From 11f21af9664ebba96c5ac80a2a6bdf047a6cd50a Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Wed, 14 Feb 2024 13:35:06 +0100 Subject: [PATCH] thank you, Mirko! --- inputs/lecture_08.tex | 11 +++++++---- inputs/lecture_09.tex | 6 +++--- inputs/lecture_11.tex | 4 ++-- inputs/lecture_12.tex | 16 +++++++++------- inputs/lecture_13.tex | 12 ++++++------ inputs/lecture_15.tex | 6 +++--- inputs/lecture_16.tex | 14 +++++++------- 7 files changed, 37 insertions(+), 32 deletions(-) diff --git a/inputs/lecture_08.tex b/inputs/lecture_08.tex index 8f3450e..8edf55c 100644 --- a/inputs/lecture_08.tex +++ b/inputs/lecture_08.tex @@ -124,9 +124,12 @@ together with the additional axiom: \begin{fact} $\BGC$ is conservative over $\ZFC$, i.e.~for all formulae $\phi$ in the language - of set theory (only set variables): + of set theory (only set variables) + we have that if $\BGC \vdash \phi$ then $\ZFC \vdash \phi$. \end{fact} + + We cannot prove this fact at this point, as the proof requires forcing. The converse is easy however, i.e.~if @@ -172,9 +175,9 @@ $\ZFC \vdash \phi$ then $\BGC \vdash \phi$. \end{lemma} \begin{proof} Suppose such sequence exists. - Then $\{x_n : n < \omega\}$ - (this exists as by definition sequence of the $x_n$ is a function - and this set is the range of that function) + Then $\{x_n : n < \omega\}$% + \footnote{This exists as by definition the sequence $(x_n)$ is a function + $f\colon \omega \to V$ and this set is the image of $f$.} violates \AxFund. For the other direction let $M \neq \emptyset$ be some set. diff --git a/inputs/lecture_09.tex b/inputs/lecture_09.tex index b58e438..a8141dd 100644 --- a/inputs/lecture_09.tex +++ b/inputs/lecture_09.tex @@ -4,12 +4,12 @@ \begin{definition} Let $R$ be a binary relation. $R$ is called \vocab{well-founded} - if for all classes $X$, + iff for all classes $X$, there is an $R$-least $y$ such that there is no $z \in X$ with $(z,y) \in R$. \end{definition} -\begin{theorem}[Induction (again, but now with classes)] +\begin{theorem}[Induction (again, but now for classes)] Suppose that $R$ is a well-founded relation. Let $X$ be a class such that for all sets $x$, \[ @@ -22,7 +22,7 @@ Consider $Y = \{x : x \not\in X\} \neq \emptyset$. By hypothesis, there is some $x \in Y$ such that $(y,x) \not\in R$ for all $y \in Y$. - In other words, if $(y,x) \not\in R$, + In other words, if $(y,x) \in R$, then $x \not\in Y$, i.e.~$x \in X$. Thus $\{y: (y,x) \in R\} \subseteq X$. Hence $x \in X \lightning$. diff --git a/inputs/lecture_11.tex b/inputs/lecture_11.tex index 2f81f92..cc55151 100644 --- a/inputs/lecture_11.tex +++ b/inputs/lecture_11.tex @@ -52,7 +52,7 @@ We often write $\kappa, \lambda, \ldots$ for cardinals. Let us now assume that for all $\kappa \in X$ there is some $\lambda \in X$ with $\lambda > \kappa$. - Suppose that $\sup(X)$ is no a cardinal + Suppose that $\sup(X)$ is not a cardinal and write $\mu = |\sup(X)|$. Then $\mu \in \sup(X)$, since $\sup(X)$ is an ordinal. @@ -96,7 +96,7 @@ so $|a| = \aleph_\beta$ for some $\beta \le |\alpha|$. \end{notation} \begin{notation} - Let $\leftindex^a b \coloneqq \{f : f \text{ is a function}, \dom(f) = \alpha, \ran(f) \subseteq b\}$. + Let $\leftindex^a b \coloneqq \{f : f \text{ is a function}, \dom(f) = a, \ran(f) \subseteq b\}$. \end{notation} \begin{definition}[Cardinal arithmetic] diff --git a/inputs/lecture_12.tex b/inputs/lecture_12.tex index b9d0fec..1979880 100644 --- a/inputs/lecture_12.tex +++ b/inputs/lecture_12.tex @@ -215,15 +215,17 @@ cf.~\yaref{def:inaccessible}. \end{IEEEeqnarray*} \begin{itemize} \item First case: $\beta \ge \alpha+1$. - Then + Note that for all $\gamma \le \beta$ + we have \[ - \aleph_{a+1}^{\aleph_\beta} \le \aleph_{\beta}^{\aleph_\beta} - \le \left( 2^{\aleph_{\beta}} \right)^{\aleph_\beta} - = 2^{\aleph_{\beta} \cdot \aleph_\beta} = 2^{\aleph_\beta} \le \aleph_{\alpha+1}^{\aleph_\beta}. + \aleph_{\gamma}^{\aleph_\beta} \le \aleph_\beta^{\aleph_\beta} + \le \left( 2^{\aleph_\beta} \right)^{\aleph_\beta} + = 2^{\aleph_\beta \cdot \aleph_\beta} = 2^{\aleph_{\beta}} + \le \aleph_\gamma^{\aleph_\beta}. \] - Also $\aleph_\alpha^{\aleph_{\beta}} = 2^{\aleph_\beta}$ - in this case (by the same argument), - so + So in this case $\aleph_\alpha^{\aleph_{\beta}} = 2^{\aleph_\beta}$ + and $\aleph_{\alpha+1}^{\aleph_\beta} = 2^{\aleph_{\beta}}$. + Thus \[ \aleph_{\alpha+1}^{\aleph_{\beta}} = 2^{\aleph_{\beta}} = \aleph_{\alpha}^{\aleph_\beta} = \aleph_{\alpha}^{\aleph_\beta} \cdot \aleph_{\alpha+1}. diff --git a/inputs/lecture_13.tex b/inputs/lecture_13.tex index d5c4feb..e3d8b35 100644 --- a/inputs/lecture_13.tex +++ b/inputs/lecture_13.tex @@ -4,7 +4,7 @@ There are many well-orders on $\omega$. Let $W$ be the set of all such well-orders. For $R, S \in W$, - write $R \le S$ if $R$ is isomorphic to + write $R \le S$ iff $R$ is isomorphic to an initial segment of $S$. Consider $\faktor{W}{\sim}$, where $R \sim S :\iff R \le S \land S \le R$. @@ -15,7 +15,7 @@ Suppose that $\{R_n : n \in \omega\} \subseteq W$ is such that $R_{n+1} < R_n$. Then there exist $n_i \in \omega$ - such that $R_i \cong R_0\defon{x <_{R_0} n_i}$ + such that $R_i \cong R_0\defon{\{x : x <_{R_0} n_i\}}$ and these form a $<_{R_0}$ strictly decreasing sequence. So $(\faktor{W}{\sim})$ @@ -116,13 +116,13 @@ then $\cf(2^{\kappa}) = 2^{\kappa} > \kappa$, since successor cardinals are regular. - Suppose $\cf(2^{k}) \le \kappa$ is a limit cardinal. + Suppose $\cf(2^{\kappa}) \le \kappa$ is a limit cardinal. Then there is some cofinal $f\colon \kappa\to 2^{\kappa}$. Write $\kappa_i = f(i)$ (replacing $f(i)$ by $|f(i)|^{+}$ we may assume that every $\kappa_i$ is a cardinal). - For $i \in \kappa$, write $\lambda_i = 2^{k}$. + For $i \in \kappa$, write $\lambda_i = 2^{\kappa}$. By \yaref{thm:koenig}, \[ \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} @@ -171,14 +171,14 @@ Relevant concepts to prove this theorem: \item We say that $A \subseteq \alpha$ is \vocab{unbounded} (in $\alpha$), iff for all $\beta < \alpha$, - there is some $\gamma \in \alpha$ + there is some $\gamma \in A$ such that $\beta < \gamma$. \item We say that $A \subseteq \alpha$ is \vocab{closed}, iff it is closed with respect to the order topology on $\alpha$, i.e.~for all $\beta < \alpha$, \[ - \sup(A \cap \beta) \in A. + \sup(A \cap \beta) \in A \cup \{0\} . \] \item $A$ is \vocab{club} (\emph{cl}osed \emph{un}bounded) iff it is closed and unbounded. diff --git a/inputs/lecture_15.tex b/inputs/lecture_15.tex index 5c651c2..46fca95 100644 --- a/inputs/lecture_15.tex +++ b/inputs/lecture_15.tex @@ -77,7 +77,7 @@ Recall the following: of arbitrary sizes. How do we do this? Let $\phi$ be a formula. - A \vocab{Skolem-function} + A \vocab{Skolem-function} over $V_\theta$ for $\phi$ is a function \[ @@ -136,7 +136,7 @@ Let's do a second proof of \yaref{thm:fodor}. \] Note that $|X_{\xi}| = |X_{\xi + 1}|$ - but the size is increased at limits. + but the size may increase at limits. It is easy to see inductively that $|X_{\xi}| < \kappa$ for every $\xi < \kappa$, while $X_\xi \subsetneq X_{\xi'}$ @@ -158,7 +158,7 @@ Let's do a second proof of \yaref{thm:fodor}. Let $\zeta < \kappa$. Let us define a strictly increasing sequence $ \langle \xi_n : n < \omega \rangle$ - a follows. + as follows. Set $\xi_0 \coloneqq \zeta$. Suppose $\xi_n$ has been chosen. Look at $X_{\xi_n} \cap \kappa$. diff --git a/inputs/lecture_16.tex b/inputs/lecture_16.tex index 8891ee3..275f4b9 100644 --- a/inputs/lecture_16.tex +++ b/inputs/lecture_16.tex @@ -41,7 +41,7 @@ where $f(\alpha) = \gamma$. \gist{% Recall that $F \subseteq \cP(\kappa)$ is a filter if $X,Y \in F \implies X \cap Y \in F$, -$X \in X, X \subseteq Y \subseteq \kappa \implies Y \in F$ +$X \in F, X \subseteq Y \subseteq \kappa \implies Y \in F$ and $\emptyset \not\in F, \kappa \in F$. \todo{Move this to the definition of filter?} }{} @@ -86,7 +86,7 @@ one cofinality. If $S \subseteq \kappa$ is stationary, there is a sequence $\langle S_i : i < \kappa \rangle$ - of pairwise disjoint stationary sets of $\kappa$ + of pairwise disjoint stationary subsets of $\kappa$ such that $S = \bigcup S_i$. \end{theorem} \begin{corollary} @@ -109,12 +109,12 @@ one cofinality. % TODO: Look at this again and think about it. % TODO TODO TODO - We will only proof this for $\aleph_1$. + We will only prove this for $\aleph_1$. Fix $S \subseteq \aleph_1$ stationary. For each $0 < \alpha < \omega_1$, either $\alpha$ is a successor ordinal - or $\alpha$ is a limit ordinal and $\cf(\alpha) = \omega_1$. + or $\alpha$ is a limit ordinal and $\cf(\alpha) = \omega$. Let $S^\ast \coloneqq \{\alpha \in S \setminus \{0\} : \alpha \text{ is a limit ordinal}\} $. $S^\ast$ is still stationary: @@ -163,7 +163,7 @@ one cofinality. Let $C' \coloneqq C \setminus (\delta^\ast + 1)$. $C'$ is still club. - As $\delta^\ast$ is stationary, + As $S^\ast$ is stationary, we may pick some $\alpha \in S^\ast \cap C'$. But then $\gamma_n^{\alpha} > \delta^\ast$ for $n$ large enough @@ -285,9 +285,9 @@ However we can say something about singular cardinals: Recall that $\CH$ says that $2^{\aleph_0} = \aleph_1$. So $\GCH \implies \CH$. -\yalabel{thm:silver} says that if $\GCH$ is true below $\kappa$, +\yaref{thm:silver} says that if $\GCH$ is true below $\kappa$, then it is true at $\kappa$. -The proof of \yalabel{thm:silver} is quite elementary, +The proof of \yaref{thm:silver} is quite elementary, so we will do it now, but the statement can only be fully appreciated later. }{}