From cac1563244c2993dbef9f23ece3f7d8716188733 Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Wed, 14 Feb 2024 18:44:32 +0100 Subject: [PATCH 1/3] silver gist --- .gitea/workflows/build.yaml | 2 +- inputs/lecture_11.tex | 2 +- inputs/lecture_17.tex | 19 +++++++++++++++---- 3 files changed, 17 insertions(+), 6 deletions(-) diff --git a/.gitea/workflows/build.yaml b/.gitea/workflows/build.yaml index 4ba83f4..f1f6861 100644 --- a/.gitea/workflows/build.yaml +++ b/.gitea/workflows/build.yaml @@ -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: diff --git a/inputs/lecture_11.tex b/inputs/lecture_11.tex index cc55151..d304051 100644 --- a/inputs/lecture_11.tex +++ b/inputs/lecture_11.tex @@ -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} diff --git a/inputs/lecture_17.tex b/inputs/lecture_17.tex index 3cc3123..e7098ff 100644 --- a/inputs/lecture_17.tex +++ b/inputs/lecture_17.tex @@ -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 $\{\beta\} = g_Y''T$, 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} From 858e017a4a9eccfce14ef07fc059bf12b2e54f59 Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Wed, 14 Feb 2024 23:19:23 +0100 Subject: [PATCH 2/3] gist complete --- inputs/lecture_01.tex | 2 ++ inputs/lecture_03.tex | 2 +- inputs/lecture_04.tex | 2 +- inputs/lecture_08.tex | 4 ++-- inputs/lecture_15.tex | 2 +- inputs/lecture_18.tex | 52 ++++++++++++++++++++++++++++++++++++++++++- inputs/lecture_22.tex | 2 +- 7 files changed, 59 insertions(+), 7 deletions(-) diff --git a/inputs/lecture_01.tex b/inputs/lecture_01.tex index 0f5a565..0299e94 100644 --- a/inputs/lecture_01.tex +++ b/inputs/lecture_01.tex @@ -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} diff --git a/inputs/lecture_03.tex b/inputs/lecture_03.tex index 06256d1..c0a5343 100644 --- a/inputs/lecture_03.tex +++ b/inputs/lecture_03.tex @@ -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. diff --git a/inputs/lecture_04.tex b/inputs/lecture_04.tex index 25845cb..56241f1 100644 --- a/inputs/lecture_04.tex +++ b/inputs/lecture_04.tex @@ -5,7 +5,7 @@ \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 diff --git a/inputs/lecture_08.tex b/inputs/lecture_08.tex index 8edf55c..2a5a0d5 100644 --- a/inputs/lecture_08.tex +++ b/inputs/lecture_08.tex @@ -108,10 +108,10 @@ 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 $\BG$ diff --git a/inputs/lecture_15.tex b/inputs/lecture_15.tex index 46fca95..a4af044 100644 --- a/inputs/lecture_15.tex +++ b/inputs/lecture_15.tex @@ -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 $(X,\in ) \prec (V_{\theta})$} diff --git a/inputs/lecture_18.tex b/inputs/lecture_18.tex index 2d99308..7499fbd 100644 --- a/inputs/lecture_18.tex +++ b/inputs/lecture_18.tex @@ -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] diff --git a/inputs/lecture_22.tex b/inputs/lecture_22.tex index 12abc76..5af0e30 100644 --- a/inputs/lecture_22.tex +++ b/inputs/lecture_22.tex @@ -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} From f4626be14bdd7767d155102625c0d4b5eaa1d9f8 Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Thu, 15 Feb 2024 04:44:48 +0100 Subject: [PATCH 3/3] some changes --- inputs/lecture_02.tex | 10 ++++++++++ inputs/lecture_03.tex | 4 ++-- inputs/lecture_04.tex | 12 ++++++++---- inputs/lecture_05.tex | 8 +++++--- inputs/lecture_06.tex | 40 +++++++++++++++++++++++++++++++++++++++- inputs/lecture_07.tex | 7 ++++++- inputs/lecture_08.tex | 5 ++++- inputs/lecture_14.tex | 2 +- inputs/lecture_17.tex | 4 ++-- 9 files changed, 77 insertions(+), 15 deletions(-) diff --git a/inputs/lecture_02.tex b/inputs/lecture_02.tex index 1075c18..d7c6879 100644 --- a/inputs/lecture_02.tex +++ b/inputs/lecture_02.tex @@ -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} diff --git a/inputs/lecture_03.tex b/inputs/lecture_03.tex index c0a5343..b740841 100644 --- a/inputs/lecture_03.tex +++ b/inputs/lecture_03.tex @@ -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 $(a,b) \cap P$ diff --git a/inputs/lecture_04.tex b/inputs/lecture_04.tex index 56241f1..82d3624 100644 --- a/inputs/lecture_04.tex +++ b/inputs/lecture_04.tex @@ -12,6 +12,7 @@ $\ZFC$ stands for \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}] diff --git a/inputs/lecture_05.tex b/inputs/lecture_05.tex index 33c2d28..c068139 100644 --- a/inputs/lecture_05.tex +++ b/inputs/lecture_05.tex @@ -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} @@ -167,7 +168,6 @@ \[ x \in b \land \forall y \in b.~y \le x. \] - In a similar way we define \vocab[Minimal element]{minimal elements} and the \vocab{minimum} of $b$. @@ -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 diff --git a/inputs/lecture_06.tex b/inputs/lecture_06.tex index 245b7ef..ee18c61 100644 --- a/inputs/lecture_06.tex +++ b/inputs/lecture_06.tex @@ -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} diff --git a/inputs/lecture_07.tex b/inputs/lecture_07.tex index 9c0441e..9bb4ee5 100644 --- a/inputs/lecture_07.tex +++ b/inputs/lecture_07.tex @@ -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} +}{} diff --git a/inputs/lecture_08.tex b/inputs/lecture_08.tex index 2a5a0d5..70e07d6 100644 --- a/inputs/lecture_08.tex +++ b/inputs/lecture_08.tex @@ -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: diff --git a/inputs/lecture_14.tex b/inputs/lecture_14.tex index 0e3f3c5..1b21f56 100644 --- a/inputs/lecture_14.tex +++ b/inputs/lecture_14.tex @@ -298,7 +298,7 @@ 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: $\xi_0 \coloneqq \min C$, diff --git a/inputs/lecture_17.tex b/inputs/lecture_17.tex index e7098ff..0b0da9b 100644 --- a/inputs/lecture_17.tex +++ b/inputs/lecture_17.tex @@ -252,7 +252,7 @@ We will only proof \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 $\{\beta\} = g_Y''T$, i.e.~$\overline{f}_Y(\alpha) < \aleph_{\beta}$ for $Y\in A_{S,T}, \alpha \in T$. + \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 $Y,Y' \in A_{S,T,\tilde{f}} \implies \forall \alpha \in T.~f_Y(\alpha) = f_{Y'}\left( \alpha \right) \overset{T \text{ unbounded}}{\implies} Y = Y'$ @@ -270,7 +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}