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2
inputs/lecture_04.tex
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@ -2,7 +2,7 @@
% Model-theoretic concepts and ultraproducts
\section{$\ZFC$}
\section{\texorpdfstring{$\ZFC$}{ZFC}}
% 1900, Russel's paradox
\todo{Russel's Paradox}

4
inputs/lecture_05.tex
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@ -104,12 +104,12 @@
We write $f\colon d \to b$.
The set of all function from $d$ to $b$
is denoted by ${}^d b$ or $b^d$.
is denoted by $\leftindex^d b$ or $b^d$.
\end{definition}
\begin{fact}
Given sets $d, b$ then
${}^d b$ exists.
$\leftindex^d b$ exists.
\end{fact}
\begin{proof}
Apply again \AxAus over $\cP(d \times b)$.

7
inputs/lecture_08.tex
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@ -103,7 +103,7 @@ Furthermore $F\,''a \coloneqq \{y : \exists x \in a .~(x,y) \in F\}$.
and $\phi$ does not have quantifiers
ranging over classes.%
\footnote{If one removes the restriction regarding
quantifiers another theory, called
quantifiers, another theory, called
\vocab{Morse-Kelly} set theory, is obtained.}
\end{axiom}
@ -113,7 +113,8 @@ Furthermore $F\,''a \coloneqq \{y : \exists x \in a .~(x,y) \in F\}$.
\todo{the following was actually done in lecture 9}
$\BGC$ (in German often NBG) is defined to be $\BG$
$\BGC$ (in German often NBG\footnote{\vocab{Neumann-Bernays-Gödel}})
is defined to be $\BG$
together with the additional axiom:
\begin{axiom}[Choice]
\[
@ -212,7 +213,7 @@ $\ZFC \vdash \phi$ then $\BGC \vdash \phi$.
&&~ ~\overline{g}(0) = x \land \forall m \in \omega.~(m+1 \in \dom(\overline{g}) \implies \overline{g}(m+1) = f(\overline{g}(m)))\}.
\end{IEEEeqnarray*}
$G$ exists as it can be obtained by \AxAus
from ${}^{< \omega}M$.
from $\leftindex^{< \omega}M$.
By induction,
for every $n \in \omega$,
there is a $\overline{g} \in G$

34
inputs/lecture_09.tex
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@ -48,6 +48,7 @@ An alternative way of formulating this is
\end{definition}
\begin{theorem}
\yalabel{Recursion Theorem}{recursion}{thm:recursion}
Let $R$ be a well-founded
and set-like relation on $A$ (i.e.~$R \subseteq A \times A$).
@ -59,11 +60,12 @@ An alternative way of formulating this is
Then there is a unique function $f$ on $A$
such that for all $x \in A$,
\[
(F\defon{ \{y \in A : (y,x) \in R\}}, x, F(x)) \in D.
(F\defon{ \{y \in A : (y,x) \in R\}}, x, F(x)) \in D\gist{,}{.}
\]
I.e.~$F(x)$ is computed from $F\defon{\{y \in A: (y,x) \in R\}}$.
\gist{i.e.~$F(x)$ is computed from $F\defon{\{y \in A: (y,x) \in R\}}$.}{}
\end{theorem}
\begin{proof}
\gist{%
Uniqueness:
Let $F, F'$ be two such functions.
@ -106,7 +108,7 @@ An alternative way of formulating this is
If $x \in \dom(F)$ and $y \in A$ with $(y,x) \in R$,
then $y \in \dom(F)$
and $(F\defon{\{y : (y,x) \in R\}}, x,F(x)) \in D$.
and \[(F\defon{\{y : (y,x) \in R\}}, x,F(x)) \in D.\]
We need to show that $\dom(F) = A$.
This holds by induction:
@ -121,8 +123,26 @@ An alternative way of formulating this is
But then $f \cup (x,z)$,
where $z$ is unique such that $(f\defon{\{y : (y,x) \in R\}}, x, z) \in D$,
is good $\lightning$.
}{
\begin{itemize}
\item Uniqueness:
Consider $\overline{A} \coloneqq \{x \in A : F(x) \neq F'(x)\}$
for two such functions. If $\overline{A} \neq \emptyset$,
there is a minimal element $\lightning$.
\item Existence:
\begin{itemize}
\item call set-function $f$ \emph{good} iff:
\begin{itemize}
\item $\dom(f) \subseteq A$,
\item $x \in \dom(f), y <_R x \implies y \in \dom(f)$,
\item $\forall x \in \dom(f).~(f\defon{\{y \in A : y <_R x\}}, x, f(x)) \in D$.
\end{itemize}
\item $F \coloneqq \bigcup \{f : f \text{ good}\}$ (comprehension)
\item $\dom F = A$ (induction).
\end{itemize}
\end{itemize}
}
\end{proof}

98
inputs/lecture_10.tex
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@ -1,6 +1,5 @@
\lecture{10}{}{} % Mirko
Applications of induction and recursion:
\subsubsection{Applications of induction and recursion}
\begin{fact}
For every set $x$ there is a transitive set $t$
such that $x \in t$.
@ -14,7 +13,7 @@ Applications of induction and recursion:
Once we have such a function,
$\{x\} \cup \bigcup \ran(F)$
is a set as desired.
To get this $F$ using the recursion theorem,
To get this $F$ using the \yaref{thm:recursion},
pick $D$ such that
\[
(\emptyset, 0, \{x\}) \in D
@ -23,7 +22,7 @@ Applications of induction and recursion:
\[
(f, n+1, \bigcup\bigcup \ran(f)) \in D.
\]
The recursion theorem then gives a function
The \yaref{thm:recursion} then gives a function
such that
\begin{IEEEeqnarray*}{rCl}
F(0) &=& \{x\},\\
@ -51,21 +50,30 @@ Applications of induction and recursion:
}{}
\begin{lemma}
There is a function $F\colon \OR \to V$
such that $F(\alpha) = \bigcup \{\cP(F(\beta)): \beta < \alpha\}$.
such that
\gist{$F(\alpha) = \bigcup \{\cP(F(\beta)): \beta < \alpha\}$.}{%
$F(\alpha) = \bigcup_{\beta < \alpha} \cP(F(\beta))$.
}
\gist{}{%
$F(\alpha)$ is denoted by $V_\alpha$.
These are called the \vocab{rank initial segments} of $V$.
}
\end{lemma}
\begin{proof}
Use the recursion theorem with $R = \in $
Use the \yaref{thm:recursion} with $R = \in $
and $(w,x,y) \in D$ iff
\[
y = \bigcup \{\cP(\overline{y}) : \overline{y} \in \ran(w)\}.
\]
This function has the following properties:
\begin{IEEEeqnarray*}{rCl}
F(0) &=& \bigcup \emptyset = \emptyset,\\
F(1) &=& \bigcup \{\cP(\emptyset)\} = \bigcup \{\{\emptyset\} \} = \{\emptyset\},\\
F(2) &=& \bigcup \{\cP(\emptyset), \cP(\{\emptyset\})\} = \bigcup \{\{\emptyset\}, \{\emptyset, \{\emptyset\} \} \} = \{\emptyset, \{\emptyset\} \},\\
\ldots
\end{IEEEeqnarray*}
\gist{%
This function has the following properties:
\begin{IEEEeqnarray*}{rCl}
F(0) &=& \bigcup \emptyset = \emptyset,\\
F(1) &=& \bigcup \{\cP(\emptyset)\} = \bigcup \{\{\emptyset\} \} = \{\emptyset\},\\
F(2) &=& \bigcup \{\cP(\emptyset), \cP(\{\emptyset\})\} = \bigcup \{\{\emptyset\}, \{\emptyset, \{\emptyset\} \} \} = \{\emptyset, \{\emptyset\} \},\\
\ldots
\end{IEEEeqnarray*}
}{}
It is easy to prove by induction:
\begin{enumerate}[(a)]
@ -76,16 +84,23 @@ Applications of induction and recursion:
for $\lambda \in \OR$ a limit.
\end{enumerate}
\end{proof}
\gist{%
\begin{notation}
Usually, one writes $V_\alpha$ for $F(\alpha)$.
Usually, one writes $V_\alpha$ for $F(\alpha)$.
They are called the \vocab{rank initial segments} of $V$.
\end{notation}
}{}
\begin{lemma}
\gist{%
If $x$ is any set, then there is some $\alpha \in \OR$
such that $x \in V_\alpha$,
i.e.~$V = \bigcup \{V_{\alpha} : \alpha \in \OR\}$.
}{
$V = \bigcup_{\alpha \in \OR} V_\alpha$.
}
\end{lemma}
\begin{proof}
\gist{%
We use induction on the well-founded $\in$-relation.
Let $A = \bigcup \{V_\alpha : \alpha \in \OR\}$.
We need to show that $A = V$.
@ -103,6 +118,7 @@ Usually, one writes $V_\alpha$ for $F(\alpha)$.
In other words $x \subseteq V_\alpha$,
hence $x \in V_{\alpha+1}$.
}{Induction on $\in$.}
\end{proof}
\begin{lemma}[\vocab{Transitive collapse}/\vocab{Mostowski collapse}]
@ -118,29 +134,43 @@ Usually, one writes $V_\alpha$ for $F(\alpha)$.
with $x \in y \iff (F(x),F(y)) \in R$ for $x,y \in B$.
\end{lemma}
\begin{proof}
``$\impliedby$'' Suppose that $R$ is ill-founded
(i.e.~not well-founded).
Then there is some $(y_n : n < \omega)$ such that $y_n \in A$
and $(y_{n+1}, y_n) \in R$ for all $n < \omega$.
But then if $F$ is an isomorphism as above,
\[
F^{-1}(Y_{n+1}) \in F^{-1}(Y_n)
\]
for all $n < \omega$ $\lightning$
``$\impliedby$''
\gist{%
Suppose that $R$ is ill-founded
(i.e.~not well-founded).
Then there is some $(y_n : n < \omega)$ such that $y_n \in A$
and $(y_{n+1}, y_n) \in R$ for all $n < \omega$.
But then if $F$ is an isomorphism as above,
\[
F^{-1}(Y_{n+1}) \in F^{-1}(Y_n)
\]
for all $n < \omega$ $\lightning$
}{%
Trivial, since $\in$ is well-founded.
}
``$\implies$ '' Suppose that $R$ is well-founded.
We want a transitive class $B$ and a function $F\colon B \leftrightarrow A$
such that
\[
x \in y \iff (F(x), F(y)) \in R.
\]
Equivalently $G\colon A \leftrightarrow B$
with $(x,y) \in R$ iff $G(x) \in G(y)$ for all $x,y \in A$.
``$\implies$ ''
\gist{%
Suppose that $R$ is well-founded.
We want a transitive class $B$ and a function $F\colon B \leftrightarrow A$
such that
\[
x \in y \iff (F(x), F(y)) \in R.
\]
Equivalently $G\colon A \leftrightarrow B$
with $(x,y) \in R$ iff $G(x) \in G(y)$ for all $x,y \in A$.
In other words, $G(y) = \{G(x) : (x,y) \in R\}$.
Such a function $G$ and class $B$ exist by the recursion theorem.
In other words, $G(y) = \{G(x) : (x,y) \in R\}$.
Such a function $G$ and class $B$ exist by the \yaref{thm:recursion}.
}{Use recursion to define $F(y) \coloneqq \{F(x) : (x,y) \in R\}$.}
\end{proof}
As a consequence of the \yaref{lem:mostowski},
we get that if $<$ is a well-order on a set $a$
then there is some transitive set $b$
with $(b, \in\defon{b}) \cong (a, <)$.
\begin{lemma}[\vocab{Rank function}]
Let $R$ be a well-founded and set-like binary relation
on a class $A$.
@ -149,7 +179,7 @@ Usually, one writes $V_\alpha$ for $F(\alpha)$.
\[(x,y) \in R \implies F(x) < F(y).\]
\end{lemma}
\begin{proof}
By the recursion theorem,
By the \yaref{thm:recursion},
there is $F$ such that
\[
F(y) = \sup \{F(x) + 1 : (x,y) \in R\}.

97
inputs/lecture_11.tex
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@ -2,12 +2,6 @@
\subsection{Cardinals}
Consequence of the Mostowski collapse:
If $<$ is a well-order on a set $a$
then there is some transitive $b$
with $(b, \in\defon{b}) \cong (a, <)$.
\begin{definition}
Let $a$ be any set.
The \vocab{cardinality} of $a$
@ -16,8 +10,10 @@ with $(b, \in\defon{b}) \cong (a, <)$.
such that there is some bijection $f\colon \alpha \to a$.
An ordinal $\alpha$ is called a \vocab{cardinal},
iff there is some set $a$ with $|a| = \alpha$
(equivalently, $|\alpha| = \alpha$).
iff \gist{%
there is some set $a$ with $|a| = \alpha$
(equivalently, $|\alpha| = \alpha$).%
}{$|\alpha| = \alpha$.}
\end{definition}
We often write $\kappa, \lambda, \ldots$ for cardinals.
@ -27,23 +23,28 @@ We often write $\kappa, \lambda, \ldots$ for cardinals.
there is come cardinal $\lambda > \kappa$.
\end{lemma}
\begin{proof}
Consider the powerset of $\kappa$.
We know that there is no surjection $\kappa \twoheadrightarrow \cP(\kappa)$.
Hence $\kappa < |2^{\kappa}|$.
\gist{%
Consider the powerset of $\kappa$.
We know that there is no surjection $\kappa \twoheadrightarrow \cP(\kappa)$.
Hence $\kappa < |2^{\kappa}|$.
}{$\kappa < |2^{\kappa}|$.}
\end{proof}
\begin{definition}
For each cardinal $\kappa$,
$\kappa^+$ denotes the least cardinal $\lambda > \kappa$.
\end{definition}
\gist{%
\begin{warning}
This has nothing to do with the ordinal successor of $\kappa$.
\end{warning}
}{}
\begin{lemma}
Let $X$ be any set of cardinals.
Then $\sup X$ is a cardinal.
\end{lemma}
\begin{proof}
\gist{%
If there is some $\kappa \in X$ with $\lambda \le \kappa$
for all $\lambda \in X$,
then $\kappa = \sup(X)$ is a cardinal.
@ -62,8 +63,19 @@ We often write $\kappa, \lambda, \ldots$ for cardinals.
However, there exists $\mu \twoheadrightarrow \sup(X)$,
hence also $\mu \twoheadrightarrow \lambda$
(which is in contradiction to $\lambda$ being a cardinal).
}{%
\begin{itemize}
\item If $\sup(X) \in X$ this is trivial.
\item Let $\sup(X) \not\in X$, $\mu \coloneqq |\sup(X)|$.
\begin{itemize}
\item Suppose that $\sup(X)$ is not a cardinal.
Then $\mu \in \sup(X)$, since $\sup(X) \in \OR$.
\item $\exists \lambda \in X.~\lambda > \mu$ $\lightning$.
\end{itemize}
\end{itemize}
}
\end{proof}
We may now use the recursion theorem
We may now use the \yaref{lem:recursion}
to define a sequence $\langle \aleph_\alpha : \alpha \in \OR \rangle$
with the following properties:
\begin{IEEEeqnarray*}{rCl}
@ -72,7 +84,7 @@ with the following properties:
\aleph_{\lambda} &=& \sup \{\aleph_\alpha : \alpha < \lambda\}.
\end{IEEEeqnarray*}
Each $\aleph_\alpha$ is a cardinal.
Also, a trivial induction show that $\alpha \le \aleph_\alpha$.
Also, a trivial induction shows that $\alpha \le \aleph_\alpha$.
In particular $|\alpha| \le \aleph_{\alpha}$.
Therefore the $\aleph_\alpha$ are all the infinite cardinals:
If $a$ is any infinite set, then $|a| \le \aleph_{|a|}$,
@ -84,7 +96,7 @@ so $|a| = \aleph_\beta$ for some $\beta \le |\alpha|$.
\end{notation}
\begin{notation}
Let ${}^a b \coloneqq \{f : f \text{ is a function}, \dom(f) = \alpha, \ran(f) \subseteq b\}$.
Let $\leftindex^a b \coloneqq \{f : f \text{ is a function}, \dom(f) = \alpha, \ran(f) \subseteq b\}$.
\end{notation}
\begin{definition}[Cardinal arithmetic]
@ -93,40 +105,44 @@ so $|a| = \aleph_\beta$ for some $\beta \le |\alpha|$.
\begin{IEEEeqnarray*}{rCl}
\kappa + \lambda &\coloneqq & |\{0\} \times \kappa \cup \{1\} \times \lambda|,\\
\kappa \cdot \lambda &\coloneqq & |\kappa \times \lambda|,\\
\kappa^{\lambda} &\coloneqq & |{}^{\lambda}\kappa|.
\kappa^{\lambda} &\coloneqq & |\leftindex^{\lambda}\kappa|.
\end{IEEEeqnarray*}
\end{definition}
\gist{%
\begin{warning}
This is very different from ordinal arithmetic!
\end{warning}
}{}
\begin{theorem}[Hessenberg]
\label{thm:hessenberg}
For all $\alpha$ we have
\[
\aleph_\alpha \cdot \aleph_\alpha = \aleph_\alpha.
\]
\]
\end{theorem}
\begin{corollary}
For all $\alpha, \beta$ it is
\[
\aleph_\alpha + \aleph_\beta = \aleph_\alpha \cdot \aleph_\beta = \max \{\aleph_\alpha, \aleph_\beta\}.
\]
\]
\end{corollary}
\begin{proof}
Wlog.~$\alpha \le \beta$.
Trivially $\aleph_\alpha \le \aleph_\beta$.
It is also clear that
\[
\aleph_{\beta} \le \aleph_\alpha + \aleph_{\beta} \le \aleph_\alpha \cdot \aleph_\beta \le \aleph_\beta \cdot \aleph_\beta = \aleph_\beta.
\]
\gist{%
Trivially $\aleph_\alpha \le \aleph_\beta$.
It is also clear that
\[
\aleph_{\beta} \le \aleph_\alpha + \aleph_{\beta} \le \aleph_\alpha \cdot \aleph_\beta \le \aleph_\beta \cdot \aleph_\beta = \aleph_\beta.
\]
}{Then $\aleph_{\beta} \le \aleph_\alpha + \aleph_{\beta} \le \aleph_\alpha \cdot \aleph_\beta \le \aleph_\beta \cdot \aleph_\beta = \aleph_\beta$.}
\end{proof}
\begin{refproof}{thm:hessenberg}
\gist{%
Define a well-order $<^\ast$ on $\OR \times \OR$ by setting
\[
(\alpha,\beta) <^\ast (\gamma,\delta)
\]
\]
iff
\begin{itemize}
\item $\max(\alpha,\beta) < \max(\gamma,\delta)$ or
@ -139,15 +155,15 @@ so $|a| = \aleph_\beta$ for some $\beta \le |\alpha|$.
There is an isomorphism
\[
(\OR, <) \cong^{\Gamma^{-1}} (\OR \times \OR, <^\ast).
\]
\]
$\Gamma$ is called the \vocab{Gödel pairing function}.
\begin{claim}
For all $\alpha$ it is
$\ran(\Gamma\defon{\aleph_\alpha \times \aleph_\alpha})) = \aleph_\alpha$,
$\ran(\Gamma\defon{\aleph_\alpha \times \aleph_\alpha}) = \aleph_\alpha$,
i.e.
\[
\aleph_\alpha = \{\xi: \exists \eta, \eta' < \aleph_\alpha.~\xi = \Gamma((\eta,\eta'))\}.
\]
\]
\end{claim}
\begin{subproof}
We use induction of $\alpha$.
@ -173,7 +189,7 @@ so $|a| = \aleph_\beta$ for some $\beta \le |\alpha|$.
If $(\gamma,\delta) <^\ast (\eta, \eta')$,
then $\max \{\gamma,\delta\} \le \max \{\eta, \eta'\}$.
Say $\eta \le \eta' < \aleph_\alpha$
and let $\aleph_\alpha = |\eta'|$.
and let $\aleph_\beta = |\eta'|$.
There is a surjection
\[f\colon \underbrace{(\eta +1)}_{ \le \aleph_\beta} \times \underbrace{(\eta' + 1)}_{\sim \aleph_\beta} \twoheadrightarrow \aleph_\alpha.\]
@ -181,15 +197,38 @@ so $|a| = \aleph_\beta$ for some $\beta \le |\alpha|$.
The inductive hypothesis then produces a surjection
$f^\ast\colon \aleph_\beta \to \aleph_\alpha \lightning$.
\end{subproof}
}{
\begin{itemize}
\item Define wellorder $<^\ast \subseteq \OR \times \OR$,
$(\alpha,\beta) <^\ast (\gamma,\delta)$ iff
\begin{itemize}
\item $\max(\alpha,\beta) < \max(\gamma,\delta)$ or
\item $\max(\alpha,\beta) = \max(\gamma,\delta)$ and $\alpha < \gamma$ or
\item $\max(\alpha,\beta) = \max(\gamma,\delta)$ and $\alpha = \gamma$ and $\beta < \delta$.
\end{itemize}
\item Gödel pairing function $\Gamma : (\OR \times \OR, <^\ast) \xrightarrow{\cong} (\OR, <)$.
\item $\forall \alpha.~\ran(\Gamma\defon{\aleph_\alpha \times \aleph_\alpha}) = \aleph_\alpha$.
\begin{itemize}
\item Induction on $\alpha$.
\item Clearly $\aleph_\alpha \subseteq \ran(\Gamma\defon{\aleph_\alpha \times \aleph_\alpha})$
($\aleph_\alpha$ is a cardinal).
\item Suppose $\aleph_\alpha \subsetneq \ran(\Gamma\defon{\aleph_\alpha \times \aleph_\alpha})$,
We get a surjection $\aleph_\beta \xrightarrow{\text{induction}} \aleph_\beta \times \aleph_\beta \to \aleph_\alpha$
for some $\beta < \alpha$.
\end{itemize}
\end{itemize}
}
\end{refproof}
\gist{%
However, exponentiation of cardinals is far from trivial:
\begin{observe}
$2^{\kappa} = |\cP(\kappa)|$,
since ${}^{\kappa} \{0,1\} \leftrightarrow\cP(\kappa)$.
since $\leftindex^{\kappa} \{0,1\} \leftrightarrow\cP(\kappa)$.
Hence by Cantor $2^{\kappa}\ge \kappa^+$.
\end{observe}
This is basically all we can say.
}{}
The \vocab{continuum hypothesis} states that $2^{\aleph_0} = \aleph_1$.

53
inputs/lecture_12.tex
View file

@ -26,17 +26,20 @@ and
\beta^{\lambda} &\coloneqq & \sup_{\alpha < \lambda} \beta^{\alpha} &~ ~\text{for limit ordinals $\lambda$}.
\end{IEEEeqnarray*}
\gist{%
\begin{example}
\leavevmode
\begin{itemize}
\item $2+ 2 =4$,
\item $196883 + 1 = 196884$,
\item $1 + \omega = \sup_{n < \omega} 1 + n = \omega \neq \omega +1$,
\item $2 \cdot \omega = \sup_{n < \omega} 2\cdot n = \omega$,
\item $\omega \cdot 2 =\omega \cdot 1 + \omega = \omega + \omega$..
\item $\omega \cdot 2 =\omega \cdot 1 + \omega = \omega + \omega$.
\end{itemize}
\end{example}
}{}
\gist{%
\begin{warning}
Cardinal arithmetic and ordinal arithmetic are very different!
The symbols are the same, but usually we will distinguish
@ -46,6 +49,7 @@ and
\end{warning}
We will very rarely use ordinal arithmetic.
}{}
\subsection{Cofinality}
@ -100,7 +104,7 @@ We will very rarely use ordinal arithmetic.
\begin{itemize}
\item $\cf(\aleph_\omega) = \omega$.
In fact $\cf(\aleph_{\lambda}) \le \lambda$
for limit ordinals $\lambda$
for limit ordinals $\lambda \neq 0$
(consider $\alpha \mapsto \aleph_\alpha$).
\item $\cf(\aleph_{\omega + \omega}) = \omega$.
\end{itemize}
@ -110,20 +114,24 @@ We will very rarely use ordinal arithmetic.
$\cf(\beta)$ is a cardinal.
\end{lemma}
\begin{proof}
\gist{%
Let $f\colon \alpha \to \beta$ be cofinal.
Then $\tilde{f}\colon |\alpha| \to \beta$,
the composition with $\alpha \leftrightarrow|\alpha|$
is cofinal as well and $|\alpha| \le \alpha$.
}{Trivial.}
\end{proof}
\gist{%
\begin{question}
How does one imagine ordinals with
cofinality $> \omega$?
\end{question}
No idea.
}{}
\begin{definition}
An ordinal $\beta$ is \vocab{regular}
An ordinal $\beta$ is \vocab{regular}
iff $\cf(\beta) = \beta$.
Otherwise $\beta$ is called \vocab{singular}.
\end{definition}
@ -172,6 +180,7 @@ In particular, a regular ordinal is always a cardinal.
by \yaref{thm:hessenberg}.
\end{proof}
\gist{%
\begin{itemize}
\item $\aleph_0, \aleph_1, \aleph_2, \ldots$ are regular,
\item $\aleph_\omega$ is singular,
@ -184,11 +193,13 @@ In particular, a regular ordinal is always a cardinal.
\item $\aleph_{\omega_1 + 1}, \ldots$ is regular,
\item $\aleph_{\omega_2}$ is singular.
\end{itemize}
}{}
\begin{question}[Hausdorff]
Is there a regular limit cardinal?
\end{question}
Maybe. This is independent of $\ZFC$.
Maybe. This is independent of $\ZFC$,
cf.~\yaref{def:inaccessible}.
\begin{theorem}[Hausdorff]
@ -197,9 +208,10 @@ Maybe. This is independent of $\ZFC$.
\]
\end{theorem}
\begin{proof}
\gist{%
Recall that
\begin{IEEEeqnarray*}{rCl}
\aleph_{\alpha+1}^{\aleph_\beta} &=& |{}^{\aleph_\beta} \aleph_{\alpha+1}|.
\aleph_{\alpha+1}^{\aleph_\beta} &=& |\leftindex^{\aleph_\beta} \aleph_{\alpha+1}|.
\end{IEEEeqnarray*}
\begin{itemize}
\item First case: $\beta \ge \alpha+1$.
@ -209,8 +221,8 @@ Maybe. This is independent of $\ZFC$.
\le \left( 2^{\aleph_{\beta}} \right)^{\aleph_\beta}
= 2^{\aleph_{\beta} \cdot \aleph_\beta} = 2^{\aleph_\beta} \le \aleph_{\alpha+1}^{\aleph_\beta}.
\]
Also $\aleph_\alpha^{\aleph_{\beta} = 2^{\aleph_\beta}}$
in this case,
Also $\aleph_\alpha^{\aleph_{\beta}} = 2^{\aleph_\beta}$
in this case (by the same argument),
so
\[
\aleph_{\alpha+1}^{\aleph_{\beta}} = 2^{\aleph_{\beta}}
@ -224,14 +236,35 @@ Maybe. This is independent of $\ZFC$.
is unbounded.
So
\[
{}^{\aleph_{\beta}}\aleph_{\alpha+1} = \bigcup_{\xi < \aleph_{\alpha+1}} {}^{\aleph_\beta} \xi
\leftindex^{\aleph_{\beta}}\aleph_{\alpha+1} = \bigcup_{\xi < \aleph_{\alpha+1}} \leftindex^{\aleph_\beta} \xi
\]
for each $\xi < \aleph_{\alpha+1}$, $|\xi| \le \aleph_\alpha$,
hence
\[
|{}^{\aleph_{\beta}}\xi| \le \aleph_\alpha^{\aleph_\beta}
|\leftindex^{\aleph_{\beta}}\xi| \le \aleph_\alpha^{\aleph_\beta}
\]
for each $\xi < \aleph_{\alpha+1}$.
Therefore, \[\aleph_{\alpha+1}^{\aleph_\beta} \le \aleph_{\alpha+1} \cdot \aleph_{\alpha}^{\aleph_{\beta}} \le \aleph_{\alpha+1}^{\aleph_\beta}.\]
\end{itemize}
}{%
$\ge$ is trivial.
\begin{itemize}
\item First case: $\beta \ge \alpha + 1$.
Note that $\gamma \le \beta \implies \aleph_{\gamma}^{\aleph_\beta} = 2^{\aleph_\beta}$,
so
\[
\aleph_{\alpha+1}^{\aleph_\beta} \overset{\alpha+1 \le \beta}{=}
2^{\aleph_\beta}
\overset{\alpha < \beta}{=} \aleph_\alpha^{\aleph_\beta}
\le \aleph_\alpha^{\aleph_\beta} \cdot \aleph_{\alpha+1}.
\]
\item Second case: $\beta < \alpha+1$:
\begin{itemize}
\item $\aleph_{\alpha+1}$ is regular, so all $f\colon \aleph_\beta \to \aleph_{\alpha+1}$ are bounded.
\item Thus $\leftindex^{\aleph_\beta}\aleph_{\alpha+1} = \bigcup_{\xi < \aleph_{\alpha+1}} \leftindex^{\aleph_\beta} \xi$ for all $\xi < \aleph_{\alpha+1}$.
\item So $\aleph_{\alpha+1}^{\aleph_\beta} \le \aleph_{\alpha+1} \aleph_\alpha^{\aleph_\beta}$.
\end{itemize}
\end{itemize}
}
\end{proof}

31
inputs/lecture_13.tex
View file

@ -1,5 +1,5 @@
\lecture{13}{2023-11-30}{}
\gist{%
\begin{remark}[``Constructive'' approach to $\omega_1$ ]
There are many well-orders on $\omega$.
Let $W$ be the set of all such well-orders.
@ -41,7 +41,9 @@
Thus $\otp(\faktor{W}{\sim}, <) = \omega_1$.
\todo{move this}
\end{remark}
}{}
\gist{%
\begin{notation}
Let $I \neq \emptyset$
and let $\{\kappa_i : i \in I\}$
@ -57,13 +59,13 @@
\]
where
\[
\bigtimes_{i \in I} A_i \coloneqq \{f : f \text{is a function with domain $I$ and $f(i) \in A_i$}\}.
\bigtimes_{i \in I} A_i \coloneqq \{f : f \text{ is a function}, \dom(f) = I, \forall i.~f(i) \in A_i\}.
\]
\end{notation}
\begin{remark}
\AxC is equivalent to $\forall i \in I.~A_i \neq \emptyset \implies \bigtimes_{i \in I} A_i \neq \emptyset$.
\end{remark}
}{}
\begin{theorem}[K\H{o}nig]
\yalabel{K\H{o}nig's Theorem}{K\H{o}nig}{thm:koenig}
Let $I \neq \emptyset$.
@ -78,6 +80,7 @@
\]
\end{theorem}
\begin{proof}
\gist{%
Consider a function $F\colon \bigcup_{i \in I} (\kappa_i \times \{i\}) \to \bigtimes_{i \in I}\lambda_i$.
We want to show that $F$ is not surjective.
@ -92,6 +95,14 @@
be defined by $i \mapsto \xi_i$.
Then $f \not\in \ran(F)$.
}{%
\begin{itemize}
\item Consider $F\colon \bigcup_{i \in I}(\kappa_i \times \{i\}) \to \bigtimes_{i \in I} \lambda_i$. Want: $F$ is not surjective.
\item Let $\xi_i$ minimal such that $\underbrace{F((\eta,i))(i)}_{\in \lambda_i} \neq \xi$
for all $\eta < \kappa_i$.
\item $(i \mapsto \eta_i) \not\in \ran(F)$.
\end{itemize}
}
\end{proof}
\begin{corollary}
@ -99,6 +110,7 @@
it is $\cf(2^{\kappa}) > \kappa$.
\end{corollary}
\begin{proof}
\gist{%
If $2^{\kappa}$ is a successor cardinal,
then $\cf(2^{\kappa}) = 2^{\kappa} > \kappa$,
since successor cardinals are regular.
@ -115,6 +127,19 @@
\sup \{\kappa_i : i < \kappa\} \le \sum_{i \in \kappa} \kappa_i < \prod_{i \in \kappa} \lambda_i = \left( 2^{\kappa} \right)^{\kappa} = 2^{\kappa \cdot \kappa} = 2^{\kappa}
\]
and $f$ is not cofinal.
}{%
\begin{itemize}
\item Trivial for successors ($\cf(2^{\kappa}) \overset{\text{regular}}{=} 2^{\kappa} > \kappa$).
\item Suppose $2^\kappa$ is a limit, $\exists f\colon \kappa \to 2^{\kappa}$ cofinal.
Wlog.~$f(i) \in \Card$.
But
\[
\sup \{f(i)) : i < \kappa\} \le \sum_{i \in \kappa} f(i)
\overset{\yaref{thm:koenig}}{<} \prod_{i \in \kappa} 2^{\kappa}
= (2^{\kappa})^{\kappa} = 2^{\kappa}. ~ ~\qedhere
\]
\end{itemize}
}
\end{proof}
\begin{fact}

95
inputs/lecture_14.tex
View file

@ -1,4 +1,5 @@
\lecture{14}{2023-12-04}{}
\gist{%
\begin{abuse}
Sometimes we say club
instead of club in $\kappa$.
@ -17,6 +18,7 @@
is club in $\kappa$.
\end{itemize}
\end{example}
}{}
\begin{lemma}
\label{lem:clubintersection}
Let $\kappa$ be regular and uncountable.
@ -34,12 +36,12 @@
Let $C_{\beta} \coloneqq \{\xi : \xi > \beta\}$.
Clearly this is club
but $\bigcap_{\beta < \kappa} C_\beta = \emptyset$.
\end{warning}
\begin{refproof}{lem:clubintersection}
\gist{%
First let $\alpha = 2$.
Let $C, D \subseteq \kappa$
be a club.
be club.
$C \cap D$ is trivially closed:
Let $\beta < \kappa$. Suppose that $(C \cap D) \cap \beta$
@ -84,11 +86,21 @@
= \sup_{i < \omega} \gamma_{\alpha \cdot n + \beta + 1} \in \bigcap_{\beta < \alpha} C_\beta$,
where we have used that.
$\cf(\kappa) > \alpha \cdot \omega$.
}{%
\begin{itemize}
\item Trivially closed.
\item Recursively define $\langle \gamma_i : i \le \alpha \cdot \omega \rangle$, by
$\gamma_0 \coloneqq \gamma$,
\[\gamma_{\alpha \cdot n + \beta + 1} = \min C_{\beta} \setminus (\gamma_{\alpha \cdot n + \beta} + 1)\]
and $\sup$ at limits.
\item Then $\sup_{i < \alpha\cdot \omega} \gamma_i \in \bigcap_{\beta < \alpha} C_\beta$
(we used $\cf(\kappa) > \alpha\cdot \omega$).
\end{itemize}
}
\end{refproof}
\begin{definition}
$F \subseteq \cP(a)$ is a \vocab{filter}
$F \subseteq \cP(a)$ is a \vocab{filter}
iff
\begin{enumerate}[(a)]
\item $X,Y \in F \implies X \cap Y \in F$,
@ -106,14 +118,16 @@
iff for all $\gamma < \alpha$ and $\{X_\beta : \beta < \gamma\} \subseteq F$
then $\bigcap \{X_\beta : \beta < \gamma\} \in F$.
\end{definition}
\gist{%
Intuitively, a filter is a collection of ``big'' subsets of $a$.
}{}
\begin{definition}
Let $\kappa$ be regular and uncountable.
The \vocab{club filter} is defined as
\[
\cF_{\kappa} \coloneqq \{X \subseteq \kappa : \exists \text{ club } C \subseteq \kappa .~ C \subseteq X\}.
\]
\]
\end{definition}
Clearly this is a filter.
@ -126,7 +140,7 @@ We have shown (assuming \AxC to choose contained clubs):
Clearly $\emptyset \not\in \cF_\kappa$,
$\kappa \in \cF_\kappa$,
and $A \in \cF_{\kappa}, A \subseteq B \in \kappa \implies B \in \cF_\kappa$.
In \autoref{lem:clubintersection} we are going to show that the intersection
In \autoref{lem:clubintersection} showed that the intersection
of $< \kappa$ many clubs is club.
\end{proof}
@ -138,7 +152,7 @@ We have shown (assuming \AxC to choose contained clubs):
\[
\diagi_{\beta < \alpha} A_{\beta} \coloneqq
\{\xi < \alpha : \xi \in \bigcap \{A_{\beta} : \beta < \xi\} \}.
\]
\]
\end{definition}
\begin{lemma}
\label{lem:diagiclub}
@ -146,9 +160,11 @@ We have shown (assuming \AxC to choose contained clubs):
If $\langle C_{\beta} : \beta < \kappa \rangle$
is a sequence of club subsets of $\kappa$,
then $\diagi_{\beta < \kappa} C_{\beta}$
contains a club. % TODO: contains or is?
contains a club.
\end{lemma}
\begin{proof}
\begin{refproof}{lem:diagiclub}
% TODO THINK
\gist{
Let us fix $\langle C_{\beta} : \beta < \alpha \rangle$.
Write $D_{\beta} \coloneqq \bigcap \{C_{\gamma} : \gamma \le \beta\} $
for $\beta < \kappa$.
@ -192,7 +208,7 @@ We have shown (assuming \AxC to choose contained clubs):
\begin{subproof}
Fix $\gamma < \kappa$.
We need to find $\delta > \gamma$
with $\gamma \in \diagi_{\beta < \kappa} D_\beta$.
with $\delta \in \diagi_{\beta < \kappa} D_\beta$.
Define $\langle \gamma_n : n < \omega \rangle$
as follows:
@ -212,20 +228,57 @@ We have shown (assuming \AxC to choose contained clubs):
for some $n < \omega$.
For $m \ge n$, $\gamma_{m+1} \in D_{\gamma_m} \subseteq D_{\gamma_n} \subseteq D_{\overline{\gamma}}$.
So $D_{\overline{\gamma}} \cap \delta$ is unbounded
in $\gamma$, hence $\delta \in D_{\overline{\gamma}}$.
in $\delta$, hence $\delta \in D_{\overline{\gamma}}$.
\end{subproof}
\end{proof}
}{%
\begin{itemize}
\item Fix $\langle C_\beta : \beta < \alpha \rangle$.
Set $D_{\beta} \coloneqq \bigcap_{\gamma \le \beta} D_{\gamma}$.
It suffices to analyze $D_{\beta}$.
\item $\diagi_{\beta < \kappa} D_{\beta}$ is closed in $\kappa$:
\begin{itemize}
\item Let $\gamma < \kappa$ such that $\left( \diagi_{\beta < \kappa} D_{\beta}\right) \cap \gamma$ unbounded in $\gamma$.
Want $\gamma \in \diagi_{\beta < \kappa} D_\beta$.
\item Let $\beta_0 < \gamma$. Want $\gamma \in D_{\beta_0}$.
\item $D_{\beta_0} \cap \gamma$ is unbounded in $\gamma$
($D_{\beta_0} \setminus \beta_0 \supseteq \diagi_{\beta < \kappa} D_{\beta} \setminus \beta_0$)
$\overset{D_{\beta_0} \text{ closed}}{\implies} \gamma \in D_{\beta_0}$.
\end{itemize}
\item $\diagi_{\beta < \kappa} D_\beta$ is unbounded in $\kappa$:
\begin{itemize}
\item Fix $\gamma < \kappa$. We need to find $\diagi_{\beta < \kappa} D_\beta \ni \delta > \gamma$.
\item Define $\langle \gamma_n : n < \omega \rangle$
by $\gamma_0 \coloneqq \gamma$,
$\gamma_{n+1} \coloneqq \min D_{\gamma_n} \setminus (\gamma_n + 1)$,
$\delta \coloneqq \sup_n \gamma_n \overset{\cf(\kappa) > \omega}{<} \kappa$
\item Want $\delta \in \diagi_{\beta < \kappa} D_\beta$,
i.e.~$\forall \epsilon < \delta.~\delta\in D_\epsilon$.
If $\epsilon < \delta$, then $\epsilon \le \gamma_n$
for $n$ large enough,
so $\gamma_{m+1} \in D_{\gamma_m} \subseteq D_{\gamma_n} \subseteq D_\epsilon$
for $m \ge n$.
Thus $\sup(D_\epsilon \cap \delta) = \delta$
$\overset{D_\epsilon \text{ closed}}{\implies} \delta \in D_{\epsilon}$.
\end{itemize}
\end{itemize}
}
\end{refproof}
\begin{remark}+
$\diagi_{\beta < \kappa} D_{\beta}$ actually
$\diagi_{\beta < \kappa} C_{\beta}$ actually
\emph{is} a club:
It suffices to show that $\diagi_{\beta < \kappa} D_\beta$ is closed.
Let $\lambda < \kappa$ be a limit ordinal.
Suppose that $\lambda \not\in \diagi_{\beta < \kappa} D_\beta$.
Then there exists $\alpha < \lambda$ such that
$\lambda \not\in D_\alpha$.
Since $D_\alpha$ is closed,
we get $\sup(D_{\alpha} \cap \lambda) < \lambda$.
In particular $\sup (\lambda \cap\diagi_{\beta < \kappa} D_{\beta}) \le \alpha \cup \sup(D_\alpha \cap \lambda) < \lambda$.
It suffices to show that $\diagi_{\beta < \kappa} C_\beta$ is closed.
This can be shown in the same way as for $\diagi_{\beta < \kappa} D_\beta$.
% Let $\lambda < \kappa$ be a limit ordinal.
% Suppose that $\lambda \not\in \diagi_{\beta < \kappa} D_\beta$.
% Then there exists $\alpha < \lambda$ such that
% $\lambda \not\in D_\alpha$.
% Since $D_\alpha$ is closed,
% we get $\sup(D_{\alpha} \cap \lambda) < \lambda$.
% In particular $\sup (\lambda \cap\diagi_{\beta < \kappa} D_{\beta}) \le \alpha \cup \sup(D_\alpha \cap \lambda) < \lambda$.
\end{remark}
\begin{definition}

16
inputs/lecture_15.tex
View file

@ -28,6 +28,7 @@
$f(\alpha) = \nu$ for all $\alpha \in T$.
\end{theorem}
\begin{proof}
\gist{%
Let $S, f$ be given.
For $\nu < \kappa$ set $S_\nu \coloneqq \{\alpha \in S : f(\alpha) = \nu\}$. % f^{-1}(\nu)$.
We aim to show that one of the $S_\nu$ is stationary.
@ -41,6 +42,16 @@
Hence $f(\alpha) \neq \nu$ for all $\nu < \alpha$,
so $f(\alpha) \ge \alpha$.
But $f$ is regressive $\lightning$
}{%
\begin{itemize}
\item For $\nu < \kappa$ set $S_\nu \coloneqq \{\alpha \in S : f(\alpha) = \nu\}$.
\item Suppose none of the $S_\nu$ is stationary,
i.e.~ $\forall \nu < \kappa.~\exists C_\nu \text{ club}.~C_\nu \cap S_\nu = \neq$.
\item $\diagi_{\nu < \alpha} C_\nu$ is club.
\item Pick $\alpha \in C \cap S$.
But then $\forall \nu < \alpha.~\alpha \in C_\nu$, i.e.~$f(\alpha) \ge \alpha$.
\end{itemize}
}
\end{proof}
\subsection{Some model theory and a second proof of Fodor's Theorem}
@ -70,7 +81,7 @@ Recall the following:
over $V_\theta$ for $\phi$
is a function
\[
f\colon {}^k V_\theta \to V_\theta,
f\colon \leftindex^k V_\theta \to V_\theta,
\]
where $k$ is the number of free variables of
$\exists v.~\phi$
@ -91,9 +102,10 @@ to be an elementary substructure of $V_\theta$.
(where $k$ is the number of free variables of $\exists v.~\phi$)
$f_\phi(x_1,\ldots,x_k) \in X$,
then $X \prec V_{\theta}$.
\end{lemma}
% TODO ANKI-MARKER
Let's do a second proof of \yaref{thm:fodor}.
\begin{refproof}{thm:fodor}
Fix $\theta > \kappa$ and look at $V_{\theta}$.

2
inputs/lecture_16.tex
View file

@ -42,6 +42,7 @@ and $\emptyset \not\in F, \kappa \in F$.
either $X \in F$ or $\kappa \setminus X \in F$.
\end{definition}
\gist{%
\begin{example}
Examples of filters:
\begin{enumerate}[(a)]
@ -55,6 +56,7 @@ and $\emptyset \not\in F, \kappa \in F$.
Then $\cF_\kappa \coloneqq \{X \subseteq \kappa: \exists C \subseteq \kappa \text{ club in $\kappa$}. C \subseteq X\}$.
\end{enumerate}
\end{example}
}{Let $\cF_\kappa \coloneqq \{X \subseteq \kappa : \exists C \subseteq \kappa \text{ club in } \kappa\}$.}
\begin{question}
Is $\cF_\kappa$ an ultrafilter?
\end{question}

15
inputs/lecture_18.tex
View file

@ -1,6 +1,7 @@
\lecture{18}{2023-12-18}{Large cardinals}
\begin{definition}
\label{def:inaccessible}
\begin{itemize}
\item A cardinal $\kappa$ is called \vocab{weakly inaccessible}
iff $\kappa$ is uncountable,\footnote{dropping this we would get that $\aleph_0$ is inaccessible}
@ -19,7 +20,7 @@
\begin{theorem}
If $\kappa$ is inaccessible,
then $V_\kappa \models \ZFC$.\footnote{More formally $(V_{\kappa}, \in ) \models\ZFC$.}
then $V_\kappa \models \ZFC$.\footnote{More formally $(V_{\kappa}, \in\defon{V_\kappa}) \models\ZFC$.}
\end{theorem}
\begin{proof}
Since $\kappa$ is regular, \AxRep works.
@ -145,11 +146,11 @@
1. $\implies$ 2.
Fix $U$.
Let ${}^{\kappa}V$ be the class of all function from $\kappa$ to $V$.
For $f,g \in {}^{\kappa}V$ define
Let $\leftindex^{\kappa}V$ be the class of all function from $\kappa$ to $V$.
For $f,g \in \leftindex^{\kappa}V$ define
$f \sim g :\iff \{\xi < \kappa : f(\xi) = g(\xi)\} \in U$.
This is an equivalence relation since $U$ is a filter.
Write $[f] = \{g \in {}^\kappa V : g \sim ~ f \land g \in V_\alpha \text{ for the least $\alpha$ such that there is some $h \in V_\alpha$ with $h \sim f$}\}$.%
Write $[f] = \{g \in \leftindex^\kappa V : g \sim ~ f \land g \in V_\alpha \text{ for the least $\alpha$ such that there is some $h \in V_\alpha$ with $h \sim f$}\}$.%
\footnote{This is know as \vocab{Scott's Trick}.
Note that by defining equivalence classes
in the usual way (i.e.~without this trick),
@ -164,7 +165,7 @@
This is independent of the choice of the representatives,
so it is well-defined.
Now write $\cF = \{[f] : f \in {}^{\kappa}V\}$
Now write $\cF = \{[f] : f \in \leftindex^{\kappa}V\}$
and look at $(\cF, \tilde{\in})$.
The key to the construction is \yaref{thm:los} (see below).
@ -188,7 +189,7 @@ Then
Let us show that $(\cF, \tilde{\in })$
is well-founded.
Otherwise there is $\langle f_n : n < \omega \rangle$
such that $f_n \in {}^\kappa V$
such that $f_n \in \leftindex^\kappa V$
and $[f_{n+1}] \tilde{\in } [f_n]$ for all $n < \omega$.
Then $X_n \coloneqq \{\xi < \kappa : f_{n+1}(\xi) \in f_n(\xi)\} \in U$,
@ -228,7 +229,7 @@ So $j(\alpha) \le \alpha$.
\begin{theorem}[\L o\'s]
\yalabel{\L o\'s's Theorem}{\L o\'s}{thm:los}
For all formulae $\phi$ and for all $f_1, \ldots, f_k \in {}^{\kappa} V$,
For all formulae $\phi$ and for all $f_1, \ldots, f_k \in \leftindex^{\kappa} V$,
\[
(\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.

1
logic.sty
View file

@ -24,6 +24,7 @@
\usepackage{multirow}
\usepackage{float}
\usepackage{scalerel}
\usepackage{leftindex}
%\usepackage{algorithmicx}
\newcounter{subsubsubsection}[subsubsection]