diff --git a/inputs/intro.tex b/inputs/intro.tex index 49c11e5..9ba46b2 100644 --- a/inputs/intro.tex +++ b/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 $\dagger$. diff --git a/inputs/lecture_01.tex b/inputs/lecture_01.tex index 319daa8..35a9042 100644 --- a/inputs/lecture_01.tex +++ b/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 ($\N < A$) diff --git a/inputs/lecture_02.tex b/inputs/lecture_02.tex index 7c76520..1075c18 100644 --- a/inputs/lecture_02.tex +++ b/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} diff --git a/inputs/lecture_03.tex b/inputs/lecture_03.tex index 4517255..170dbed 100644 --- a/inputs/lecture_03.tex +++ b/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. diff --git a/inputs/lecture_04.tex b/inputs/lecture_04.tex index fd5bf47..6c65c9f 100644 --- a/inputs/lecture_04.tex +++ b/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 + \[ + \forall u.~((u \in z) \iff (u \in x \lor u \in y)), + \] + $z = \bigcap x$ for \[ - \forall z.~(z \in x \iff \exists v.(v \in y \land z \in v)). + \forall u.~((u \in z) \iff (\forall v.~(v \in x \implies u \in v))), + \] + $z = \bigcup x$ 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*} diff --git a/inputs/lecture_05.tex b/inputs/lecture_05.tex index c421a29..de6f6f0 100644 --- a/inputs/lecture_05.tex +++ b/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} diff --git a/inputs/lecture_06.tex b/inputs/lecture_06.tex index 98ceb3b..245b7ef 100644 --- a/inputs/lecture_06.tex +++ b/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 $y + 1 \in z + 1$. diff --git a/inputs/lecture_13.tex b/inputs/lecture_13.tex index 7bd0c83..05824e2 100644 --- a/inputs/lecture_13.tex +++ b/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 $S$.