $x, y$ we write
$(x,y)$ for $\{\{x\}, \{x,y\}\}$.
@ -108,7 +109,7 @@
\end{definition}
\begin{fact}
Given sets $d, b$ then
Given sets $d, b$ then
$\leftindex^d b$ exists.
\end{fact}
\begin{proof}
@ -130,7 +131,7 @@
(In other mathematical fields, this is sometimes
denoted as $f(a)$. We don't do that here.)
\end{notation}
}{[Some boring definitions omitted.]}
\begin{definition}
A binary relation $\le $ on a set $a$
is a \vocab{partial order}
@ -168,7 +169,6 @@
x \in b \land \forall y \in b.~y \le x.
\]
In a similar way we define \vocab[Minimal element]{minimal elements}
and the \vocab{minimum} of $b$.
We say that $x $ is an \vocab{upper bound}
@ -181,6 +181,7 @@
(This does not necessarily exist.)
Similarly $\text{\vocab{$\inf$}}(b)$ is defined.
\end{definition}
\gist{
\begin{remark}+
Note that in a partial order,
a maximal element is not necessarily a maximum.
@ -200,6 +201,7 @@
We write $(a,\le_a) \cong (b, \le_b)$
if they are isomorphic.
\end{definition}
}{}
\begin{definition}
Let $(a,\le)$ be a partial order.
Then $(a,\le)$ is

40
inputs/lecture_06.tex
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@ -8,6 +8,7 @@
Then $a$ has a maximal element.
\end{theorem}
\begin{refproof}{thm:zorn}
\gist{%
Fix $(a, \le )$ as in the hypothesis.
Let $A \coloneqq \{ \{(b,x) : x \in b\} : b \subseteq a, b \neq \emptyset\}$.
Note that $A$ is a set (use separation on $\cP(\cP(a) \times \bigcup \cP(a))$).
@ -65,6 +66,36 @@
Then $B = B_{u_0}^{\le^{\ast\ast}}$.
So $\le^{\ast\ast} \in W$, but now $u_0 \in b$.
So $b$ must have a maximum.
\todo{Why does this prove the lemma?}
}{
\begin{itemize}
\item $A \coloneqq \{\{(b,x) : x \in b\}, b \subseteq a, b \neq \emptyset\}$.
\item \AxC $\leadsto$ choice function on $A$,
$f\colon \cP(a) \setminus \{\emptyset\} \to a$, $f(b) \in b$.
\item $\le^\ast$ on $a$:
\begin{itemize}
\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
\item $B_u^{\le'} = \{w \in a : w \text{ $\le $-upper bound of } \{v \in b : v <' u\} \}$.
\end{itemize}
\item $\le', \le '' \in W \implies \le' \substack{\subseteq\\\supseteq} \le''$:
\begin{itemize}
\item $(b, \le') \overset{g}{\cong} (c, \le'') (\defon{v})$.
\item $g = \id_b$:
\begin{itemize}
\item $u_0$ $\le'$ minimal with $g(u_0) \neq u_0$.
\item $\{w \in b : w <' u_0\} \overset{g \defon{\ldots}}{=} \{w \in c : w <'' g(u_0)\}$.
\item $B^{\le '}_{u_0} = B_{g(u_0)}^{\le ''} \neq \emptyset$,
so $u_0 = f(B^{\le '}_{u_0}) = f(B^{\le ''}_{g(u_0)}) = g(u_0) \lightning$
\end{itemize}
\end{itemize}
\item $\le^\ast \coloneqq \bigcup W$ is wo on $b \subseteq a$.
\item Suppose $b$ has no maximum. Then $B \cap b = \emptyset$.
\item $u_0 \coloneqq f(B)$, $\le^{\ast\ast} = \le^\ast \cup \{(u,u_0) | u \in b\} \cup \{(u_0,u_0)\}$.
\item $B = B_{u_0}^{\le^{\ast\ast}}$, so $\le^{\ast\ast} \in W$,
but $u_0 \in b \lightning$. ?
\end{itemize}
}
\end{refproof}
\begin{remark}
@ -82,6 +113,7 @@
Then $A$ contains a $\subseteq$-maximal element.
\end{corollary}
\gist{%
\begin{remark}[Cultural enrichment]
Other assertions which are equivalent
to the \yaref{ax:c}:
@ -96,12 +128,14 @@
\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}
\gist{
\begin{goal}
We want to define nice representatives of the equivalence classes
of well-orders.
@ -115,7 +149,7 @@ We can hence form the smallest inductive set
Note that $\omega$ exists, as it is a subset of the inductive
set given by \AxInf.
We call $\omega$ the set of \vocab{natural numbers}.
}{}
\begin{notation}
We write $0$ for $\emptyset$,
and $y + 1$ for $y \cup \{y\}$.
@ -147,16 +181,19 @@ We have the following principle of induction:
and for all $y, z \in x$,
we have that $y = z$, $y \in z$ or $y \ni z$.
\end{definition}
\gist{
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}
\gist{
We use \yaref{lem:induction}.
Clearly $\emptyset$ is an ordinal.
Now let $\alpha$ be an ordinal.
@ -169,6 +206,7 @@ Clearly, the $\in$-relation is a well-order on an ordinal $x$.
since $\alpha$ is an ordinal.
Suppose $x = \alpha$.
Then either $y = x$ or $y \in \alpha = x$.
}{Induction}
\end{proof}
\begin{lemma}

7
inputs/lecture_07.tex
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@ -20,6 +20,7 @@
\end{enumerate}
\end{lemma}
\begin{refproof}{lem:7:ordinalfacts}
\gist{
We have already proved (a) before.
(b) Fix $x \in \alpha$. Then $x \subseteq \alpha$.
@ -113,6 +114,7 @@
But this violates \AxFund,
as $\alpha_0 \in \beta_0 \in \alpha_0$.
\end{subproof}
}{Long and tedious, but not many ideas.}
\end{refproof}
\begin{lemma}
@ -146,6 +148,7 @@ for example $\bigcup \omega = \omega$.
Otherwise $\alpha$ is called
a \vocab[Ordinal!limit]{limit ordinal}.
\end{definition}
\gist{
\begin{observe}
Note that $\alpha$ is a limit ordinal iff for all
$\beta \in \alpha$, $\beta + 1 \in \alpha$:
@ -158,7 +161,8 @@ for example $\bigcup \omega = \omega$.
then by definition there is some $\beta \in \alpha$,
with $\beta + 1 = \alpha$, so $\beta + 1 \not\in \alpha$.
\end{observe}
}{}
\gist{
\begin{notation}
If $\alpha, \beta$ are ordinals,
we write $\alpha < \beta$ for $\alpha \in \beta$
@ -185,3 +189,4 @@ for example $\bigcup \omega = \omega$.
\item $\omega +1 = \omega \cup \{\omega\} , \omega + 2, \ldots$,
\end{itemize}
\end{example}
}{}

9
inputs/lecture_08.tex
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@ -1,6 +1,7 @@
\lecture{08}{2023-11-13}{Induction and recursion}
\subsection{Classes}
\gist{
It is often very handy to work in a class theory rather than
in set theory.
@ -11,10 +12,11 @@ sets (denoted by lower case letters)
and classes (denoted by capital letters),
as well as one binary relation symbol $\in$
for membership.
}{}
\vocab{Bernays-Gödel class theory} (\vocab{BG})
has the following axioms:
\gist{
\begin{axiom}[Extensionality]
\yalabel{Axiom of Extensionality}{(Ext)}{ax:bg:ext}
\[
@ -54,6 +56,7 @@ has the following axioms:
\exists x .~(\emptyset \in x \land \left( \forall y \in x .~y \cup \{y\} \in x \right)).
\]
\end{axiom}
}{\AxExt, \AxFund, \AxPair, \AxUnion, \AxPow, \AxInf as from $\ZF$.}
Together with the following axioms for classes:
@ -108,10 +111,10 @@ Furthermore $F\,''a \coloneqq \{y : \exists x \in a .~(x,y) \in F\}$.
\end{axiom}
\todo{notation: $\emptyset, \cap$}
\gist{\todo{notation: $\emptyset, \cap$}}{}
\todo{the following was actually done in lecture 9}
\gist{(The following was actually done in lecture 9, but has been moved here for clarity.)}{}
$\BGC$ (in German often NBG\footnote{\vocab{Neumann-Bernays-Gödel}})
is defined to be $\BG$

2
inputs/lecture_11.tex
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@ -75,7 +75,7 @@ We often write $\kappa, \lambda, \ldots$ for cardinals.
\end{itemize}
}
\end{proof}
We may now use the \yaref{lem:recursion}
We may now use the \yaref{thm:recursion}
to define a sequence $\langle \aleph_\alpha : \alpha \in \OR \rangle$
with the following properties:
\begin{IEEEeqnarray*}{rCl}

2
inputs/lecture_14.tex
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@ -322,7 +322,7 @@ We have shown (assuming \AxC to choose contained clubs):
and $S_1 \coloneqq \{\xi < \kappa : \cf(\xi) = \omega_1\}$.
Clearly these are disjoint.
They are both stationary:
Let $c \subseteq \kappa$ be a club.
Let $C \subseteq \kappa$ be a club.
Let $(\xi_i : i \le \omega_1)$
be defined as follows:
$\xi_0 \coloneqq \min C$,

2
inputs/lecture_15.tex
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@ -60,7 +60,7 @@
Recall the following:
\begin{definition}
A substructure $X \subseteq V_\theta$\todo{make this more general. Explain why $V_\theta$ is a model}
A substructure $X \subseteq V_\theta$\gist{\todo{make this more general. Explain why $V_\theta$ is a model}}{}
is an \vocab{elementary substructure}
of $V_\theta$,
denoted $X \prec V_{\theta}$,\footnote{more formally $(X,\in ) \prec (V_{\theta})$}

21
inputs/lecture_17.tex
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@ -236,8 +236,7 @@ We will only proof
\item $X \le Y :\iff \{\alpha < \omega_1 : f_X(\alpha) \le f_Y(\alpha)\}$ stationary.
\item (2) $X\le Y \lor Y \le X$:
Suppose $X \not\le Y, Y \not\le X$. Choose witnessing clubs $C, D$.
$C \cap D$ is club, but then $f_X(\alpha) \le f_Y(\alpha)$ or
$f_X(\alpha) \ge f_Y(\alpha)$ for $\alpha \in C \cap D$.
$C \cap D$ is club, so $C\cap D \ni \alpha \implies f_X(\alpha) \substack{\nleq\\\ngeq} f_Y(\alpha) \lightning$
\item (3) $X \subseteq \aleph_{ \omega_1}$, then $|\underbrace{\{Y \subseteq \aleph_{ \omega_1} : Y \le X\}}_{A}| \le \aleph_{ \omega_1}$
\begin{itemize}
\item Suppose $|A| \ge \aleph_{ \omega_1 + 1}$.
@ -245,9 +244,21 @@ We will only proof
$2^{\aleph_1} = \aleph_2 \implies$ at most $\aleph_2$ such $S_Y$.
\item If $\forall S \in \cP( \omega_1).~ |\underbrace{\{Y \in A : S_Y = S\}}_{A_S}| < \aleph_{ \omega_1 + 1}$,
then $|A| \le \aleph_2 \cdot <\aleph_{ \omega_1 + 1}$ $\lightning$ $\aleph_{ \omega_1 + 1}$ regular.
\item So $\exists S \in \omega_1.~|A_S| = \aleph_{ \omega_1 + 1 + 1}$.
% TODO TODO TODO hier weiter!
\item So $\exists S \in \omega_1.~|A_S| = \aleph_{ \omega_1 + 1}$.
\item Fix surjection $\langle g_\alpha : \aleph_\alpha \twoheadrightarrow f_X(\alpha) + 1 : \alpha \in S \rangle$.
($f_Y(\alpha) \le f_X(\alpha) < \aleph_{\alpha+1}$)
\item $\forall Y \in A_S$ define $\overline{f}_Y \colon S \to \aleph_{ \omega_1}, \alpha \mapsto \min \{\xi : g_\alpha(\xi) = f_Y(\alpha)\}$.
\item $S^\ast$ ($S \cap$ limit ordinals) is stationary.
\item $\forall Y \in A$ define $h_Y\colon S^\ast \to \omega_1, \alpha \mapsto \min \{\beta < \alpha : \overline{f}_Y(\alpha) < \aleph_\beta\}$.
\item Apply \yaref{thm:fodor} to $h_Y, S^\ast$ to get $T_Y \subseteq S^\ast$ stationary with $h_Y\defon{T_Y}$ constant.
\item $\exists T.~|\underbrace{\{Y \in A_S : T_Y = T\}}_{A_{S,T}}| = \aleph_{ \omega_1 + 1}$.
\item Let $\{\beta\} = h_Y''T$, i.e.~$\overline{f}_Y(\alpha) < \aleph_{\beta}$ for $Y\in A_{S,T}, \alpha \in T$.
\item $\leftindex^T \aleph_\beta \le 2^{\aleph_\beta \cdot \aleph_1} = \aleph_{\beta+1} \aleph_2 < \aleph_{ \omega_1}$.
\item $\exists \tilde{f}\colon T \to \aleph_\beta .~ |\underbrace{\{Y \in A_{S,T} : \overline{f}_Y\defon{T} = \tilde{f}\} }_{A_{S,T,\tilde{f}}}| = \aleph_{ \omega_1 + 1}$.
\item $Y,Y' \in A_{S,T,\tilde{f}} \implies \forall \alpha \in T.~f_Y(\alpha) = f_{Y'}\left( \alpha \right) \overset{T \text{ unbounded}}{\implies} Y = Y'$
Thus $|A_{S,T, \tilde{f}} | \le 1 \lightning$.
\end{itemize}
\item Define sequence $\langle X_i : i < \aleph_{ \omega_1 + 1} \rangle$
of subsets of $\aleph_{\omega_1}$:
\begin{itemize}
@ -259,7 +270,7 @@ We will only proof
\item $P \coloneqq \{Y \subseteq \aleph_{ \omega_1} : \exists i < \aleph_{ \omega_1 + 1}.~Y \le X_i\}$
\end{itemize}
\item $|P| \overset{\text{in fact } =}{\le} \aleph_{ \omega_1 + 1} \aleph_{ \omega_1} = \aleph_{ \omega_1 + 1}$ by (3).
\item $X \in \cP(\aleph_{ \omega_1}) \setminus M \implies \forall i < \aleph_{ \omega_1 + 1}.~X_i \le X$
\item $X \in \cP(\aleph_{ \omega_1}) \setminus P \implies \forall i < \aleph_{ \omega_1 + 1}.~X_i \le X$
$\lightning$ (3).
Thus $P = \cP(\aleph_{ \omega_1})$.
\end{itemize}

52
inputs/lecture_18.tex
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@ -77,6 +77,7 @@
\end{enumerate}
\end{theorem}
\begin{proof}
\gist{%
2. $\implies$ 1.:
Fix $j\colon V \to M$.
Let $U = \{X \subseteq \kappa : \kappa \in j(X)\}$.
@ -221,11 +222,60 @@ such that
as otherwise $\forall \delta < \alpha. ~ \kappa \setminus X_\delta \in U$,
i.e.~$\emptyset = (\bigcap_{\delta < \alpha} \kappa \setminus X_{\delta}) \cap X \in U \lightning$.
We get $[f] = [c_{\delta}]$,
so $\beta = \sigma([f]) = \delta([c_{\delta}]) = j(\delta) = \delta$,
so $\beta = \sigma([f]) = \sigma([c_{\delta}]) = j(\delta) = \delta$,
where for the last equality we have applied the induction hypothesis.
So $j(\alpha) \le \alpha$.
For all $\eta < \kappa$, we have $\eta = \sigma([c_{\eta}]) < \sigma([c_{\id}]) < \sigma([c_{\kappa}])$,
so $j(\kappa) > \kappa$.
% It is also easy to show $j(\kappa) > \kappa$.
}{%
$2 \implies 1$:
Fix $j\colon V \to M$. $U \coloneqq \{X \subseteq \kappa : \kappa \in j(X)\}$ is an UF:
\begin{itemize}
\item $M \models j(X \cap Y) = j(X) \cap j(Y)$,
hence $\kappa \in j(X) \cap j(Y) \implies \kappa \in j(X \cap Y)$
\item $M \ni X \subseteq Y \implies Y \in M$, $\emptyset \not\in M$ (same argument)
\item $\kappa \in U$:
\begin{itemize}
\item $\forall \alpha \in \OR.~ j(\alpha) \in \Ord, j(\alpha) \ge \alpha$:
\begin{itemize}
\item Write $\alpha\in \OR$ and use $\alpha \in \OR \iff M \models j(\alpha) \in \OR$.
\item Suppose $j(\alpha) < \alpha$, $\alpha$ minimal,
but $M \models j(j(\alpha)) < j(\alpha) \implies j(j(\alpha)) < j(\alpha)$.
\end{itemize}
\item $j(\kappa) \neq \kappa \implies j(\kappa) > \kappa \implies \kappa \in j(\kappa)$.
\end{itemize}
\item Ultrafilter: $\kappa \in j(\kappa) = j(X \cup (\kappa \setminus X)) = j(X) \cup j(\kappa \setminus X)$.
\item $< \kappa$ closed: For $\theta < \kappa$, $X_i \in U$:
$\kappa \in \bigcap_{i < \theta} j(X_i) = j\left( \bigcap_{i < \theta} X_i \right) \in U$.
($j(\theta) = \theta$, so $j\left( \langle X_i : i < \theta \rangle \right) = \langle j(X_i) : i < \theta \rangle$.
\item Not principal:
$\xi < \kappa \implies j(\{\xi\}) = \{\xi\} \not\ni \kappa$.
\end{itemize}
$1 \implies 2$:
\begin{itemize}
\item Fix $U$. Consider $\leftindex^\kappa V$.
\item $f \sim g :\iff \{\xi < \kappa: f(\xi) = g(\xi)\} \in U$.
\item $[f] \coloneqq \{g : g \sim f \land g \in V_\alpha \text{ for minimal } \alpha\}$ (Scott's Trick).
\item $[f] \tilde{\in } [g] :\iff \{\xi < \kappa: f(\xi ) \in g(\xi)\} \in U$.
\item \yaref{thm:los}: $(\cF, \tilde{\in }) \models \phi([f_1], \ldots, [f_k]) \iff \{\xi < \kappa: (V, \in ) \models \phi(f_1(\xi), \ldots, f_k(\xi))\} \in U$.
\item $\overline{j}(x) \coloneqq [ \xi \mapsto x]$.
\item $\tilde{\in }$ well-founded: lift decreasing sequences ($U$ is $<\kappa$ closed, $ \omega < \kappa$)
\item $\tilde{\in }$ set-like $\overset{\yaref{thm:mostowksi}}{\leadsto}$ $(\cF, \tilde{\in }) \overset{\sigma}{\cong} (M, \in )$.
\item $j \coloneqq \sigma \circ \overline{j}$.
\item $\alpha < \kappa \implies j(\alpha) = \alpha$ (know $j(\alpha) \ge \alpha$):
\begin{itemize}
\item Induction for $j(\alpha) \le \alpha$: Fix $\alpha$, $\sigma([f]) = \beta \in j(\alpha)$.
\item $[f] \tilde{\in }[c_\alpha]$.
\item $\exists \delta < \alpha.~[f] = [c_\delta]$ ($U $ is $<\kappa$ closed)
\item $\beta = \sigma([f]) = \sigma([c_\delta]) = j(\delta) \overset{\text{IH}}{=} \delta \in \alpha$.
\end{itemize}
\item $j(\kappa) \neq \kappa$:
$\forall \eta < \kappa.~\eta = \sigma([c_{\eta}]) < \sigma([\id]) < \sigma([\kappa])$.
\end{itemize}
}
\end{proof}
\begin{theorem}[\L o\'s]

2
inputs/lecture_22.tex
View file

@ -137,7 +137,7 @@ is well founded.
so $M[g] \models \text{``$\{x,y\}$ is the pair of $x$ and $y$''}$.
Hence $M[g] \models \AxPair$.
\item \AxUnion:
Similar to \AxPair.\todo{Exercise}
Similar to \AxPair.\gist{\todo{Exercise}}{}
\end{itemize}
\end{proof}