Suggestions by Mirko (pages 1 - 18), thank you!
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@ -1,4 +1,4 @@
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These are my notes on the lecture Probability Theory,
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These are my notes on the lecture Logic II
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taught by \textsc{Ralf Schindler}
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taught by \textsc{Ralf Schindler}
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in winter 23/24 at the University Münster.
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in winter 23/24 at the University Münster.
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@ -12,6 +12,6 @@ If you find errors or want to improve something,
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please send me a message:\\
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please send me a message:\\
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\texttt{lecturenotes@jrpie.de}.
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\texttt{lecturenotes@jrpie.de}.
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This notes follow the way the material was presented in the lecture rather
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These notes follow the way the material was presented in the lecture rather
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closely. Additions (e.g.~from exercise sheets)
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closely. Additions (e.g.~from exercise sheets)
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and slight modifications have been marked with $\dagger$.
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and slight modifications have been marked with $\dagger$.
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@ -31,7 +31,7 @@ Literature
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\end{definition}
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\end{definition}
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\begin{lemma}
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\begin{lemma}
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If $A \le B$,
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If $A \le B$,
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then there is a surjection $g\colon B \twoheadrightarrow B$.
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then there is a surjection $g\colon B \twoheadrightarrow A$.
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\end{lemma}
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\end{lemma}
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\begin{proof}
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\begin{proof}
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Fix $f\colon A \hookrightarrow B$.
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Fix $f\colon A \hookrightarrow B$.
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@ -150,7 +150,8 @@ Literature
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\begin{definition}
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\begin{definition}
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The \vocab{continuum hypothesis} ($\CH$)
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The \vocab{continuum hypothesis} ($\CH$)
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says that there is no set $A$ such that
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says that there is no set $A$ such that
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$\N < A < \R$.
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$\N < A < \R$,
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i.e.~every uncountable subset $A \subseteq \R$ is in bijection with $\R$.
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$\CH$ is equivalent to the statement that there is no set $A \subset \R$
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$\CH$ is equivalent to the statement that there is no set $A \subset \R$
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which is uncountable ($\N < A$)
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which is uncountable ($\N < A$)
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@ -69,9 +69,16 @@
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But then $x \in A' \subseteq A \lightning$.
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But then $x \in A' \subseteq A \lightning$.
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\end{refproof}
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\end{refproof}
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\begin{definition}
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\begin{definition}
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$P \subseteq \R$ is called \vocab{perfect}
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$P \subseteq \R$ (or, more generally, a subset of any topological space)
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is called \vocab{perfect}
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iff $P \neq \emptyset$ and $P = P'$.
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iff $P \neq \emptyset$ and $P = P'$.
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\end{definition}
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\end{definition}
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\begin{example}+
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Note that being perfect depends on the surrounding topological space:
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For example, $[0,1] \cap \Q$
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is perfect as a subset of $\Q$,
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but not perfect as a subset of $\R$.
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\end{example}
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We want to prove two things:
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We want to prove two things:
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\begin{itemize}
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\begin{itemize}
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@ -122,7 +129,7 @@ We want to prove two things:
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If $t \neq t' \in \{0,1\}^{\omega}$,
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If $t \neq t' \in \{0,1\}^{\omega}$,
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then there is some $n$ such that $t\defon{n} \neq t'\defon{n}$,
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then there is some $n$ such that $t\defon{n} \neq t'\defon{n}$,
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hence $f(t) \in [a_{t\defon{n}}, b_{t\defon{n}}]$
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hence $f(t) \in [a_{t\defon{n}}, b_{t\defon{n}}]$
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and $f(t') \in [a_{t\defon{n}}, b_{t\defon{n}}]$
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and $f(t') \in [a_{t'\defon{n}}, b_{t'\defon{n}}]$
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which are disjoint.
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which are disjoint.
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Thus $f(t) \neq f(t')$, i.e.~$f$ is injective.
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Thus $f(t) \neq f(t')$, i.e.~$f$ is injective.
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\end{proof}
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\end{proof}
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@ -58,7 +58,7 @@ all condensation points are accumulation points.
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Then
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Then
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\begin{IEEEeqnarray*}{rCl}
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\begin{IEEEeqnarray*}{rCl}
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A \setminus P &=& \bigcap_{x \in A \setminus P} (a_x, b_x) \cap A.
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A \setminus P &=& \bigcup_{x \in A \setminus P} (a_x, b_x) \cap A.
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\end{IEEEeqnarray*}
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\end{IEEEeqnarray*}
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$\subseteq $ holds by the choice of $a_x$ and $b_x$.
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$\subseteq $ holds by the choice of $a_x$ and $b_x$.
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For $\supseteq$ let $y$ be an element of the RHS.
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For $\supseteq$ let $y$ be an element of the RHS.
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@ -25,9 +25,24 @@ $\ZFC$ stands for
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y \in x \land z \in x \land \forall a .~(a \in x \implies a = y \lor a = z).
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y \in x \land z \in x \land \forall a .~(a \in x \implies a = y \lor a = z).
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\]
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\]
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Let $x = \bigcup y$ denote
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We write $z = x \cap y$ for
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\[\forall u.~((u \in z) \implies u \in x \land u \in y),\]
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$z = x \cup y$ for
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\[
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\[
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\forall z.~(z \in x \iff \exists v.(v \in y \land z \in v)).
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\forall u.~((u \in z) \iff (u \in x \lor u \in y)),
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\]
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$z = \bigcap x$ for
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\[
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\forall u.~((u \in z) \iff (\forall v.~(v \in x \implies u \in v))),
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\]
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$z = \bigcup x$ for
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\[
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\forall u.~((u \in z) \iff \exists v.~(v \in x \land u \in v))
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\]
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and
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$z = x \setminus y$ for
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\[
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\forall u.~((u \in z) \iff (u \in x \land u \not\in y)).
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\]
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\]
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\end{notation}
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\end{notation}
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$\ZFC$ consists of the following axioms:
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$\ZFC$ consists of the following axioms:
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@ -86,7 +101,7 @@ $\ZFC$ consists of the following axioms:
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% (PWA)
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% (PWA)
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We write $x = \cP(y)$
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We write $x = \cP(y)$
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for
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for
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$\forall z.~(z \in x \iff x \subseteq z)$.
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$\forall z.~(z \in x \iff z \subseteq x)$.
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The power set axiom states
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The power set axiom states
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\[
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\[
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\forall x.~\exists y.~y=\cP(x).
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\forall x.~\exists y.~y=\cP(x).
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@ -103,7 +118,9 @@ $\ZFC$ consists of the following axioms:
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\yalabel{Axiom of Separation}{(Aus)}{ax:aus}
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\yalabel{Axiom of Separation}{(Aus)}{ax:aus}
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Let $\phi$ be some fixed
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Let $\phi$ be some fixed
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fist order formula in $\cL_\in$.
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fist order formula in $\cL_\in$
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with free variables $x, v_1, \ldots, v_p$.
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Let $b$ be a variable that is not free in $\phi$.
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Then $\AxAus_{\phi}$
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Then $\AxAus_{\phi}$
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states
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states
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\[
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\[
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@ -115,16 +132,11 @@ $\ZFC$ consists of the following axioms:
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for $\forall x.~(x \in b \iff x \in a \land f(x))$.
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for $\forall x.~(x \in b \iff x \in a \land f(x))$.
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Then \AxAus can be formulated as
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Then \AxAus can be formulated as
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\[
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\[
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\forall a.~\exists b.~(b = \{x \in a; \phi(x)\}).
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\forall a.~\exists b.~(b = \{x \in a | \phi(x)\}).
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\]
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\]
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\end{axiomschema}
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\end{axiomschema}
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\begin{notation}
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\todo{$\cap, \setminus, \bigcap$}
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% We write $z = x \cap y$ for $\forall u.~((u \in z) \implies u \in x \land u \in y)$,
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% $Z = x \setminus y$ for ...
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% $x = \bigcap y$ for ...
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\end{notation}
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\begin{remark}
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\begin{remark}
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\AxAus proves that
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\AxAus proves that
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\begin{itemize}
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\begin{itemize}
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\end{remark}
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\end{remark}
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\begin{axiomschema}[\vocab{Replacement} (Fraenkel)]
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\begin{axiomschema}[\vocab{Replacement} (Fraenkel)]
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\yalabel{Axiom of Replacement}{(Rep)}{ax:rep}
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\yalabel{Axiom of Replacement}{(Rep)}{ax:rep}
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Let $\phi$ be some $\cL_{\in }$ formula.
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Let $\phi$ be some $\cL_{\in }$ formula
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with free variables $x, y$.\todo{Allow more variables}
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Then
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Then
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\[
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\[
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\forall v_1 .~\exists b.~\forall y.~(y \in b \iff \exists x .~(x \in a \land \phi(x,y,v_1,v_p))).
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\forall x \in a.~\exists !y \phi(x,y) \implies \exists b.~\forall x \in a.~\exists y \in b.~\phi(x,y).
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% \forall v_1 \ldots \forall v_p .~\forall x.~ \exists y'.~\forall y.~(y = y' \iff \phi(x))
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% \implies \forall a.~\exists b.~\forall y.~(y \in b \iff \exists x.~(x \in a \land \phi(x))
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\]
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\]
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\end{axiomschema}
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\end{axiomschema}
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\begin{axiom}[\vocab{Choice}]
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\begin{axiom}[\vocab{Choice}]
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\yalabel{Axiom of Choice}{(C)}{ax:c}
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\yalabel{Axiom of Choice}{(C)}{ax:c}
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Every family of non-empty sets has a \vocab{choice set}:
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Every family of pairwise disjoint non-empty sets has a \vocab{choice set}:
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\begin{IEEEeqnarray*}{rCl}
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\begin{IEEEeqnarray*}{rCl}
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\forall x .~&(&\\
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\forall x .~&(&\\
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&& ((\forall y \in x.~y \neq \emptyset) \land (\forall y \in x .~\forall y' \in x .~(y \neq y' \implies y \cap y' = \emptyset))\\
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&& ((\forall y \in x.~y \neq \emptyset) \land (\forall y \in x .~\forall y' \in x .~(y \neq y' \implies y \cap y' = \emptyset)))\\
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&& \implies\exists z.~\forall y \in x.~\exists u.~(z \cap y = \{u\})\\
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&& \implies\exists z.~\forall y \in x.~\exists u.~(z \cap y = \{u\})\\
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&)&
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&)&
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\end{IEEEeqnarray*}
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\end{IEEEeqnarray*}
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$x \le y \land x \le y \implies x = y$,
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$x \le y \land x \le y \implies x = y$,
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and
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and
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\item \vocab{transitive},
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\item \vocab{transitive},
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i.e.~$x \le y \land x \le z \implies x \le z$.
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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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\end{itemize}
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If additionally $\forall x,y.~(x\le y \lor y \le x)$,
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If additionally $\forall x,y.~(x\le y \lor y \le x)$,
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of $b$
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of $b$
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iff
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iff
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\[
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\[
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x \in b \land \lnot \exists y \in b .~(y > x).b .~(y > x).
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x \in b \land \lnot \exists y \in b .~(y > x).
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\]
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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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\]
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In a similar way we define \vocab[Minimal element]{minimal elements}
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In a similar way we define \vocab[Minimal element]{minimal elements}
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of $b$.
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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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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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of $b$ if $\forall y \in b.~(x \ge y)$.
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Similarly \vocab[Lower bound]{lower bounds}
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Similarly \vocab[Lower bound]{lower bounds}
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are defined.
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are defined.
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We say $x = \sup(b)$ if $x$ is the minimum
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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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of the set of upper bounds of $b$.
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(This does not necessarily exist.)
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(This does not necessarily exist.)
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Similarly $\inf(b)$ is defined.
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Similarly $\text{\vocab{$\inf$}}(b)$ is defined.
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\end{definition}
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\end{definition}
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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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\begin{definition}
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Let $(a, \le_a)$ and $(b, \le_b)$
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Let $(a, \le_a)$ and $(b, \le_b)$
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be two partial orders.
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be two partial orders.
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Then a function $f\colon a\to b$ is caled
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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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\vocab{order-preserving}
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iff
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iff
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\[
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\[
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\forall x,y \in a.~(x \le_a y) \iff f(x) \le_b f(y).
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\forall x,y \in a.~(x \le_a y) \iff f(x) \le_b f(y).
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\]
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\]
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An order preserving bijection
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An order-preserving bijection
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is called an isomorphism.
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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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We write $(a,\le_a) \cong (b, \le_b)$
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if they are isomorphic.
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if they are isomorphic.
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\end{definition}
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\end{definition}
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\begin{lemma}
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\begin{lemma}
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Let $(a, \le)$ be a well-order.
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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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Let $f\colon a \to a$
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be an order preserving map.
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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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Then $f(x) \ge x$ for all $x \in a$.
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\end{lemma}
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\end{lemma}
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\begin{proof}
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\begin{proof}
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\end{proof}
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\end{proof}
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\begin{lemma}
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\begin{lemma}
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If $(a, \le )$ is a well order
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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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and $f\colon (a, \le) \leftrightarrow (a, \le)$
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is an isomorphism,
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is an isomorphism,
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then $f$ is the identity.
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then $f$ is the identity.
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\end{lemma}
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\end{lemma}
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\begin{proof}
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\begin{proof}
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Let $f,g$ be isomorphisms
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Let $f,g$ be isomorphisms
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and consider $g^{-1}\circ f \colon (a, \le ) \xrightarrow{\cong} (a, \le )$.
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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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We have already shown that $g^{-1}\circ f$ must be the identity,
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so $g = f$.
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so $g = f$.
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\end{proof}
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\end{proof}
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then write $(a, \le )\defon{x}$
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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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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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\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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\begin{theorem}
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Let $(a,\le_a)$ and $(b,\le_b)$ be well-orders.
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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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Then exactly one of the following three holds:
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so $r$ is an injective function
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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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from a subset of $a$ to a subset of $b$.
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\begin{claim}
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\begin{claim}
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$r$ is order preserving:
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$r$ is order-preserving:
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\end{claim}
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\end{claim}
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\begin{subproof}
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\begin{subproof}
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If $x <_a x'$, then consider the unique $y'$
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If $x <_a x'$, then consider the unique $y'$
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Let $x \coloneqq \min(a \setminus \dom(r))$
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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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and $y \coloneqq \min(b\setminus \ran(r))$.
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Then $(a,\le)\defon{x} \cong (b, \le)\defon{y}$.
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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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But now $(x,y) \in r$ which is a contradiction.
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\end{subproof}
|
\end{subproof}
|
||||||
\end{proof}
|
\end{proof}
|
||||||
|
|
|
@ -3,13 +3,13 @@
|
||||||
\begin{theorem}[Zorn]
|
\begin{theorem}[Zorn]
|
||||||
\yalabel{Zorn's Lemma}{Zorn}{thm:zorn}
|
\yalabel{Zorn's Lemma}{Zorn}{thm:zorn}
|
||||||
Let $(a, \le )$ be a partial order with $a \neq \emptyset$.
|
Let $(a, \le )$ be a partial order with $a \neq \emptyset$.
|
||||||
Assume that $b \le a$ with $b \neq \emptyset$
|
Assume that for all $b \subseteq a$ with $b \neq \emptyset$
|
||||||
and $ b$ linearly ordered, $b$ has an upper bound,
|
and $ b$ linearly ordered, $b$ has an upper bound.
|
||||||
Then $a$ has a maximal element.
|
Then $a$ has a maximal element.
|
||||||
\end{theorem}
|
\end{theorem}
|
||||||
\begin{refproof}{thm:zorn}
|
\begin{refproof}{thm:zorn}
|
||||||
Fix $(a, \le )$ as in the hypothesis.
|
Fix $(a, \le )$ as in the hypothesis.
|
||||||
Let $A \coloneqq \{ \{(b,x) : x in b\} : b \le a, b \neq \emptyset\}$.
|
Let $A \coloneqq \{ \{(b,x) : x \in b\} : b \subseteq a, b \neq \emptyset\}$.
|
||||||
Note that $A$ is a set (use separation on $\cP(\cP(a) \times \bigcup \cP(a))$).
|
Note that $A$ is a set (use separation on $\cP(\cP(a) \times \bigcup \cP(a))$).
|
||||||
Note further that if $b_1 \neq b_2$,
|
Note further that if $b_1 \neq b_2$,
|
||||||
then $\{(b_1, x) : x \in b_1\} $
|
then $\{(b_1, x) : x \in b_1\} $
|
||||||
|
@ -51,7 +51,7 @@
|
||||||
Then $u_0 = f(B_{u_0}^{\le'}) = f(B_{g(u_0)}^{\le''}) = g(u_0)$.
|
Then $u_0 = f(B_{u_0}^{\le'}) = f(B_{g(u_0)}^{\le''}) = g(u_0)$.
|
||||||
Thus $g$ is the identity.
|
Thus $g$ is the identity.
|
||||||
\end{subproof}
|
\end{subproof}
|
||||||
Given the claim, we can now see that $\bigcup W$ is a well order $\le^{\ast\ast}$
|
Given the claim, we can now see that $\bigcup W$ is a well-order $\le^{\ast\ast}$
|
||||||
of $a$.
|
of $a$.
|
||||||
Let $B = \{w \in a | w \text{ is a $\le$-upper bound of $b$}\}$
|
Let $B = \{w \in a | w \text{ is a $\le$-upper bound of $b$}\}$
|
||||||
(this is not empty by the hypothesis).
|
(this is not empty by the hypothesis).
|
||||||
|
@ -63,7 +63,7 @@
|
||||||
\le^{\ast\ast} = \le^{\ast} \cup \{(u,u_0) | u \in b\} \cup \{(u_0,u_0)\}.
|
\le^{\ast\ast} = \le^{\ast} \cup \{(u,u_0) | u \in b\} \cup \{(u_0,u_0)\}.
|
||||||
\]
|
\]
|
||||||
Then $B = B_{u_0}^{\le^{\ast\ast}}$.
|
Then $B = B_{u_0}^{\le^{\ast\ast}}$.
|
||||||
So $\le^{\ast\ast} \in W$, but now $n_0 \in b$.
|
So $\le^{\ast\ast} \in W$, but now $u_0 \in b$.
|
||||||
So $b$ must have a maximum.
|
So $b$ must have a maximum.
|
||||||
\end{refproof}
|
\end{refproof}
|
||||||
|
|
||||||
|
@ -83,7 +83,7 @@
|
||||||
\end{corollary}
|
\end{corollary}
|
||||||
|
|
||||||
\begin{remark}[Cultural enrichment]
|
\begin{remark}[Cultural enrichment]
|
||||||
Other assertion which are equivalent
|
Other assertions which are equivalent
|
||||||
to the \yaref{ax:c}:
|
to the \yaref{ax:c}:
|
||||||
\begin{itemize}
|
\begin{itemize}
|
||||||
\item Every infinite family of non-empty sets
|
\item Every infinite family of non-empty sets
|
||||||
|
@ -139,7 +139,7 @@ We have the following principle of induction:
|
||||||
|
|
||||||
\begin{definition}
|
\begin{definition}
|
||||||
A set $x$ is \vocab{transitive},
|
A set $x$ is \vocab{transitive},
|
||||||
if $\forall y \in x.~y \subseteq x$.
|
iff $\forall y \in x.~y \subseteq x$.
|
||||||
\end{definition}
|
\end{definition}
|
||||||
\begin{definition}
|
\begin{definition}
|
||||||
A set $x$ is called an \vocab{ordinal} (or \vocab{ordinal number})
|
A set $x$ is called an \vocab{ordinal} (or \vocab{ordinal number})
|
||||||
|
@ -196,7 +196,7 @@ Clearly, the $\in$-relation is a well-order on an ordinal $x$.
|
||||||
\begin{enumerate}[(a)]
|
\begin{enumerate}[(a)]
|
||||||
\item $\phi(0,0)$
|
\item $\phi(0,0)$
|
||||||
\item $\forall z \in \omega. ~\phi(0, z) \implies \phi(0,z+1)$.
|
\item $\forall z \in \omega. ~\phi(0, z) \implies \phi(0,z+1)$.
|
||||||
\item $\forall y \in \omega.((\forall z' \in \omega.~\phi(y,z')) \implies (\forall z \in \omega.~\phi(y+1, z')))$.
|
\item $\forall y \in \omega.~((\forall z' \in \omega.~\phi(y,z')) \implies (\forall z \in \omega.~\phi(y+1, z)))$.
|
||||||
\end{enumerate}
|
\end{enumerate}
|
||||||
(a) and (b) are trivial.
|
(a) and (b) are trivial.
|
||||||
Fix $y \in \omega$ and
|
Fix $y \in \omega$ and
|
||||||
|
@ -209,7 +209,8 @@ Clearly, the $\in$-relation is a well-order on an ordinal $x$.
|
||||||
so $\phi(y+1, 0)$ is true, since $\phi$ is symmetric.
|
so $\phi(y+1, 0)$ is true, since $\phi$ is symmetric.
|
||||||
Now if $\phi(y+1,z)$ is true,
|
Now if $\phi(y+1,z)$ is true,
|
||||||
we want to show $\phi(y+1,z+1)$ is true as well.
|
we want to show $\phi(y+1,z+1)$ is true as well.
|
||||||
We have $y + 1 \in z \lor y + 1 = z \lor y + 1 \ni z$
|
We have
|
||||||
|
\[(y + 1 \in z) \lor (y + 1 = z) \lor (y + 1 \ni z)\]
|
||||||
by assumption.
|
by assumption.
|
||||||
\begin{itemize}
|
\begin{itemize}
|
||||||
\item If $y + 1 \in z \lor y+1 = z$, then clearly $y + 1 \in z + 1$.
|
\item If $y + 1 \in z \lor y+1 = z$, then clearly $y + 1 \in z + 1$.
|
||||||
|
|
|
@ -2,7 +2,7 @@
|
||||||
|
|
||||||
\begin{remark}[``Constructive'' approach to $\omega_1$ ]
|
\begin{remark}[``Constructive'' approach to $\omega_1$ ]
|
||||||
There are many well-orders on $\omega$.
|
There are many well-orders on $\omega$.
|
||||||
Let $W$ be the set of all such well orders.
|
Let $W$ be the set of all such well-orders.
|
||||||
For $R, S \in W$,
|
For $R, S \in W$,
|
||||||
write $R \le S$ if $R$ is isomorphic to
|
write $R \le S$ if $R$ is isomorphic to
|
||||||
an initial segment of $S$.
|
an initial segment of $S$.
|
||||||
|
|
Loading…
Reference in a new issue