\lecture{05}{2023-10-30}{}

\begin{definition}
    Zermelo:
    \[\Zermelo \coloneqq \AxExt + \AxFund + \AxPair + \AxUnion + \AxPow + \AxInf + \AxAus_{\phi}\]

    Zermelo and Fraenkl:
    \[
    {\ZF} \coloneqq  Z + \AxRep_{\phi} % TODO fix parenthesis
    \]

    \[
    {\ZFC} \coloneqq  \ZF + \AxC
    \]

    Variants:

    \[
    {\ZFC^{-}} \coloneqq  \ZFC \setminus \AxPow.
    \]
    \[
    {\ZFC^{-\infty}}  \coloneqq \ZFC \setminus \AxInf
    \]
\end{definition}

\gist{%
\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 \AxAus 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 $\leftindex^d b$ or $b^d$.
\end{definition}

\begin{fact}
   Given sets $d, b$ then
    $\leftindex^d b$ exists.
\end{fact}
\begin{proof}
    Apply again \AxAus 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}
}{[Some boring definitions omitted.]}
\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.~$x \le y \land y \le z \implies x \le z$.
    \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
    \[
    x \in  b \land \lnot \exists  y \in b .~(y > x).
    \] 
    We say that $x$ is the \vocab{maximum} of $b$,
    $x = \text{\vocab{$\max$}}(b)$,
    iff
    \[
    x \in b \land \forall y \in b.~y \le x.
    \]

    In a similar way we define \vocab[Minimal element]{minimal elements} 
    and the \vocab{minimum} of $b$.
    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 $x = \text{\vocab{$\sup$}}(b)$ if $x$ is the minimum
    of the set of upper bounds of $b$.
    (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.
    However for linear orders these notions coincide.
\end{remark}
\begin{definition}
    Let $(a, \le_a)$ and $(b, \le_b)$
    be two partial orders.
    Then a function $f\colon a\to b$ is called
    \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 \vocab{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 $g^{-1}\circ f \colon (a, \le_a) \xrightarrow{\cong} (a, \le_a )$.
    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{abuse}+
    For a partial order $(a, \le_a)$
    we %\footnote{i.e.~the lazy author of these notes}
    sometimes just write $a$.
\end{abuse}
\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 $\ran(r) \subsetneq b$.

        Let $x \coloneqq  \min(a \setminus \dom(r))$
        and $y \coloneqq  \min(b\setminus \ran(r))$.
        Then $(a,\le_a)\defon{x} \cong (b, \le_b)\defon{y}$.
        But now $(x,y) \in r$ which is a contradiction.
    \end{subproof}
\end{proof}