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@ -1,4 +1,4 @@
These are my notes on the lecture Probability Theory, These are my notes on the lecture Logic II
taught by \textsc{Ralf Schindler} taught by \textsc{Ralf Schindler}
in winter 23/24 at the University Münster. in winter 23/24 at the University Münster.
@ -12,6 +12,6 @@ If you find errors or want to improve something,
please send me a message:\\ please send me a message:\\
\texttt{lecturenotes@jrpie.de}. \texttt{lecturenotes@jrpie.de}.
This notes follow the way the material was presented in the lecture rather These notes follow the way the material was presented in the lecture rather
closely. Additions (e.g.~from exercise sheets) closely. Additions (e.g.~from exercise sheets)
and slight modifications have been marked with $\dagger$. and slight modifications have been marked with $\dagger$.

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@ -31,7 +31,7 @@ Literature
\end{definition} \end{definition}
\begin{lemma} \begin{lemma}
If $A \le B$, If $A \le B$,
then there is a surjection $g\colon B \twoheadrightarrow B$. then there is a surjection $g\colon B \twoheadrightarrow A$.
\end{lemma} \end{lemma}
\begin{proof} \begin{proof}
Fix $f\colon A \hookrightarrow B$. Fix $f\colon A \hookrightarrow B$.
@ -150,7 +150,8 @@ Literature
\begin{definition} \begin{definition}
The \vocab{continuum hypothesis} ($\CH$) The \vocab{continuum hypothesis} ($\CH$)
says that there is no set $A$ such that says that there is no set $A$ such that
$\N < A < \R$. $\N < A < \R$,
i.e.~every uncountable subset $A \subseteq \R$ is in bijection with $\R$.
$\CH$ is equivalent to the statement that there is no set $A \subset \R$ $\CH$ is equivalent to the statement that there is no set $A \subset \R$
which is uncountable ($\N < A$) which is uncountable ($\N < A$)

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@ -69,9 +69,16 @@
But then $x \in A' \subseteq A \lightning$. But then $x \in A' \subseteq A \lightning$.
\end{refproof} \end{refproof}
\begin{definition} \begin{definition}
$P \subseteq \R$ is called \vocab{perfect} $P \subseteq \R$ (or, more generally, a subset of any topological space)
is called \vocab{perfect}
iff $P \neq \emptyset$ and $P = P'$. iff $P \neq \emptyset$ and $P = P'$.
\end{definition} \end{definition}
\begin{example}+
Note that being perfect depends on the surrounding topological space:
For example, $[0,1] \cap \Q$
is perfect as a subset of $\Q$,
but not perfect as a subset of $\R$.
\end{example}
We want to prove two things: We want to prove two things:
\begin{itemize} \begin{itemize}
@ -122,7 +129,7 @@ We want to prove two things:
If $t \neq t' \in \{0,1\}^{\omega}$, If $t \neq t' \in \{0,1\}^{\omega}$,
then there is some $n$ such that $t\defon{n} \neq t'\defon{n}$, then there is some $n$ such that $t\defon{n} \neq t'\defon{n}$,
hence $f(t) \in [a_{t\defon{n}}, b_{t\defon{n}}]$ hence $f(t) \in [a_{t\defon{n}}, b_{t\defon{n}}]$
and $f(t') \in [a_{t\defon{n}}, b_{t\defon{n}}]$ and $f(t') \in [a_{t'\defon{n}}, b_{t'\defon{n}}]$
which are disjoint. which are disjoint.
Thus $f(t) \neq f(t')$, i.e.~$f$ is injective. Thus $f(t) \neq f(t')$, i.e.~$f$ is injective.
\end{proof} \end{proof}

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@ -58,7 +58,7 @@ all condensation points are accumulation points.
Then Then
\begin{IEEEeqnarray*}{rCl} \begin{IEEEeqnarray*}{rCl}
A \setminus P &=& \bigcap_{x \in A \setminus P} (a_x, b_x) \cap A. A \setminus P &=& \bigcup_{x \in A \setminus P} (a_x, b_x) \cap A.
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
$\subseteq $ holds by the choice of $a_x$ and $b_x$. $\subseteq $ holds by the choice of $a_x$ and $b_x$.
For $\supseteq$ let $y$ be an element of the RHS. For $\supseteq$ let $y$ be an element of the RHS.

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@ -25,9 +25,24 @@ $\ZFC$ stands for
y \in x \land z \in x \land \forall a .~(a \in x \implies a = y \lor a = z). y \in x \land z \in x \land \forall a .~(a \in x \implies a = y \lor a = z).
\] \]
Let $x = \bigcup y$ denote We write $z = x \cap y$ for
\[\forall u.~((u \in z) \implies u \in x \land u \in y),\]
$z = x \cup y$ for
\[ \[
\forall z.~(z \in x \iff \exists v.(v \in y \land z \in v)). \forall u.~((u \in z) \iff (u \in x \lor u \in y)),
\]
$z = \bigcap x$ for
\[
\forall u.~((u \in z) \iff (\forall v.~(v \in x \implies u \in v))),
\]
$z = \bigcup x$ for
\[
\forall u.~((u \in z) \iff \exists v.~(v \in x \land u \in v))
\]
and
$z = x \setminus y$ for
\[
\forall u.~((u \in z) \iff (u \in x \land u \not\in y)).
\] \]
\end{notation} \end{notation}
$\ZFC$ consists of the following axioms: $\ZFC$ consists of the following axioms:
@ -86,7 +101,7 @@ $\ZFC$ consists of the following axioms:
% (PWA) % (PWA)
We write $x = \cP(y)$ We write $x = \cP(y)$
for for
$\forall z.~(z \in x \iff x \subseteq z)$. $\forall z.~(z \in x \iff z \subseteq x)$.
The power set axiom states The power set axiom states
\[ \[
\forall x.~\exists y.~y=\cP(x). \forall x.~\exists y.~y=\cP(x).
@ -103,7 +118,9 @@ $\ZFC$ consists of the following axioms:
\yalabel{Axiom of Separation}{(Aus)}{ax:aus} \yalabel{Axiom of Separation}{(Aus)}{ax:aus}
Let $\phi$ be some fixed Let $\phi$ be some fixed
fist order formula in $\cL_\in$. fist order formula in $\cL_\in$
with free variables $x, v_1, \ldots, v_p$.
Let $b$ be a variable that is not free in $\phi$.
Then $\AxAus_{\phi}$ Then $\AxAus_{\phi}$
states states
\[ \[
@ -115,16 +132,11 @@ $\ZFC$ consists of the following axioms:
for $\forall x.~(x \in b \iff x \in a \land f(x))$. for $\forall x.~(x \in b \iff x \in a \land f(x))$.
Then \AxAus can be formulated as Then \AxAus can be formulated as
\[ \[
\forall a.~\exists b.~(b = \{x \in a; \phi(x)\}). \forall a.~\exists b.~(b = \{x \in a | \phi(x)\}).
\] \]
\end{axiomschema} \end{axiomschema}
\begin{notation}
\todo{$\cap, \setminus, \bigcap$}
% We write $z = x \cap y$ for $\forall u.~((u \in z) \implies u \in x \land u \in y)$,
% $Z = x \setminus y$ for ...
% $x = \bigcap y$ for ...
\end{notation}
\begin{remark} \begin{remark}
\AxAus proves that \AxAus proves that
\begin{itemize} \begin{itemize}
@ -135,19 +147,22 @@ $\ZFC$ consists of the following axioms:
\end{remark} \end{remark}
\begin{axiomschema}[\vocab{Replacement} (Fraenkel)] \begin{axiomschema}[\vocab{Replacement} (Fraenkel)]
\yalabel{Axiom of Replacement}{(Rep)}{ax:rep} \yalabel{Axiom of Replacement}{(Rep)}{ax:rep}
Let $\phi$ be some $\cL_{\in }$ formula. Let $\phi$ be some $\cL_{\in }$ formula
with free variables $x, y$.\todo{Allow more variables}
Then Then
\[ \[
\forall v_1 .~\exists b.~\forall y.~(y \in b \iff \exists x .~(x \in a \land \phi(x,y,v_1,v_p))). \forall x \in a.~\exists !y \phi(x,y) \implies \exists b.~\forall x \in a.~\exists y \in b.~\phi(x,y).
% \forall v_1 \ldots \forall v_p .~\forall x.~ \exists y'.~\forall y.~(y = y' \iff \phi(x))
% \implies \forall a.~\exists b.~\forall y.~(y \in b \iff \exists x.~(x \in a \land \phi(x))
\] \]
\end{axiomschema} \end{axiomschema}
\begin{axiom}[\vocab{Choice}] \begin{axiom}[\vocab{Choice}]
\yalabel{Axiom of Choice}{(C)}{ax:c} \yalabel{Axiom of Choice}{(C)}{ax:c}
Every family of non-empty sets has a \vocab{choice set}: Every family of pairwise disjoint non-empty sets has a \vocab{choice set}:
\begin{IEEEeqnarray*}{rCl} \begin{IEEEeqnarray*}{rCl}
\forall x .~&(&\\ \forall x .~&(&\\
&& ((\forall y \in x.~y \neq \emptyset) \land (\forall y \in x .~\forall y' \in x .~(y \neq y' \implies y \cap y' = \emptyset))\\ && ((\forall y \in x.~y \neq \emptyset) \land (\forall y \in x .~\forall y' \in x .~(y \neq y' \implies y \cap y' = \emptyset)))\\
&& \implies\exists z.~\forall y \in x.~\exists u.~(z \cap y = \{u\})\\ && \implies\exists z.~\forall y \in x.~\exists u.~(z \cap y = \{u\})\\
&)& &)&
\end{IEEEeqnarray*} \end{IEEEeqnarray*}

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@ -144,7 +144,7 @@
$x \le y \land x \le y \implies x = y$, $x \le y \land x \le y \implies x = y$,
and and
\item \vocab{transitive}, \item \vocab{transitive},
i.e.~$x \le y \land x \le z \implies x \le z$. i.e.~$x \le y \land y \le z \implies x \le z$.
\end{itemize} \end{itemize}
If additionally $\forall x,y.~(x\le y \lor y \le x)$, If additionally $\forall x,y.~(x\le y \lor y \le x)$,
@ -159,32 +159,44 @@
of $b$ of $b$
iff iff
\[ \[
x \in b \land \lnot \exists y \in b .~(y > x).b .~(y > x). x \in b \land \lnot \exists y \in b .~(y > x).
\]
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} In a similar way we define \vocab[Minimal element]{minimal elements}
of $b$. and the \vocab{minimum} of $b$.
We say that $x $ is an \vocab{upper bound} We say that $x $ is an \vocab{upper bound}
of $b$ if $\forall y \in b.~(x \ge y)$. of $b$ if $\forall y \in b.~(x \ge y)$.
Similarly \vocab[Lower bound]{lower bounds} Similarly \vocab[Lower bound]{lower bounds}
are defined. are defined.
We say $x = \sup(b)$ if $x$ is the minimum We say $x = \text{\vocab{$\sup$}}(b)$ if $x$ is the minimum
of the set of upper bounds of $b$. of the set of upper bounds of $b$.
(This does not necessarily exist.) (This does not necessarily exist.)
Similarly $\inf(b)$ is defined. Similarly $\text{\vocab{$\inf$}}(b)$ is defined.
\end{definition} \end{definition}
\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} \begin{definition}
Let $(a, \le_a)$ and $(b, \le_b)$ Let $(a, \le_a)$ and $(b, \le_b)$
be two partial orders. be two partial orders.
Then a function $f\colon a\to b$ is caled Then a function $f\colon a\to b$ is called
\vocab{order preserving} \vocab{order-preserving}
iff iff
\[ \[
\forall x,y \in a.~(x \le_a y) \iff f(x) \le_b f(y). \forall x,y \in a.~(x \le_a y) \iff f(x) \le_b f(y).
\] \]
An order preserving bijection An order-preserving bijection
is called an isomorphism. is called an \vocab{isomorphism}.
We write $(a,\le_a) \cong (b, \le_b)$ We write $(a,\le_a) \cong (b, \le_b)$
if they are isomorphic. if they are isomorphic.
\end{definition} \end{definition}
@ -211,7 +223,7 @@
\begin{lemma} \begin{lemma}
Let $(a, \le)$ be a well-order. Let $(a, \le)$ be a well-order.
Let $f\colon a \to a$ Let $f\colon a \to a$
be an order preserving map. be an order-preserving map.
Then $f(x) \ge x$ for all $x \in a$. Then $f(x) \ge x$ for all $x \in a$.
\end{lemma} \end{lemma}
\begin{proof} \begin{proof}
@ -221,7 +233,7 @@
\end{proof} \end{proof}
\begin{lemma} \begin{lemma}
If $(a, \le )$ is a well order If $(a, \le )$ is a well-order
and $f\colon (a, \le) \leftrightarrow (a, \le)$ and $f\colon (a, \le) \leftrightarrow (a, \le)$
is an isomorphism, is an isomorphism,
then $f$ is the identity. then $f$ is the identity.
@ -239,7 +251,7 @@
\end{lemma} \end{lemma}
\begin{proof} \begin{proof}
Let $f,g$ be isomorphisms Let $f,g$ be isomorphisms
and consider $g^{-1}\circ f \colon (a, \le ) \xrightarrow{\cong} (a, \le )$. 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, We have already shown that $g^{-1}\circ f$ must be the identity,
so $g = f$. so $g = f$.
\end{proof} \end{proof}
@ -250,6 +262,11 @@
then write $(a, \le )\defon{x}$ then write $(a, \le )\defon{x}$
for $(\{y \in a | y \le x\}, \le \cap \{y \in a | y \le x\}^2)$. for $(\{y \in a | y \le x\}, \le \cap \{y \in a | y \le x\}^2)$.
\end{definition} \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} \begin{theorem}
Let $(a,\le_a)$ and $(b,\le_b)$ be well-orders. Let $(a,\le_a)$ and $(b,\le_b)$ be well-orders.
Then exactly one of the following three holds: Then exactly one of the following three holds:
@ -269,7 +286,7 @@
so $r$ is an injective function so $r$ is an injective function
from a subset of $a$ to a subset of $b$. from a subset of $a$ to a subset of $b$.
\begin{claim} \begin{claim}
$r$ is order preserving: $r$ is order-preserving:
\end{claim} \end{claim}
\begin{subproof} \begin{subproof}
If $x <_a x'$, then consider the unique $y'$ If $x <_a x'$, then consider the unique $y'$
@ -287,7 +304,7 @@
Let $x \coloneqq \min(a \setminus \dom(r))$ Let $x \coloneqq \min(a \setminus \dom(r))$
and $y \coloneqq \min(b\setminus \ran(r))$. and $y \coloneqq \min(b\setminus \ran(r))$.
Then $(a,\le)\defon{x} \cong (b, \le)\defon{y}$. Then $(a,\le_a)\defon{x} \cong (b, \le_b)\defon{y}$.
But now $(x,y) \in r$ which is a contradiction. But now $(x,y) \in r$ which is a contradiction.
\end{subproof} \end{subproof}
\end{proof} \end{proof}

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@ -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 $y + 1 \in z + 1$.

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@ -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 $S$.