diff --git a/inputs/lecture_05.tex b/inputs/lecture_05.tex new file mode 100644 index 0000000..dd4db06 --- /dev/null +++ b/inputs/lecture_05.tex @@ -0,0 +1,293 @@ +\lecture{05}{2023-10-30}{} + +\begin{definition} + Zermelo: + \[\Zermelo \coloneqq \AoE + \AoF + \AoP + \AoU + \Pow + \AoI + \Aus_{\phi}\] + + Zermelo and Fraenkl: + \[ + {\ZF} \coloneqq Z + (\Rep_{\phi}) + \] + + \[ + {\ZFC} \coloneqq \ZF + \Choice + \] + + Variants: + + \[ + {\ZFC^{-}} \coloneqq \ZFC \setminus \Pow. + \] + \[ + {\ZFC^{-\infty}} \coloneqq \ZFC \setminus \Inf + \] +\end{definition} + +\begin{definition} + For sets $x, y$ we write + $(x,y)$ for $\{\{x\}, \{x,y\}\}$. +\end{definition} +\begin{remark} + Note that $(x,y) = (a,b) \iff x = a \land y = b$. + $\ZFC$ proves that $(x,y)$ always exists. +\end{remark} + +\begin{definition} + For sets $x_1,\ldots, x_{n+1}$ + we write + \[ + (x_1,\ldots,x_{n+1}) \coloneqq ((x_1,\ldots,x_n), x_{n+1}) + \] + where we assume that $(x_1,\ldots,x_{n})$ + is already defined. +\end{definition} +\begin{definition} + The \vocab{cartesian product} + $a \times b$ of two sets $a$ and $b$ + is defined to be $a \times b \coloneqq \{(x,y) | x \in a \land y \in b\}$. +\end{definition} +\begin{fact} + $a \times b$ exists. +\end{fact} +\begin{proof} + Use $\Aus$ over $\cP(\cP(a \cup b))$. +\end{proof} + +\begin{definition} + For $a_1,\ldots, a_n$ + we define + \[ + a_1 \times \ldots \times a_n \coloneqq \left( a_1 \times \ldots\times a_{n-1} \right) \times a_n. + \] + recursively. + + For $a = a_1 = \ldots = a_n$, + we write $a^n$ for $a_1 \times \ldots \times a_n$. +\end{definition} + +\begin{remark} + The fact that $\ZFC$ can be used + to encode all of mathematics, + should not be overestimated. + It is clumsy to do it that way. + Nobody cares anymore. + There are better foundations. + What makes $\ZFC$ special + is that it allows to investigate infinity. +\end{remark} + +\begin{definition} + An \vocab{$n$-ary relation} $R$ is a subset + of $a_1 \times \ldots \times a_n$ + for some sets $a_1,\ldots,a_n$. + + For a \vocab{binary relation} $R $ (i.e.~$n = 2$) + we define + \[ + \dom(R) \coloneqq \{ x | \exists y.~(x,y) \in R\} + \] + and + \[ + \ran(R) \coloneqq \{ y | \exists x.~(x,y) \in R\}. + \] +\end{definition} +\begin{definition} + A binary relation $R$ + is a \vocab{function} + iff + \[ + \forall x \in \dom(R).~\exists y.~\forall y'.~(y' = y \iff xRy'). + \] + + A function $f$ is a function from $d$ to $b$ + iff $d = \dom(f)$ and $\ran(f) \subseteq b$. + + We write $f\colon d \to b$. + The set of all function from $d$ to $b$ + is denoted by ${}^d b$ or $b^d$. +\end{definition} + +\begin{fact} + Given sets $d, b$ then + ${}^d b$ exists. +\end{fact} +\begin{proof} + Apply again $\Aus$ over $\cP(d \times b)$. +\end{proof} + +\begin{definition} + We all know how + \vocab{injective}, \vocab{surjective}, \vocab{bijective}, $\ldots$ + are defined. + % TODO +\end{definition} + +\begin{notation} + For $f\colon d \to b$ and $a \subseteq d$ + we write $f''a \coloneqq \{f(x) : x \in a\}$ + (the \vocab{pointwise image} of $a$ under $f$). + + (In other mathematical fields, this is sometimes + denoted as $f(a)$. We don't do that here.) +\end{notation} + +\begin{definition} + A binary relation $\le $ on a set $a$ + is a \vocab{partial order} + iff $\le $ is + \begin{itemize} + \item \vocab{reflexive}, + i.e.~$x \le x$, + \item \vocab{antisymmetric} (sometimes + this is also called \vocab{symmetric}), + i.e.~ + $x \le y \land x \le y \implies x = y$, + and + \item \vocab{transitive}, + i.e.~$x \le y \land x \le z \implies x \le z$. + \end{itemize} + + If additionally $\forall x,y.~(x\le y \lor y \le x)$, + $\le $ is called a \vocab{linear order} + (or \vocab{total order}). +\end{definition} + +\begin{definition} + Let $(a, \le )$ be a partial order. + Let $b \subseteq a$. + We say that $x$ is a \vocab{maximal element} + of $b$ + iff + \[ + x \in b \land \lnot \exists y \in b .~(y > x).b .~(y > x). + \] + + In a similar way we define \vocab[Minimal element]{minimal elements} + of $b$. + We say that $x $ is an \vocab{upper bound} + of $b$ if $\forall y \in b.~(x \ge y)$. + Similarly \vocab[Lower bound]{lower bounds} + are defined. + + We say $x = \sup(b)$ if $x$ is the minimum + of the set of upper bounds of $b$. + (This does not necessarily exist.) + Similarly $\inf(b)$ is defined. +\end{definition} +\begin{definition} + Let $(a, \le_a)$ and $(b, \le_b)$ + be two partial orders. + Then a function $f\colon a\to b$ is caled + \vocab{order preserving} + iff + \[ + \forall x,y \in a.~(x \le_a y) \iff f(x) \le_b f(y). + \] + An order preserving bijection + is called an isomorphism. + 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 + a \vocab{well-order}, + iff + \[ + \forall b \subseteq a.~b\neq \emptyset \implies \min(b) \text{ exists}. + \] +\end{definition} + +\begin{fact} + Let $(a, \le )$ be a well-order, + then $(a, \le )$ is total. +\end{fact} +\begin{proof} + For $x,y \in a$ + consider $\{x,y\}$. + Then $\min(\{x,y\}) \le x,y$. +\end{proof} + +\begin{lemma} + Let $(a, \le)$ be a well-order. + Let $f\colon a \to a$ + be an order preserving map. + Then $f(x) \ge x$ for all $x \in a$. +\end{lemma} +\begin{proof} + Consider $x_0 \coloneqq \min(\{x \in a | f(x) < x\})$. +% Then $y_0 \coloneqq f(x_0) < x_0$, +% so $f(f(y_0)) < f(x_0) < x_0 = y_0$. +\end{proof} + +\begin{lemma} + If $(a, \le )$ is a well order + and $f\colon (a, \le) \leftrightarrow (a, \le)$ + is an isomorphism, + then $f$ is the identity. +\end{lemma} +\begin{proof} + By the last lemma, we know that + $f(x) \ge x$ and $f^{-1}(x) \ge x$. +\end{proof} + +\begin{lemma} + Suppose $(a, \le_a)$ and $(b, \le_b)$ are well-orderings + such that $(a, \le_a) \cong (b, \le_b)$. + Then there is a unique isomorphism + $f\colon a \to b$. +\end{lemma} +\begin{proof} + Let $f,g$ be isomorphisms + and consider $g^{-1}\circ f \colon (a, \le ) \xrightarrow{\cong} (a, \le )$. + We have already shown that $g^{-1}\circ f$ must be the identity, + so $g = f$. +\end{proof} + +\begin{definition} + If $(a, \le )$ is a partial order + and if $x \in a$, + then write $(a, \le )\defon{x}$ + for $(\{y \in a | y \le x\}, \le \cap \{y \in a | y \le x\}^2)$. +\end{definition} +\begin{theorem} + Let $(a,\le_a)$ and $(b,\le_b)$ be well-orders. + Then exactly one of the following three holds: + \begin{enumerate}[(i)] + \item $a \cong b$, + \item $\exists x \in b.~a \cong b\defon{x}$, + \item $\exists x \in a.~a\defon{x} \cong b$. + \end{enumerate} +\end{theorem} +\begin{proof} + Let us define a relation $r \subseteq a \times b$ as follows: + Let $(x,y) \in r$ iff $a\defon{x} \cong b\defon{y}$. + By the previous lemma, + for each $x \in a$, there is at most one $y \in b$ + such that $(x,y) \in r$ + and vice versa, + so $r$ is an injective function + from a subset of $a$ to a subset of $b$. + \begin{claim} + $r$ is order preserving: + \end{claim} + \begin{subproof} + If $x <_a x'$, then consider the unique $y'$ + such that $a\defon{x'} \cong b\defon{y'}$. + The isomorphism restricts to $a\defon{x} \cong b\defon{y}$ + for some $y <_b y'$. + \end{subproof} + + \begin{claim} + $\dom(r) = a \lor \ran(r) = b$. + \end{claim} + \begin{subproof} + Suppose that $\dom(r) \subsetneq a$ + and $\ran(r) \susbsetneq b$. + + Let $x \coloneqq \min(a \setminus \dom(r))$ + and $y \coloneqq \min(b\setminus \ran(r))$. + Then $(a,\le)\defon{x} \cong (b, \le)\defon{y}$. + But now $(x,y) \in r$ which is a contradiction. + \end{subproof} +\end{proof} diff --git a/logic.sty b/logic.sty index 76dd601..8b9710c 100644 --- a/logic.sty +++ b/logic.sty @@ -104,6 +104,8 @@ \DeclareSimpleMathOperator{Con} + +\DeclareMathOperator{\Zermelo}{Z} \DeclareSimpleMathOperator{ZF} \DeclareSimpleMathOperator{ZFC} \DeclareSimpleMathOperator{HOD} @@ -115,6 +117,15 @@ \DeclareSimpleMathOperator{Rep} \DeclareSimpleMathOperator{Pow} +\DeclareSimpleMathOperator{AoE} +\DeclareSimpleMathOperator{AoF} +\DeclareSimpleMathOperator{AoP} +\DeclareSimpleMathOperator{AoU} +\DeclareSimpleMathOperator{AoI} +\DeclareSimpleMathOperator{Choice} +\DeclareSimpleMathOperator{Inf} + + \renewcommand{\Aus}{\text{Aus}} % \DeclareSimpleMathOperator{Aus} \DeclareSimpleMathOperator{Infinity} diff --git a/logic2.tex b/logic2.tex index f0624db..888ff01 100644 --- a/logic2.tex +++ b/logic2.tex @@ -27,6 +27,8 @@ \input{inputs/lecture_01} \input{inputs/lecture_02} \input{inputs/lecture_03} +% \input{inputs/lecture_04} +\input{inputs/lecture_05} \cleardoublepage