Merge branch 'master' of https://git.abstractnonsen.se/josia-notes/w23-logic-2

inputs/intro.tex

@@ -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}
 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:\\
 \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)
 and slight modifications have been marked with $\dagger$.

inputs/lecture_01.tex

@@ -31,7 +31,7 @@ Literature
 \end{definition}
 \begin{lemma}
     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}
 \begin{proof}
     Fix $f\colon A \hookrightarrow B$.
@@ -150,7 +150,8 @@ Literature
 \begin{definition}
     The \vocab{continuum hypothesis} ($\CH$)
     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 $A \subset \R$
     which is uncountable ($\N < A$)

inputs/lecture_02.tex

@@ -69,9 +69,16 @@
     But then $x \in A' \subseteq A \lightning$.
 \end{refproof}
 \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 $P = P'$.
 \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:
 \begin{itemize}
@@ -122,7 +129,7 @@ We want to prove two things:
     If $t \neq t' \in \{0,1\}^{\omega}$,
     then there is some $n$ such that $t\defon{n} \neq  t'\defon{n}$,
     hence $f(t) \in [a_{t\defon{n}}, b_{t\defon{n}}]$
-    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.
     Thus $f(t) \neq f(t')$, i.e.~$f$ is injective.
 \end{proof}

inputs/lecture_03.tex

@@ -58,7 +58,7 @@ all condensation points are accumulation points.
 
         Then
         \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*}
         $\subseteq $ holds by the choice of $a_x$ and $b_x$.
         For $\supseteq$ let $y$ be an element of the RHS.

inputs/lecture_04.tex

@@ -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).
     \]
 
-    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}
 $\ZFC$ consists of the following axioms:
@@ -86,7 +101,7 @@ $\ZFC$ consists of the following axioms:
     % (PWA)
     We write $x = \cP(y)$
     for
-    $\forall  z.~(z \in x \iff x \subseteq z)$.
+    $\forall  z.~(z \in x \iff z \subseteq x)$.
     The power set axiom states
     \[
     \forall x.~\exists y.~y=\cP(x).
@@ -103,7 +118,9 @@ $\ZFC$ consists of the following axioms:
     \yalabel{Axiom of Separation}{(Aus)}{ax:aus}
 
     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 $\AxAus_{\phi}$
     states
     \[
@@ -115,16 +132,11 @@ $\ZFC$ consists of the following axioms:
     for $\forall x.~(x \in b \iff x \in a \land f(x))$.
     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}
 
-\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}
     \AxAus proves that
     \begin{itemize}
@@ -135,19 +147,22 @@ $\ZFC$ consists of the following axioms:
 \end{remark}
 \begin{axiomschema}[\vocab{Replacement} (Fraenkel)]
     \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
     \[
-    \forall v_1 .~\exists  b.~\forall y.~(y \in b \iff \exists  x .~(x \in a \land \phi(x,y,v_1,v_p))).
+        \forall x \in a.~\exists !y \phi(x,y) \implies \exists b.~\forall x \in a.~\exists y \in b.~\phi(x,y).
+        % \forall v_1 \ldots \forall v_p .~\forall x.~ \exists y'.~\forall y.~(y = y' \iff \phi(x))
+        % \implies \forall a.~\exists b.~\forall y.~(y \in b \iff \exists x.~(x \in a \land \phi(x))
     \]
 \end{axiomschema}
 
 \begin{axiom}[\vocab{Choice}]
     \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}
         \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\})\\
                     &)&
     \end{IEEEeqnarray*}

inputs/lecture_05.tex

@@ -144,7 +144,7 @@
             $x \le y \land x \le y \implies x = y$,
             and
         \item \vocab{transitive},
-            i.e.~$x \le y \land x \le z \implies x \le z$.
+            i.e.~$x \le y \land y \le z \implies x \le z$.
     \end{itemize}
 
     If additionally $\forall x,y.~(x\le y \lor y \le x)$,
@@ -159,32 +159,44 @@
     of $b$
     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} 
-    of $b$.
+    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 $x = \sup(b)$ if $x$ is the minimum
+    We say $x = \text{\vocab{$\sup$}}(b)$ if $x$ is the minimum
     of the set of upper bounds of $b$.
     (This does not necessarily exist.)
-    Similarly $\inf(b)$ is defined.
+    Similarly $\text{\vocab{$\inf$}}(b)$ is defined.
 \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}
     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} 
+    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 isomorphism.
+    An order-preserving bijection
+    is called an \vocab{isomorphism}.
     We write $(a,\le_a) \cong (b, \le_b)$
     if they are isomorphic.
 \end{definition}
@@ -211,7 +223,7 @@
 \begin{lemma}
     Let $(a, \le)$ be a well-order.
     Let $f\colon a \to a$
-    be an order preserving map.
+    be an order-preserving map.
     Then $f(x) \ge x$ for all $x \in a$.
 \end{lemma}
 \begin{proof}
@@ -221,7 +233,7 @@
 \end{proof}
 
 \begin{lemma}
-    If $(a, \le )$ is a well order
+    If $(a, \le )$ is a well-order
     and $f\colon (a, \le) \leftrightarrow (a, \le)$
     is an isomorphism,
     then $f$ is the identity.
@@ -239,7 +251,7 @@
 \end{lemma}
 \begin{proof}
     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,
     so $g = f$.
 \end{proof}
@@ -250,6 +262,11 @@
     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:
@@ -269,7 +286,7 @@
     so $r$ is an injective function
     from a subset of $a$ to a subset of $b$.
     \begin{claim}
-        $r$ is order preserving:
+        $r$ is order-preserving:
     \end{claim}
     \begin{subproof}
         If $x <_a x'$, then consider the unique $y'$
@@ -287,7 +304,7 @@
 
         Let $x \coloneqq  \min(a \setminus \dom(r))$
         and $y \coloneqq  \min(b\setminus \ran(r))$.
-        Then $(a,\le)\defon{x} \cong (b, \le)\defon{y}$.
+        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}

inputs/lecture_06.tex

@@ -3,13 +3,13 @@
 \begin{theorem}[Zorn]
     \yalabel{Zorn's Lemma}{Zorn}{thm:zorn}
     Let $(a, \le )$ be a partial order with $a \neq \emptyset$.
-    Assume that $b \le  a$ with $b \neq \emptyset$
-    and $ b$ linearly ordered, $b$ has an upper bound,
+    Assume that for all $b \subseteq a$ with $b \neq \emptyset$
+    and $ b$ linearly ordered, $b$ has an upper bound.
     Then $a$ has a maximal element.
 \end{theorem}
 \begin{refproof}{thm:zorn}
     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 further that if $b_1 \neq  b_2$,
     then $\{(b_1, x) : x \in  b_1\} $
@@ -51,7 +51,7 @@
         Then $u_0 = f(B_{u_0}^{\le'}) = f(B_{g(u_0)}^{\le''}) = g(u_0)$.
         Thus $g$ is the identity.
     \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 $a$.
     Let $B = \{w \in a | w \text{ is a $\le$-upper bound of $b$}\}$
     (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)\}.
     \]
     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.
 \end{refproof}
 
@@ -83,7 +83,7 @@
 \end{corollary}
 
 \begin{remark}[Cultural enrichment]
-    Other assertion which are equivalent
+    Other assertions which are equivalent
     to the \yaref{ax:c}:
     \begin{itemize}
         \item Every infinite family of non-empty sets
@@ -139,7 +139,7 @@ We have the following principle of induction:
 
 \begin{definition}
     A set $x$ is \vocab{transitive},
-    if $\forall y \in x.~y \subseteq x$.
+    iff $\forall y \in x.~y \subseteq x$.
 \end{definition}
 \begin{definition}
     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)]
         \item $\phi(0,0)$
         \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}
     (a) and (b) are trivial.
     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.
     Now if $\phi(y+1,z)$ is true,
     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.
     \begin{itemize}
         \item If $y + 1 \in z \lor y+1 = z$, then clearly $y + 1 \in z + 1$.

inputs/lecture_13.tex

@@ -2,7 +2,7 @@
 
 \begin{remark}[``Constructive'' approach to $\omega_1$ ]
     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 $R, S \in W$,
     write $R \le S$ if $R$ is isomorphic to
     an initial segment of $S$.