diff --git a/inputs/lecture_04.tex b/inputs/lecture_04.tex index 6c65c9f..25845cb 100644 --- a/inputs/lecture_04.tex +++ b/inputs/lecture_04.tex @@ -2,7 +2,7 @@ % Model-theoretic concepts and ultraproducts -\section{$\ZFC$} +\section{\texorpdfstring{$\ZFC$}{ZFC}} % 1900, Russel's paradox \todo{Russel's Paradox} diff --git a/inputs/lecture_05.tex b/inputs/lecture_05.tex index de6f6f0..33c2d28 100644 --- a/inputs/lecture_05.tex +++ b/inputs/lecture_05.tex @@ -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 $\cP(d \times b)$. diff --git a/inputs/lecture_08.tex b/inputs/lecture_08.tex index 8a7c7e2..8f3450e 100644 --- a/inputs/lecture_08.tex +++ b/inputs/lecture_08.tex @@ -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 $\overline{g} \in G$ diff --git a/inputs/lecture_09.tex b/inputs/lecture_09.tex index cbd0b52..b58e438 100644 --- a/inputs/lecture_09.tex +++ b/inputs/lecture_09.tex @@ -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} - - - - diff --git a/inputs/lecture_10.tex b/inputs/lecture_10.tex index 889313b..f144b06 100644 --- a/inputs/lecture_10.tex +++ b/inputs/lecture_10.tex @@ -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\}. diff --git a/inputs/lecture_11.tex b/inputs/lecture_11.tex index 95f5879..2f81f92 100644 --- a/inputs/lecture_11.tex +++ b/inputs/lecture_11.tex @@ -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 $2^{\aleph_0} = \aleph_1$. diff --git a/inputs/lecture_12.tex b/inputs/lecture_12.tex index 73dd365..29c8cf2 100644 --- a/inputs/lecture_12.tex +++ b/inputs/lecture_12.tex @@ -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} diff --git a/inputs/lecture_13.tex b/inputs/lecture_13.tex index c1a9012..a277354 100644 --- a/inputs/lecture_13.tex +++ b/inputs/lecture_13.tex @@ -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} diff --git a/inputs/lecture_14.tex b/inputs/lecture_14.tex index 56711e7..84abc52 100644 --- a/inputs/lecture_14.tex +++ b/inputs/lecture_14.tex @@ -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} diff --git a/inputs/lecture_15.tex b/inputs/lecture_15.tex index 878c124..1335f07 100644 --- a/inputs/lecture_15.tex +++ b/inputs/lecture_15.tex @@ -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 $V_{\theta}$. diff --git a/inputs/lecture_16.tex b/inputs/lecture_16.tex index b62940e..e6104e3 100644 --- a/inputs/lecture_16.tex +++ b/inputs/lecture_16.tex @@ -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} diff --git a/inputs/lecture_18.tex b/inputs/lecture_18.tex index 120c3ee..0320872 100644 --- a/inputs/lecture_18.tex +++ b/inputs/lecture_18.tex @@ -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. diff --git a/logic.sty b/logic.sty index 594650f..e7fd748 100644 --- a/logic.sty +++ b/logic.sty @@ -24,6 +24,7 @@ \usepackage{multirow} \usepackage{float} \usepackage{scalerel} +\usepackage{leftindex} %\usepackage{algorithmicx} \newcounter{subsubsubsection}[subsubsection]