lecture 5

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}

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}

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