diff --git a/AFL_updated.pdf b/AFL_updated.pdf index 0b5aab5..dc1952b 100644 Binary files a/AFL_updated.pdf and b/AFL_updated.pdf differ diff --git a/LogicAndSetTheory/05_set_theory.tex b/LogicAndSetTheory/05_set_theory.tex index 3069c10..254f2fc 100644 --- a/LogicAndSetTheory/05_set_theory.tex +++ b/LogicAndSetTheory/05_set_theory.tex @@ -277,7 +277,8 @@ \subsection{Transitive sets} \begin{proof} NTS that $x \subseteq \omega \; \forall x \in \omega$. \\ Form the set $z = \{y \in \omega : y \subseteq \omega\}$ by (Sep). - Check $z$ is a successor set, so $z = \omega$. + Check $z$ is a successor set, so $z = \omega$, i.e. $\omega$ transitive. + Similarly, $\{x \in \omega : \text{`$x$ is transitive'}\}$ is also a successor set ($\cup x^+ = x$) so it is $\omega$. So every element of $\omega$ is a transitive set. \end{proof} @@ -295,7 +296,7 @@ \subsection{Transitive sets} This holds as any intersection of transitive sets is transitive. \end{remark} -\underline{Idea}: If $x \subseteq y$ and y transitive then $\bigcup x \subseteq y$ so $\bigcup \bigcup x \subseteq y$, $\bigcup \bigcup \bigcup x \subseteq y$, \dots \\ +\underline{Idea}: If $x \subseteq y$ and $y$ transitive then $\bigcup x \subseteq y$ so $\bigcup \bigcup x \subseteq y$, $\bigcup \bigcup \bigcup x \subseteq y$, \dots \\ We want to form $\bigcup \{x, \bigcup x, \bigcup \bigcup x, \dots\}$, this is a set by (Rep). We need a function-class $0 \mapsto x$, $1 \mapsto \bigcup x$, $2 \mapsto \bigcup \bigcup x$, \dots @@ -327,10 +328,11 @@ \subsection{Transitive sets} Also by (Un) we can form $t = \bigcup w$. Then $x \in t$, since $x \in w$ ($\{(0, x)\}$ is an attempt). - Given $a \in t$, we have $z \in w$, $a \in z$. + Given $a \in t$, we have $a \in z$ for some $z \in w$. Then there's an attempt $f$ and $n \in w$ s.t. $z = f(n)$. \\ By $(\ast\ast)$ there's an attempt $g$ with $n^+ \in \dom g$. - Then $n \in \dom g$ so $\bigcup z = \bigcup f(n) = \bigcup g(n)$ by $(\ast)$ and $\bigcup g(n) = g(n^+) \in w$, hence $a \in t$. + Then $n \in \dom g$ so $\bigcup z = \bigcup f(n) = \bigcup g(n)$ by $(\ast)$ and $\bigcup g(n) = g(n^+) \in w$. + Thus for any $b \in a$, $b \in \cup z \in w$ so $b \in t$, i.e. $t$ transitive. \end{proof} Transitive closures allow us to pass from the large universe of sets, which is not a set itself, into a smaller world which is a set closed under $\in$ that contains the relevant sets in question. @@ -356,14 +358,14 @@ \subsection{\texorpdfstring{$\in$}{∈}-induction} If $y \in z$, then $y \in t$ (as $t$ transitive) and $y \notin u$ (by minimality), so $p(y)$. By assumption $p(z)$ holds \Lightning\ of $z \in u$. \end{proof} -The name of this theorem should be read `epsilon-induction', even though the membership relation is denoted $\in$ and not $\epsilon$ or $\varepsilon$. +The name of this theorem should be read `epsilon-induction', even though the membership relation is denoted $\in$ and not $\epsilon$. The principle of $\in$-induction is equivalent to the axiom of foundation (Fnd) in the presence of the other axioms of $\mathsf{ZF}$. -\underline{Clever Idea}: We say that $x$ is \vocab{regular} if $(\forall y)(x \in y \implies y \text{ has a $\in$-minimal element})$. +\underline{Clever Idea}: We say that $x$ is \vocab{regular} if $(\forall y)(x \in y \implies y \text{ has a $\in$-minimal element})$ (this defn is the clever part). The axiom of foundation is equivalent to the assertion that every set is regular. Given $\in$-induction, we can prove every set is regular. -Suppose $(\forall y \in x)(y \text{ is regular})$; we need to show $x$ is regular. +Fix some $x$ and suppose $(\forall y \in x)(y \text{ is regular})$; we need to show $x$ is regular. For a set $z$ with $x \in z$, if $x$ is minimal in $z$, $x$ is clearly regular as required. If $x$ is not minimal in $z$, there exists $y \in x$ s.t. $y \in z$. So $z$ has a minimal element as $y$ is regular. @@ -430,7 +432,7 @@ \subsection{Well-founded relations} \end{example} \begin{definition}[Local] - $r$ is \vocab{local} if $(\forall x)(\forall y)(\exists z)(z r x \wedge z r y\footnote{$zry = r(z, y)$})$, i.e. the $r$-predecessors of $x$ form a set. + $r$ is \vocab{local} if $(\forall x)(\exists y)(\forall z)(z \in y \wedge z r x\footnote{$zrx = r(z, x)$})$, i.e. the $r$-predecessors of $x$ form a set. \end{definition} \begin{example} @@ -453,6 +455,7 @@ \subsection{Well-founded relations} (\forall x \in a)(\forall y \in a)((\forall z \in a)(zrx \Leftrightarrow zry) \implies x = y) \end{align*} \end{definition} +This is just the axiom of extensionality applied to the relation $r$. % Therefore, $p$-induction and $p$-recursion hold for all relation-classes $p$ that are well-founded and local. % In particular, if $r$ is a well-founded relation on a set $a$, it is clearly local and hence we have $r$-induction and $r$-recursion. @@ -478,11 +481,11 @@ \subsection{Well-founded relations} \begin{proof} By $r$-recursion on $a$, there's a function class $f$ s.t. $\forall x \in a$, $f(x) = \{f(y) : y \in a \wedge y r x \}$. - Note that $f$ i s a function, not just a fcn class since $\qty{(x, f(x)) : x \in a}$ is a set by (Rep). + Note that $f$ is a function, not just a fcn class since $\qty{(x, f(x)) : x \in a}$ is a set by (Rep). Then $b = \qty{f(x) : x \in a}$ is a set by (Rep). \\ $b$ is transitive: Let $z \in b$ and $w \in z$. - There's $x \in a$ s.t. $z = f(x)$, and so $y \in a$ s.t. $y r x$ and $w = f(y) \in b$. + There's $x \in a$ s.t. $z = f(x)$, and so a $y \in a$ s.t. $y r x$ and $w = f(y) \in b$. Clearly $f$ surjective and $\forall x, y \in a$, $xry \implies f(x) \in f(y)$. @@ -538,18 +541,22 @@ \subsection{Well-founded relations} \begin{remark} (2) says that $\alpha$ really \underline{is} the set of ordinals $< \alpha$; \\ (3) says that $\in$ linearly orders the class $ON$; \\ - (4) resolves the class of notation $x^+$ in section $2$ and $5$; \\ + (4) resolves the clash of notation $x^+$ in section $2$ and $5$. + According to the definition in section $2$, $\alpha^+$ is the unique (up to order-isomorphism) well-ordered set that consists of $\alpha$ as a proper initial segment and one extra element that is a maximum. + By Mostowski, this well-ordered set is order-isomorpic to a unique ordinal (its order-type). + (4) shows that this ordinal is the successor of the set $\alpha$ as defined in this section. + In particular, this shows that the successor of an ordinal is an ordinal; \\ (5) now shows that any set of well-ordered sets has an upper bound. \end{remark} \begin{proof} \begin{enumerate} \item Let $\gamma \in \alpha$. - Then $\gamma \subseteq \alpha$ ($\alpha$ is transitive) and hence $\in$ linearly orders $\gamma$. + Then $\gamma \subseteq \alpha$ ($\alpha$ is transitive) and hence $\in$ linearly orders $\gamma$\footnote{$\in$ linearly orders $\alpha$ and $\gamma$ a subset so $\eval{\in}_\gamma$ linearly orders it.}. Given $\eta \in \delta$, $\delta \in \gamma$ then $\delta \in \alpha$ and so $\eta \in \alpha$ ($\alpha$ is transitive). - Since $\in$ is transitive on $\alpha$, we have $\eta \in \gamma$. + Since $\in$ is transitive\footnote{As $\in$ a well ordering.} on $\alpha$, we have $\eta \in \gamma$. So $\gamma$ is a transitive set, so $\gamma$ is an ordinal. - \item If $\beta \in \alpha$, then $I_\beta = \qty{\gamma \in \alpha : \gamma \in \beta} = \beta$, so $\beta < \alpha$. + \item If $\beta \in \alpha$, then $I_\beta = \qty{\gamma \in \alpha : \gamma \in\footnote{We are well-ordered by $\in$ not $<$ so we use $\in$ to define initial segments} \beta} = \beta$ as $\beta \subset \alpha$ by transitivity of $\alpha$, so $\beta < \alpha$. Any proper i.s. of $\alpha$ is of the form $I_\gamma$ for some $\gamma \in \alpha$. So $\beta < \alpha \implies \beta \in \alpha$. \item Done by (2) diff --git a/LogicAndSetTheory/06_cardinals.tex b/LogicAndSetTheory/06_cardinals.tex index e71e4c9..25b37bc 100644 --- a/LogicAndSetTheory/06_cardinals.tex +++ b/LogicAndSetTheory/06_cardinals.tex @@ -60,13 +60,13 @@ \subsection{The hierarchy of alephs} Conversely, let $\delta$ be an infinite initial ordinal. We need $\delta = \omega_\alpha$ for some $\alpha$. \\ - Easy induction show that $\alpha \leq \omega_\alpha \; \forall a$ so $\delta < \omega_{\delta + 1}$. + Easy induction show that $\alpha \leq \omega_\alpha \; \forall \alpha$ so $\delta < \omega_{\delta + 1}$. Take the least $\alpha$ s.t. $\delta < \omega_\alpha$. - Then $\alpha \neq 0$ and $a$ is not a limit, o/w $\delta < \omega$ for some $\gamma < \alpha$ \Lightning. \\ + Then $\alpha \neq 0$ and $\alpha$ is not a limit, o/w $\delta < \omega_\gamma$ for some $\gamma < \alpha$ \Lightning. \\ So $\alpha = \beta^+$ for some $\beta$. So we have $\omega_\beta \leq \delta < \omega_{\beta^+} = \gamma(\omega_\beta)$. Hence $\delta \hookrightarrow \omega_\beta$ and $\omega_\beta \hookrightarrow \delta$. - So by Schr\"oder-Bernstein, $\omega_\beta \equiv \delta$, so $\delta = \omega_\beta$ as $\delta$ is initial. + So by Schr\"oder-Bernstein, $\omega_\beta \equiv \delta$\footnote{As $\gamma(\omega_\beta)$ minimal order which doesn't inject into $\omega_\beta$}, so $\delta = \omega_\beta$ as $\delta$ is initial. \end{proof} \begin{definition}[Aleph Numbers] @@ -93,7 +93,7 @@ \subsection{Cardinal Arithmetic} Let $m, n$ be cardinals. Then, \begin{enumerate} - \item $m + n = \mathrm{card}\qty(M \amalg N)$; + \item $m + n = \mathrm{card}\qty(M \sqcup N)$; \item $m \cdot n = \mathrm{card}(M \times N)$; \item $m^n = \mathrm{card}(M^N)$; \end{enumerate} @@ -103,7 +103,7 @@ \subsection{Cardinal Arithmetic} \underline{Some Properties} \\ $m + n = n + m$ ($M \sqcup N \equiv N \sqcup M$) \\ $m \cdot n = n \cdot m$ ($M \times N \equiv N \times M$) \\ -$m (n + p) = mn + mp$ ($M \times (N \sqcup P) \equiv M \times N \sqcup M \times P$) +$m (n + p) = mn + mp$ ($M \times (N \sqcup P) \equiv M \times N \sqcup M \times P$) \\ $(m^n)^p = m^{np}$, $m^n m^p = m^{n + p}$, $m^p \cdot n^p = (mn)^p$ (not true for ordinals though) $m \leq n \implies m + p \leq n + p$, $mp \leq np$, $m^p \leq n^p$ etc. @@ -161,6 +161,8 @@ \subsection{Cardinal Arithmetic} \[ \aleph_\beta \leq \aleph_\alpha + \aleph_\beta \leq 2 \cdot \aleph_\beta \leq \aleph_\alpha \aleph_\beta \leq \aleph_\beta^2 = \aleph_\beta \] \end{proof} +Hence, for example, $X \sqcup X$ bijects with $X$ for any infinite set $X$. + \begin{note} In ZFC one can define more general infinite sums and products of cardinals. In the definitions below, as earlier, lower-case letters denote cardinals and upper-case letters denote sets with cardinality the corresponding lower-case letter. \end{note} @@ -193,12 +195,10 @@ \subsection{Cardinal Arithmetic} \begin{align*} 2^{\aleph_0}=\aleph_1 \end{align*} -P. Cohen proved in the 1960 s that if $\mathrm{ZFC}$ is consistent, then so are $\mathrm{ZFC}+\mathrm{CH}$ and $\mathrm{ZFC}+\neg \mathrm{CH}$. So $\mathrm{CH}$ is independent of $\mathrm{ZFC}$. - -Hence, for example, $X \amalg X$ bijects with $X$ for any infinite set $X$. +P. Cohen proved in the 1960s that if $\mathrm{ZFC}$ is consistent, then so are $\mathrm{ZFC}+\mathrm{CH}$ and $\mathrm{ZFC}+\neg \mathrm{CH}$. So $\mathrm{CH}$ is independent of $\mathrm{ZFC}$. +Extra stuff on cardinal exponentiation: \\ Cardinal exponentiation is not as simple as addition and multiplication. For instance, in $\mathsf{ZF}$, $2^{\aleph_0}$ need not even be an aleph number, for instance if $\mathbb R$ is not well-orderable. -In $\mathsf{ZFC}$, the statement $2^{\aleph_0} = \aleph_1$ is independent of the axioms; this is called the \vocab{continuum hypothesis}. $\mathsf{ZFC}$ does not even decide if $2^{\aleph_0} < 2^{\aleph_1}$. Even today, not all implications about cardinal exponentiation (such as $\aleph_\alpha^{\aleph_\beta}$) are known. \ No newline at end of file diff --git a/LogicAndSetTheory/logicandsettheory.pdf b/LogicAndSetTheory/logicandsettheory.pdf index 0cee45b..eaf3857 100644 Binary files a/LogicAndSetTheory/logicandsettheory.pdf and b/LogicAndSetTheory/logicandsettheory.pdf differ