Josia Pietsch
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313 lines
8.3 KiB
TeX
313 lines
8.3 KiB
TeX
\lecture{05}{2023-10-30}{}
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\begin{definition}
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Zermelo:
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\[\Zermelo \coloneqq \AxExt + \AxFund + \AxPair + \AxUnion + \AxPow + \AxInf + \AxAus_{\phi}\]
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Zermelo and Fraenkl:
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\[
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{\ZF} \coloneqq Z + \AxRep_{\phi} % TODO fix parenthesis
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\]
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\[
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{\ZFC} \coloneqq \ZF + \AxC
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\]
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Variants:
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\[
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{\ZFC^{-}} \coloneqq \ZFC \setminus \AxPow.
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\]
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\[
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{\ZFC^{-\infty}} \coloneqq \ZFC \setminus \AxInf
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\]
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\end{definition}
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\gist{%
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\begin{definition}
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For sets $x, y$ we write
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$(x,y)$ for $\{\{x\}, \{x,y\}\}$.
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\end{definition}
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\begin{remark}
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Note that $(x,y) = (a,b) \iff x = a \land y = b$.
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$\ZFC$ proves that $(x,y)$ always exists.
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\end{remark}
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\begin{definition}
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For sets $x_1,\ldots, x_{n+1}$
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we write
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\[
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(x_1,\ldots,x_{n+1}) \coloneqq ((x_1,\ldots,x_n), x_{n+1})
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\]
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where we assume that $(x_1,\ldots,x_{n})$
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is already defined.
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\end{definition}
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\begin{definition}
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The \vocab{cartesian product}
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$a \times b$ of two sets $a$ and $b$
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is defined to be $a \times b \coloneqq \{(x,y) | x \in a \land y \in b\}$.
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\end{definition}
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\begin{fact}
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$a \times b$ exists.
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\end{fact}
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\begin{proof}
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Use \AxAus over $\cP(\cP(a \cup b))$.
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\end{proof}
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\begin{definition}
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For $a_1,\ldots, a_n$
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we define
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\[
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a_1 \times \ldots \times a_n \coloneqq \left( a_1 \times \ldots\times a_{n-1} \right) \times a_n.
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\]
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recursively.
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For $a = a_1 = \ldots = a_n$,
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we write $a^n$ for $a_1 \times \ldots \times a_n$.
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\end{definition}
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\begin{remark}
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The fact that $\ZFC$ can be used
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to encode all of mathematics,
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should not be overestimated.
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It is clumsy to do it that way.
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Nobody cares anymore.
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There are better foundations.
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What makes $\ZFC$ special
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is that it allows to investigate infinity.
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\end{remark}
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\begin{definition}
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An \vocab{$n$-ary relation} $R$ is a subset
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of $a_1 \times \ldots \times a_n$
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for some sets $a_1,\ldots,a_n$.
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For a \vocab{binary relation} $R $ (i.e.~$n = 2$)
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we define
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\[
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\dom(R) \coloneqq \{ x | \exists y.~(x,y) \in R\}
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\]
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and
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\[
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\ran(R) \coloneqq \{ y | \exists x.~(x,y) \in R\}.
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\]
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\end{definition}
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\begin{definition}
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A binary relation $R$
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is a \vocab{function}
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iff
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\[
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\forall x \in \dom(R).~\exists y.~\forall y'.~(y' = y \iff xRy').
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\]
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A function $f$ is a function from $d$ to $b$
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iff $d = \dom(f)$ and $\ran(f) \subseteq b$.
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We write $f\colon d \to b$.
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The set of all function from $d$ to $b$
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is denoted by $\leftindex^d b$ or $b^d$.
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\end{definition}
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\begin{fact}
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Given sets $d, b$ then
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$\leftindex^d b$ exists.
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\end{fact}
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\begin{proof}
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Apply again \AxAus over $\cP(d \times b)$.
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\end{proof}
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\begin{definition}
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We all know how
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\vocab{injective}, \vocab{surjective}, \vocab{bijective}, $\ldots$
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are defined.
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% TODO
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\end{definition}
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\begin{notation}
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For $f\colon d \to b$ and $a \subseteq d$
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we write $f''a \coloneqq \{f(x) : x \in a\}$
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(the \vocab{pointwise image} of $a$ under $f$).
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(In other mathematical fields, this is sometimes
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denoted as $f(a)$. We don't do that here.)
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\end{notation}
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}{[Some boring definitions omitted.]}
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\begin{definition}
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A binary relation $\le $ on a set $a$
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is a \vocab{partial order}
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iff $\le $ is
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\begin{itemize}
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\item \vocab{reflexive},
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i.e.~$x \le x$,
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\item \vocab{antisymmetric} (sometimes
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this is also called \vocab{symmetric}),
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i.e.~
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$x \le y \land x \le y \implies x = y$,
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and
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\item \vocab{transitive},
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i.e.~$x \le y \land y \le z \implies x \le z$.
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\end{itemize}
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If additionally $\forall x,y.~(x\le y \lor y \le x)$,
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$\le $ is called a \vocab{linear order}
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(or \vocab{total order}).
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\end{definition}
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\begin{definition}
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Let $(a, \le )$ be a partial order.
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Let $b \subseteq a$.
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We say that $x$ is a \vocab{maximal element}
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of $b$
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iff
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\[
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x \in b \land \lnot \exists y \in b .~(y > x).
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\]
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We say that $x$ is the \vocab{maximum} of $b$,
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$x = \text{\vocab{$\max$}}(b)$,
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iff
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\[
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x \in b \land \forall y \in b.~y \le x.
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\]
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In a similar way we define \vocab[Minimal element]{minimal elements}
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and the \vocab{minimum} of $b$.
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We say that $x $ is an \vocab{upper bound}
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of $b$ if $\forall y \in b.~(x \ge y)$.
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Similarly \vocab[Lower bound]{lower bounds}
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are defined.
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We say $x = \text{\vocab{$\sup$}}(b)$ if $x$ is the minimum
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of the set of upper bounds of $b$.
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(This does not necessarily exist.)
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Similarly $\text{\vocab{$\inf$}}(b)$ is defined.
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\end{definition}
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\gist{
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\begin{remark}+
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Note that in a partial order,
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a maximal element is not necessarily a maximum.
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However for linear orders these notions coincide.
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\end{remark}
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\begin{definition}
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Let $(a, \le_a)$ and $(b, \le_b)$
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be two partial orders.
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Then a function $f\colon a\to b$ is called
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\vocab{order-preserving}
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iff
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\[
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\forall x,y \in a.~(x \le_a y) \iff f(x) \le_b f(y).
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\]
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An order-preserving bijection
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is called an \vocab{isomorphism}.
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We write $(a,\le_a) \cong (b, \le_b)$
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if they are isomorphic.
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\end{definition}
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}{}
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\begin{definition}
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Let $(a,\le)$ be a partial order.
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Then $(a,\le)$ is
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a \vocab{well-order},
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iff
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\[
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\forall b \subseteq a.~b\neq \emptyset \implies \min(b) \text{ exists}.
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\]
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\end{definition}
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\begin{fact}
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Let $(a, \le )$ be a well-order,
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then $(a, \le )$ is total.
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\end{fact}
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\begin{proof}
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For $x,y \in a$
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consider $\{x,y\}$.
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Then $\min(\{x,y\}) \le x,y$.
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\end{proof}
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\begin{lemma}
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Let $(a, \le)$ be a well-order.
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Let $f\colon a \to a$
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be an order-preserving map.
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Then $f(x) \ge x$ for all $x \in a$.
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\end{lemma}
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\begin{proof}
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Consider $x_0 \coloneqq \min(\{x \in a | f(x) < x\})$.
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% Then $y_0 \coloneqq f(x_0) < x_0$,
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% so $f(f(y_0)) < f(x_0) < x_0 = y_0$.
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\end{proof}
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\begin{lemma}
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If $(a, \le )$ is a well-order
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and $f\colon (a, \le) \leftrightarrow (a, \le)$
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is an isomorphism,
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then $f$ is the identity.
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\end{lemma}
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\begin{proof}
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By the last lemma, we know that
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$f(x) \ge x$ and $f^{-1}(x) \ge x$.
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\end{proof}
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\begin{lemma}
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Suppose $(a, \le_a)$ and $(b, \le_b)$ are well-orderings
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such that $(a, \le_a) \cong (b, \le_b)$.
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Then there is a unique isomorphism
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$f\colon a \to b$.
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\end{lemma}
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\begin{proof}
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Let $f,g$ be isomorphisms
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and consider $g^{-1}\circ f \colon (a, \le_a) \xrightarrow{\cong} (a, \le_a )$.
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We have already shown that $g^{-1}\circ f$ must be the identity,
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so $g = f$.
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\end{proof}
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\begin{definition}
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If $(a, \le )$ is a partial order
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and if $x \in a$,
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then write $(a, \le )\defon{x}$
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for $(\{y \in a | y \le x\}, \le \cap \{y \in a | y \le x\}^2)$.
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\end{definition}
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\begin{abuse}+
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For a partial order $(a, \le_a)$
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we %\footnote{i.e.~the lazy author of these notes}
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sometimes just write $a$.
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\end{abuse}
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\begin{theorem}
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Let $(a,\le_a)$ and $(b,\le_b)$ be well-orders.
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Then exactly one of the following three holds:
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\begin{enumerate}[(i)]
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\item $a \cong b$,
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\item $\exists x \in b.~a \cong b\defon{x}$,
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\item $\exists x \in a.~a\defon{x} \cong b$.
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\end{enumerate}
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\end{theorem}
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\begin{proof}
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Let us define a relation $r \subseteq a \times b$ as follows:
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Let $(x,y) \in r$ iff $a\defon{x} \cong b\defon{y}$.
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By the previous lemma,
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for each $x \in a$, there is at most one $y \in b$
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such that $(x,y) \in r$
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and vice versa,
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so $r$ is an injective function
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from a subset of $a$ to a subset of $b$.
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\begin{claim}
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$r$ is order-preserving:
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\end{claim}
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\begin{subproof}
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If $x <_a x'$, then consider the unique $y'$
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such that $a\defon{x'} \cong b\defon{y'}$.
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The isomorphism restricts to $a\defon{x} \cong b\defon{y}$
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for some $y <_b y'$.
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\end{subproof}
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\begin{claim}
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$\dom(r) = a \lor \ran(r) = b$.
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\end{claim}
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\begin{subproof}
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Suppose that $\dom(r) \subsetneq a$
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and $\ran(r) \subsetneq b$.
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Let $x \coloneqq \min(a \setminus \dom(r))$
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and $y \coloneqq \min(b\setminus \ran(r))$.
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Then $(a,\le_a)\defon{x} \cong (b, \le_b)\defon{y}$.
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But now $(x,y) \in r$ which is a contradiction.
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\end{subproof}
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\end{proof}
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