From bbae7136b17332cb443c48ea8735cf2f2a335591 Mon Sep 17 00:00:00 2001 From: Joshua Potter Date: Sun, 13 Aug 2023 10:03:41 -0600 Subject: [PATCH] Add notion of "assumption"s and rename `pair` to a more general `tuple` macro. --- Bookshelf/Enderton/Logic.tex | 53 +++- Bookshelf/Enderton/Set.tex | 572 +++++++++++++++++------------------ preamble.tex | 20 ++ 3 files changed, 354 insertions(+), 291 deletions(-) diff --git a/Bookshelf/Enderton/Logic.tex b/Bookshelf/Enderton/Logic.tex index caebefb..81b019f 100644 --- a/Bookshelf/Enderton/Logic.tex +++ b/Bookshelf/Enderton/Logic.tex @@ -35,6 +35,14 @@ An \textbf{expression} is a finite sequence of symbols. +\section{\defined{\texorpdfstring{$n$}{n}-tuple}}% +\hyperlabel{ref:n-tuple} + +An \textbf{$n$-tuple} is recursively defined as + $$\ltuple{x_1}{x_{n+1}} = \tuple{\ltuple{x_1}{x_n}, x_{n+1}}$$ + for $n > 1$. +We also define $\tuple{x} = x$. + \section{\defined{Well-Formed Formula}}% \hyperlabel{ref:well-formed-formula} @@ -63,17 +71,52 @@ \chapter{Useful Facts About Sets}% \hyperlabel{chap:useful-facts-about-sets} -\section{\sorry{Lemma 0A}}% +\section{\unverified{Lemma 0A}}% \hyperlabel{sec:lemma-0a} \begin{lemma}[0A] - Assume that $\langle x_1, \ldots, x_m \rangle = - \langle y_1, \ldots, y_m, \ldots, y_{m+k} \rangle$. - Then $x_1 = \langle y_1, \ldots, y_{k+1} \rangle$. + Assume that $\ltuple{x_1}{x_m} = \ltuple{y_1, \ldots, y_m}{y_{m+k}}$. + Then $x_1 = \ltuple{y_1}{y_{k+1}}$. \end{lemma} \begin{proof} - TODO + + For natural number $m$, let $P(m)$ be the statement: + \begin{assumption} + \hyperlabel{sec:lemma-0a-eq1} + \text{If } \ltuple{x_1}{x_m} = \ltuple{y_1, \ldots, y_m}{y_{m+k}} + \text{ then } x_1 = \ltuple{y_1}{y_{k+1}}. + \end{assumption} + \noindent + We proceed by induction on $m$. + + \paragraph{Base Case}% + + Suppose $\tuple{x_1} = \ltuple{y_1}{y_{1 + k}}$. + By definition of an \nameref{ref:n-tuple}, $\tuple{x_1} = x_1$. + Thus $x_1 = \ltuple{y_1}{y_{k + 1}}$. + Hence $P(1)$ holds true. + + \paragraph{Inductive Step}% + + Suppose for $m \geq 1$ that $P(m)$ is true and assume + \begin{equation} + \hyperlabel{sec:lemma-0a-eq2} + \ltuple{x_1}{x_{m+1}} = \ltuple{y_1, \ldots, y_{m+1}}{y_{m+1+k}}. + \end{equation} + By definition of an \nameref{ref:n-tuple}, we can decompose + \eqref{sec:lemma-0a-eq2} into the following two identities + \begin{align*} + x_{m+1} & = y_{m+1+k} \\ + \ltuple{x_1}{x_m} & = \ltuple{y_1}{y_{m+k}}. + \end{align*} + By \eqref{sec:lemma-0a-eq1}, $P(m)$ implies $x_1 = \ltuple{y_1}{y_{k+1}}$. + Hence $P(m+1)$ holds true. + + \paragraph{Conclusion}% + + By induction, $P(m)$ holds true for all $m \geq 1$. + \end{proof} \chapter{Sentential Logic}% diff --git a/Bookshelf/Enderton/Set.tex b/Bookshelf/Enderton/Set.tex index 751653f..e4b704e 100644 --- a/Bookshelf/Enderton/Set.tex +++ b/Bookshelf/Enderton/Set.tex @@ -6,7 +6,6 @@ \input{../../preamble} \makecode{../..} -\newcommand{\pair}[1]{\left< #1 \right>} \newcommand{\ineq}{\,\mathop{\underline{\in}}\,} \begin{document} @@ -86,7 +85,7 @@ \hyperlabel{ref:composition} The \textbf{composition} of sets $F$ and $G$ is - $$F \circ G = \{\pair{u, v} \mid \exists t(uGt \land tFv)\}.$$ + $$F \circ G = \{\tuple{u, v} \mid \exists t(uGt \land tFv)\}.$$ \code{Bookshelf/Enderton/Set/Relation}{Set.Relation.comp} @@ -104,7 +103,7 @@ \hyperlabel{ref:domain} The \textbf{domain} of set $R$, denoted $\dom{R}$, is given by - $$x \in \dom{R} \iff \exists y \pair{x, y} \in R.$$ + $$x \in \dom{R} \iff \exists y \tuple{x, y} \in R.$$ \code{Bookshelf/Enderton/Set/Relation}{Set.Relation.dom} @@ -261,7 +260,7 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \hyperlabel{ref:inverse} The \textbf{inverse} of a set $F$ is the set - $$F^{-1} = \{\pair{u, v} \mid vFu\}.$$ + $$F^{-1} = \{\tuple{u, v} \mid vFu\}.$$ \code{Bookshelf/Enderton/Set/Relation}{Set.Relation.inv} @@ -335,7 +334,7 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \section{\defined{Ordered Pair}}% \hyperlabel{ref:ordered-pair} - For any sets $u$ and $v$, the \textbf{ordered pair} $\pair{u, v}$ is + For any sets $u$ and $v$, the \textbf{ordered pair} $\tuple{u, v}$ is the set $\{\{u\}, \{u, v\}\}$. \code*{Bookshelf/Enderton/Set/OrderedPair}{OrderedPair} @@ -446,7 +445,7 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \hyperlabel{ref:range} The \textbf{range} of set $R$, denoted $\ran{R}$, is given by - $$x \in \ran{R} \iff \exists t \pair{t, x} \in R.$$ + $$x \in \ran{R} \iff \exists t \tuple{t, x} \in R.$$ \code{Bookshelf/Enderton/Set/Relation}{Set.Relation.ran} @@ -475,7 +474,7 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \hyperlabel{ref:restriction} The \textbf{restriction} of a set $F$ to set $A$ is the set - $$F \restriction A = \{\pair{u, v} \mid uFv \land u \in A\}.$$ + $$F \restriction A = \{\tuple{u, v} \mid uFv \land u \in A\}.$$ \code{Bookshelf/Enderton/Set/Relation}{Set.Relation.restriction} @@ -2546,7 +2545,7 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and For any sets $x$, $y$, $u$, and $v$, \begin{equation} \hyperlabel{sub:theorem-3a-eq1} - \pair{u, v} = \pair{x, y} \iff u = x \land v = y. + \tuple{u, v} = \tuple{x, y} \iff u = x \land v = y. \end{equation} \end{theorem} @@ -2562,7 +2561,7 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \paragraph{($\Rightarrow$)}% - Suppose $\pair{u, v} = \pair{x, y}$. + Suppose $\tuple{u, v} = \tuple{x, y}$. Then, by definition of an \nameref{ref:ordered-pair}, \begin{equation} \hyperlabel{sub:theorem-3a-eq2} @@ -2616,7 +2615,7 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \hyperlabel{sub:lemma-3b} \begin{lemma}[3B] - If $x \in C$ and $y \in C$, then $\pair{x, y} \in \powerset{\powerset{C}}$. + If $x \in C$ and $y \in C$, then $\tuple{x, y} \in \powerset{\powerset{C}}$. \end{lemma} \code{Bookshelf/Enderton/Set/Chapter\_3} @@ -2628,7 +2627,7 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and $\{x\}$ and $\{x, y\}$ are members of $\powerset{C}$. Likewise, $\{\{x\}, \{x, y\}\}$ is a member of $\powerset{\powerset{C}}$. By definition of an \nameref{ref:ordered-pair}, - $\pair{x, y} = \{\{x\}, \{x, y\}\}$. + $\tuple{x, y} = \{\{x\}, \{x, y\}\}$. This concludes our proof. \end{proof} @@ -2637,7 +2636,7 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \begin{theorem}[3C] For any sets $A$ and $B$, there is a set whose members are exactly the - pairs $\pair{x, y}$ with $x \in A$ and $y \in B$. + pairs $\tuple{x, y}$ with $x \in A$ and $y \in B$. \end{theorem} \lean{Mathlib/SetTheory/ZFC/Basic}{Set.prod} @@ -2646,16 +2645,16 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and Define $C = A \cup B$. Then for all $x \in A$ and for all $y \in B$, $x$ and $y$ are both in $C$. By \nameref{sub:lemma-3b}, it follows that - $\pair{x, y} \in \powerset{\powerset{C}}$. + $\tuple{x, y} \in \powerset{\powerset{C}}$. The \nameref{ref:power-set-axiom} indicates $\powerset{\powerset{C}}$ is indeed a set. Therefore the \nameref{ref:subset-axioms} are applicable. This implies the existence of a set $D$ such that $$\forall z, (z \in D \iff z \in \powerset{\powerset{C}} \land (\exists x, \exists y, x \in A \land y \in B \land - z = \pair{x, y})).$$ + z = \tuple{x, y})).$$ By construction $D$ is the set whose members are exactly the pairs - $\pair{x, y}$ with $x \in A$ and $y \in B$. + $\tuple{x, y}$ with $x \in A$ and $y \in B$. \end{proof} \section{Relations}% @@ -2665,16 +2664,16 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \hyperlabel{sub:theorem-3d} \begin{theorem}[3D] - If $\pair{x, y} \in A$, then $x$ and $y$ belong to $\bigcup\bigcup A$. + If $\tuple{x, y} \in A$, then $x$ and $y$ belong to $\bigcup\bigcup A$. \end{theorem} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.theorem\_3d} \begin{proof} - Let $A$ be a set and $\pair{x, y} \in A$. + Let $A$ be a set and $\tuple{x, y} \in A$. By definition of an \nameref{ref:ordered-pair}, - $$\pair{x, y} = \{\{x\}, \{x, y\}\}.$$ + $$\tuple{x, y} = \{\{x\}, \{x, y\}\}.$$ By \nameref{sub:exercise-2.3}, $\{\{x\}, \{x, y\}\} \subseteq \bigcup A$. Then $\{x, y\} \in \bigcup A$. Another application of \nameref{sub:exercise-2.3} implies @@ -2709,9 +2708,9 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \paragraph{(i)}% By definition of the \nameref{ref:domain}, $x \in \dom{(F^{-1})}$ if and - only if there exists some $y$ such that $\pair{x, y} \in F^{-1}$. + only if there exists some $y$ such that $\tuple{x, y} \in F^{-1}$. By definition of the \nameref{ref:inverse} of a set, - $\pair{y, x} \in F$. + $\tuple{y, x} \in F$. By definition of the \nameref{ref:range}, $x \in \ran{F}$. Since each step holds biconditionally, it follows $\dom{(F^{-1})} = \ran{F}$ as expected. @@ -2719,9 +2718,9 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \paragraph{(ii)}% By definition of the \nameref{ref:range}, $x \in \ran{(F^{-1})}$ if and - only if there exists some $t$ such that $\pair{t, x} \in F^{-1}$. + only if there exists some $t$ such that $\tuple{t, x} \in F^{-1}$. By definition of the \nameref{ref:inverse} of a set, - $\pair{x, t} \in F$. + $\tuple{x, t} \in F$. By definition of the \nameref{ref:domain}, $x \in \dom{F}$. Since each step holds biconditionally, it follows $\ran{(F^{-1})} = \dom{F}$. @@ -2731,8 +2730,8 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and By definition of the \nameref{ref:inverse} of a set, \begin{align*} (F^{-1})^{-1} - & = \{\pair{u, v} \mid \pair{v, u} \in F^{-1}\} \\ - & = \{\pair{u, v} \mid \pair{u, v} \in F\} \\ + & = \{\tuple{u, v} \mid \tuple{v, u} \in F^{-1}\} \\ + & = \{\tuple{u, v} \mid \tuple{u, v} \in F\} \\ & = F. \end{align*} @@ -2766,22 +2765,22 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and Suppose $F^{-1}$ is a \nameref{ref:function}. By definition, for each $x \in \dom{(F^{-1})}$, there is only one $y$ - such that $\pair{x, y} \in F^{-1}$. + such that $\tuple{x, y} \in F^{-1}$. By definition of the \nameref{ref:inverse} of $F$, - $F^{-1} = \{\pair{u, v} \mid vFu\}$. + $F^{-1} = \{\tuple{u, v} \mid vFu\}$. Then for each $x \in \ran{F}$, there exists exactly one $y$ such that - $\pair{y, x} \in F$. + $\tuple{y, x} \in F$. This definitionally means $F$ is single-rooted. \subparagraph{($\Leftarrow$)}% Suppose $F$ is single-rooted. By definition, for each $x \in \ran{F}$, there is only one $t$ such that - $\pair{t, x} \in F$. + $\tuple{t, x} \in F$. By definition of the \nameref{ref:inverse} of $F$, - $F^{-1} = \{\pair{u, v} \mid vFu\}$. + $F^{-1} = \{\tuple{u, v} \mid vFu\}$. Then for each $x \in \dom{(F^{-1})}$ there exists exactly one $t$ such - that $\pair{x, t} \in F^{-1}$. + that $\tuple{x, t} \in F^{-1}$. This definitionally means $F^{-1}$ is a function. \paragraph{(ii)}% @@ -2822,11 +2821,11 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and By hypothesis, $F$ is one-to-one. This means it is single-rooted, i.e. for all $x \in \ran{F}$, there exists - exactly one $t$ such that $\pair{t, x} \in F$. + exactly one $t$ such that $\tuple{t, x} \in F$. By definition of the \nameref{ref:inverse} of $F$, - $\pair{x, t} \in F^{-1}$. + $\tuple{x, t} \in F^{-1}$. But then for all $x \in \dom{(F^{-1})}$, there exists exactly one $t$ such - that $\pair{x, t} \in F^{-1}$. + that $\tuple{x, t} \in F^{-1}$. Thus $F^{-1}$ is a function. \paragraph{(ii)}% @@ -2834,11 +2833,11 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and By hypothesis, $F$ is single-valued. That is, for all $x \in \dom{F}$, there exists exactly one $y$ such that - $\pair{x, y} \in F$. + $\tuple{x, y} \in F$. By definition of the \nameref{ref:inverse} of $F$, - $\pair{y, x} \in F^{-1}$. + $\tuple{y, x} \in F^{-1}$. But then for all $x \in \ran{(F^{-1})}$, there exists exactly one $y$ such - that $\pair{y, x} \in F^{-1}$. + that $\tuple{y, x} \in F^{-1}$. Thus $F^{-1}$ is single-rooted. \paragraph{Conclusion}% @@ -2867,11 +2866,11 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and Suppose $F$ is a one-to-one \nameref{ref:function}. Then \nameref{sub:one-to-one-inverse} indicates $F^{-1}$ is a one-to-one function with domain $\ran{F}$ and range $\dom{F}$. - For all $x \in \dom{F}$, $\pair{x, F(x)} \in F$. - Then $\pair{F(x), x} \in F^{-1}$. + For all $x \in \dom{F}$, $\tuple{x, F(x)} \in F$. + Then $\tuple{F(x), x} \in F^{-1}$. Since $F^{-1}$ is single-valued, $F^{-1}(F(x)) = x$. - For all $y \in \ran{F}$, $\pair{y, F^{-1}(y)} \in F^{-1}$. - Then $\pair{F^{-1}(y), y} \in F$. + For all $y \in \ran{F}$, $\tuple{y, F^{-1}(y)} \in F^{-1}$. + Then $\tuple{F^{-1}(y), y} \in F$. Since $F$ is single-valued, $F(F^{-1}(y)) = y$. \end{proof} @@ -2899,12 +2898,12 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and By definition of the \nameref{ref:composition} of $F$ and $G$, \begin{equation} \hyperlabel{sub:theorem-3h-eq2} - F \circ G = \{\pair{u, v} \mid \exists t(uGt \land tFv)\}. + F \circ G = \{\tuple{u, v} \mid \exists t(uGt \land tFv)\}. \end{equation} By construction, $F \circ G$ is a relation. By the definition of the \nameref{ref:domain} of a relation, $x \in \dom{(F \circ G)}$ if and only if there exists some $y$ such that - $\pair{x, y} \in F \circ G$. + $\tuple{x, y} \in F \circ G$. We prove that (i) $F \circ G$ is a function with domain satisfying \eqref{sub:theorem-3h-eq1}, and (ii) $(F \circ G)(x) = F(G(x))$. @@ -2912,22 +2911,22 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \hyperlabel{par:theorem-3h-i} By \eqref{sub:theorem-3h-eq2}, there exists some $t$ such that - $\pair{x, t} \in G$ and $\pair{t, y} \in F$. + $\tuple{x, t} \in G$ and $\tuple{t, y} \in F$. Since $G$ is single-valued, $t$ is uniquely determined by $x$. Since $F$ is single-valued, $y$ is uniquely determined by $t$. Therefore, by transitivity, $y$ is uniquely determined by $x$. Thus $F \circ G$ is single-valued, i.e. $F \circ G$ is a function. Furthermore, by definition of function application, $t = G(x)$. - Thus $$\pair{x, G(x)} \in G \quad\text{and}\quad \pair{G(x), y} \in F.$$ + Thus $$\tuple{x, G(x)} \in G \quad\text{and}\quad \tuple{G(x), y} \in F.$$ This immediately implies \eqref{sub:theorem-3h-eq1} holds true. \paragraph{(ii)}% Let $x \in \dom{(F \circ G)}$. - By definition, $\pair{x, (F \circ G)(x)} \in F \circ G$. + By definition, $\tuple{x, (F \circ G)(x)} \in F \circ G$. Then \eqref{sub:theorem-3h-eq2} implies $(F \circ G)(x)$ satisfies - $\pair{G(x), (F \circ G)(x)} \in F$. + $\tuple{G(x), (F \circ G)(x)} \in F$. This is equivalent to saying $F(G(x)) = (F \circ G)(x)$ as expected. \end{proof} @@ -2944,13 +2943,13 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \begin{proof} By definition of the \nameref{ref:composition} of $F$ and $G$, - $$F \circ G = \{\pair{u, v} \mid \exists t(uGt \land tFv)\}.$$ + $$F \circ G = \{\tuple{u, v} \mid \exists t(uGt \land tFv)\}.$$ By definition of the \nameref{ref:inverse} of a function, \begin{align*} (F \circ G)^{-1} - & = \{\pair{u, v} \mid \exists t (vGt \land tFu)\} \\ - & = \{\pair{u, v} \mid \exists t (tFu \land vGt)\} \\ - & = \{\pair{u, v} \mid + & = \{\tuple{u, v} \mid \exists t (vGt \land tFu)\} \\ + & = \{\tuple{u, v} \mid \exists t (tFu \land vGt)\} \\ + & = \{\tuple{u, v} \mid \exists t \left[ u(F^{-1})t \land t(G^{-1})v \right]\} \\ & = G^{-1} \circ F^{-1}. \end{align*} @@ -2985,8 +2984,8 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and All that remains is to prove $F$ is single-rooted. Let $y \in \ran{F}$. By definition of the \nameref{ref:range} of a function, there exists - some $x_1$ such that $\pair{x_1, y} \in F$. - Suppose there exists a set $x_2$ such that $\pair{x_2, y} \in F$. + some $x_1$ such that $\tuple{x_1, y} \in F$. + Suppose there exists a set $x_2$ such that $\tuple{x_2, y} \in F$. By hypothesis, $G(F(x_1)) = G(F(x_2))$ implies $I_A(x_1) = I_A(x_2)$. Thus $x_1 = x_2$. Therefore $F$ must be single-rooted. @@ -3026,14 +3025,14 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and Suppose $F$ maps $A$ \textit{onto} $B$. By definition of maps onto, $\ran{F} = B$. Then for all $y \in B$, there exists some $x \in A$ such that - $\pair{x, y} \in F$. + $\tuple{x, y} \in F$. Notice though that $F^{-1}[\{y\}]$ may not be a singleton set. Then there is no obvious way to \textit{choose} an element from each preimage to form a function. By the \nameref{ref:axiom-of-choice-1}, there exists a function $H \subseteq F^{-1}$ such that $\dom{H} = \dom{F^{-1}} = B$. - For all $y \in B$, $\pair{y, H(y)} \in H \subseteq F^{-1}$ - meaning $\pair{H(y), y} \in F$. + For all $y \in B$, $\tuple{y, H(y)} \in H \subseteq F^{-1}$ + meaning $\tuple{H(y), y} \in F$. Thus $F(H(y)) = y$ as expected. \end{proof} @@ -3224,8 +3223,8 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and By definition of the difference of two sets, $v \in \img{F}{A}$ and $v \not\in \img{F}{B}$. By definition of the \nameref{ref:image} of a set, there exists a set - $u \in A$ such that $\pair{u, v} \in F$. - Likewise, $\forall w \in B, \pair{w, v} \not\in F$. + $u \in A$ such that $\tuple{u, v} \in F$. + Likewise, $\forall w \in B, \tuple{w, v} \not\in F$. Thus $u \not\in B$, since otherwise we get an immediate contradiction. Therefore $u \in A - B$ meaning $v \in \img{F}{A - B}$. @@ -3430,13 +3429,13 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and $F \colon A \rightarrow A$. Suppose $F$ is \nameref{ref:compatible} with $R$. Next define \nameref{ref:relation} $\hat{F}$ to be - $$\hat{F} = \{\pair{[x]_R, [F(x)]_R} \mid x \in A\}.$$ + $$\hat{F} = \{\tuple{[x]_R, [F(x)]_R} \mid x \in A\}.$$ By construction $\hat{F}$ has domain $A / R$ and $\ran{\hat{F}} \subseteq A / R$. All that remains is proving $\hat{F}$ is single-valued. Let $[x_1]_R, [x_2]_R \in \dom{\hat{F}}$ such that $[x_1]_R = [x_2]_R$. - By definition of $\hat{F}$, $\pair{[x_1]_R, [F(x_1)]_R} \in \hat{F}$ - and $\pair{[x_2]_R, [F(x_2)]_R} \in \hat{F}$. + By definition of $\hat{F}$, $\tuple{[x_1]_R, [F(x_1)]_R} \in \hat{F}$ + and $\tuple{[x_2]_R, [F(x_2)]_R} \in \hat{F}$. By \nameref{sub:lemma-3n}, $[x_1]_R = [x_2]_R$ implies $x_1Rx_2$. Since $F$ is compatible, $F(x_1)RF(x_2)$. Another application of \nameref{sub:lemma-3n} implies that @@ -3484,7 +3483,7 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and Let $x \in A$. By definition, $R$ is \nameref{ref:trichotomous}. Then only one of $xRx$ and $x = x$ can hold. - Since $x = x$ obviously holds, it follows $\pair{x, x} \not\in R$. + Since $x = x$ obviously holds, it follows $\tuple{x, x} \not\in R$. \paragraph{(ii)}% @@ -3503,10 +3502,10 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and Suppose that we attempted to generalize the Kuratowski definitions of ordered pairs to ordered triples by defining - $$\pair{x, y, z}^* = \{\{x\}, \{x, y\}, \{x, y, z\}\}.$$ + $$\tuple{x, y, z}^* = \{\{x\}, \{x, y\}, \{x, y, z\}\}.$$ Show that this definition is unsuccessful by giving examples of objects $u$, $v$, $w$, $x$, $y$, $z$ with - $\pair{x, y, z}^* = \pair{u, v, w}^*$ but with either + $\tuple{x, y, z}^* = \tuple{u, v, w}^*$ but with either $y \neq v$ or $z \neq w$ (or both). \code*{Bookshelf/Enderton/Set/Chapter\_3} @@ -3517,19 +3516,19 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and Let $u = 1$, $v = 2$, and $w = 2$. Then \begin{align*} - \pair{x, y, z}^* + \tuple{x, y, z}^* & = \{\{x\}, \{x, y\}, \{x, y, z\}\} \\ & = \{\{1\}, \{1, 1\}, \{1, 1, 2\}\} \\ & = \{\{1\}, \{1, 2\}\}. \end{align*} Likewise \begin{align*} - \pair{u, v, w}^* + \tuple{u, v, w}^* & = \{\{u\}, \{u, v\}, \{u, v, w\}\} \\ & = \{\{1\}, \{1, 2\}, \{1, 2, 2\}\} \\ & = \{\{1\}, \{1, 2\}\}. \end{align*} - Thus $\pair{x, y, z}^* = \pair{u, v, w}^*$ but $y \neq v$. + Thus $\tuple{x, y, z}^* = \tuple{u, v, w}^*$ but $y \neq v$. \end{proof} \subsection{\verified{Exercise 3.2a}}% @@ -3545,13 +3544,13 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and Then by \nameref{sub:corollary-3c} and the definition of the union of sets, \begin{align*} A \times (B \cup C) - & = \{ \pair{x, y} \mid x \in A \land y \in (B \cup C) \} \\ - & = \{ \pair{x, y} \mid + & = \{ \tuple{x, y} \mid x \in A \land y \in (B \cup C) \} \\ + & = \{ \tuple{x, y} \mid x \in A \land (y \in B \lor y \in C) \} \\ - & = \{ \pair{x, y} \mid + & = \{ \tuple{x, y} \mid (x \in A \land y \in B) \lor (x \in A \land y \in C) \} \\ - & = \{ \pair{x, y} \mid (x \in A \land y \in B) \} \cup - \{ \pair{x, y} \mid (x \in A \land y \in C) \} \\ + & = \{ \tuple{x, y} \mid (x \in A \land y \in B) \} \cup + \{ \tuple{x, y} \mid (x \in A \land y \in C) \} \\ & = (A \times B) \cup (A \times C). \end{align*} \end{proof} @@ -3568,9 +3567,9 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and Let $A$, $B$, and $C$ be arbitrary sets such that $A \neq \emptyset$. By \nameref{sub:corollary-3c}, \begin{align} - A \times B & = \{ \pair{x, y} \mid x \in A \land y \in B \} + A \times B & = \{ \tuple{x, y} \mid x \in A \land y \in B \} & \hyperlabel{sub:exercise-3.2b-eq1} \\ - A \times C & = \{ \pair{x, y} \mid x \in A \land y \in C \}. + A \times C & = \{ \tuple{x, y} \mid x \in A \land y \in C \}. & \hyperlabel{sub:exercise-3.2b-eq2} \end{align} There are two cases to consider: @@ -3579,11 +3578,11 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and Suppose $B \neq \emptyset$. Then $A \times B \neq \emptyset$ and $A \times C \neq \emptyset$. - Let $\pair{x, y} \in A \times B$. + Let $\tuple{x, y} \in A \times B$. By \eqref{sub:exercise-3.2b-eq1}, $x \in A$ and $y \in B$. By the \nameref{ref:extensionality-axiom}, - $$\pair{x, y} \in A \times B \iff \pair{x, y} \in A \times C.$$ - Therefore $\pair{x, y} \in A \times C$. + $$\tuple{x, y} \in A \times B \iff \tuple{x, y} \in A \times C.$$ + Therefore $\tuple{x, y} \in A \times C$. By \eqref{sub:exercise-3.2b-eq2}, $x \in A$ and $y \in C$. Since membership of $y$ in $B$ and in $C$ holds biconditionally, the \nameref{ref:extensionality-axiom} indicates $B = C$. @@ -3591,9 +3590,9 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \paragraph{Case 2}% Suppose $B = \emptyset$. - Then there is no $\pair{x, y}$ such that $x \in A$ and $y \in B$. + Then there is no $\tuple{x, y}$ such that $x \in A$ and $y \in B$. Thus $A \times B = \emptyset$ and $A \times C = \emptyset$. - But then there cannot exist an $\pair{x, y}$ such that $x \in A$ + But then there cannot exist an $\tuple{x, y}$ such that $x \in A$ and $y \in C$ either. Since $A \neq \emptyset$, it must be the case that $C = \emptyset$. Thus $B = C$. @@ -3614,11 +3613,11 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and By \nameref{sub:corollary-3c} and the definition of the union of sets, \begin{align*} A \times \bigcup\mathscr{B} - & = \{ \pair{x, y} \mid + & = \{ \tuple{x, y} \mid x \in A \land y \in \bigcup\mathscr{B} \} \\ - & = \{ \pair{x, y} \mid + & = \{ \tuple{x, y} \mid x \in A \land (\exists b \in \mathscr{B}), y \in b \} \\ - & = \{ \pair{x, y} \mid + & = \{ \tuple{x, y} \mid (\exists b \in \mathscr{B}), x \in A \land y \in b \} \\ & = \bigcup\; \{ A \times X \mid X \in \mathscr{B} \}. \end{align*} @@ -3632,7 +3631,7 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \begin{proof} For the sake of contradiction, suppose there exists a set $A$ to which every ordered pair belongs. - That is, for all sets $x$ and $y$, $\pair{x, y} = \{\{x\}, \{x, y\}\}$ + That is, for all sets $x$ and $y$, $\tuple{x, y} = \{\{x\}, \{x, y\}\}$ is a member of $A$. By the \nameref{ref:union-axiom}, it follows that $\bigcup\bigcup A$ is the set to which every set belongs. @@ -3667,7 +3666,7 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and Thus, by the \nameref{ref:subset-axioms}, the following is also a set: $$C = \{ y \in \powerset{(A \times B)} \mid \exists a \in A, \forall x, \left[ x \in y \iff - \exists b \in B, x = \pair{a, b} \right] \}.$$ + \exists b \in B, x = \tuple{a, b} \right] \}.$$ We now show that $C$ satisfies \eqref{sub:exercise-3.5a-eq1}. \paragraph{($\Rightarrow$)}% @@ -3675,12 +3674,12 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and Suppose $y \in C$. Then there exists some $a \in A$ such that $$\forall x, \left[ x \in y \iff - \exists b \in B, x = \pair{a, b} \right].$$ + \exists b \in B, x = \tuple{a, b} \right].$$ By the \nameref{ref:extensionality-axiom}, \begin{align*} y - & = \{ \pair{a, b} \mid b \in B \} \\ - & = \{ \pair{x, b} \mid x \in \{a\} \land b \in B \} \\ + & = \{ \tuple{a, b} \mid b \in B \} \\ + & = \{ \tuple{x, b} \mid x \in \{a\} \land b \in B \} \\ & = \{ \{a\} \times B \}. \end{align*} @@ -3688,9 +3687,9 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and Suppose $y = \{a\} \times B$ for some $a \in A$. By \nameref{sub:corollary-3c}, $x \in \{a\} \times B$ if and only if - $\exists b \in B$ such that $x = \pair{a, b}$. + $\exists b \in B$ such that $x = \tuple{a, b}$. But then $x \in y$ if and only if $\exists b \in B$ such that - $x = \pair{a, b}$. + $x = \tuple{a, b}$. This immediately proves $y \in C$. \end{proof} @@ -3720,7 +3719,7 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and Let $c \in A \times B$. Then there exists some $a \in A$ and $b \in B$ such that - $c = \pair{a, b}$. + $c = \tuple{a, b}$. Thus $c \in \{a\} \times B$. We also note $\{a\} \times B \in \{\{x\} \times B \mid x \in A\}$, specifically when $x = a$. @@ -3763,10 +3762,10 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and We show for all $a \in A$, $a \in \dom{A} \times \ran{A}$. Let $a \in A$. Since $A$ is a relation, $a$ is an ordered pair. - Then there exists some sets $x$ and $y$ such that $a = \pair{x, y}$. + Then there exists some sets $x$ and $y$ such that $a = \tuple{x, y}$. By the definition of the \nameref{ref:domain} and \nameref{ref:range} of $A$, $x \in \dom{A}$ and $y \in \ran{A}$. - Thus $a = \pair{x, y} \in \dom{A} \times \ran{A}$ as well. + Thus $a = \tuple{x, y} \in \dom{A} \times \ran{A}$ as well. This proves $A \subseteq \dom{A} \times \ran{A}$. \paragraph{($\Leftarrow$)}% @@ -3798,11 +3797,11 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and That is, $x \in \dom{R}$ or $x \in \ran{R}$. If $x \in \dom{R}$, then there exists some $y$ such that - $\pair{x, y} = \{\{x\}, \{x, y\}\} \in R$. + $\tuple{x, y} = \{\{x\}, \{x, y\}\} \in R$. Then $\{x\} \in \bigcup R$ and $x \in \bigcup\bigcup R$. On the other hand, if $x \in \ran{R}$, then there exists some $t$ such that - $\pair{t, x} = \{\{t\}, \{t, x\}\} \in R$. + $\tuple{t, x} = \{\{t\}, \{t, x\}\} \in R$. Then $\{t, x\} \in \bigcup R$ and $x \in \bigcup\bigcup R$. \paragraph{(ii)}% @@ -3811,7 +3810,7 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and Let $t \in \bigcup\bigcup R$. Then there exists some member $T \in \bigcup R$ such that $t \in T$. Likewise there exists some member $T' \in R$ such that $T \in T'$. - By definition of a relation, $T' = \pair{x, y} = \{\{x\}, \{x, y\}\}$ for + By definition of a relation, $T' = \tuple{x, y} = \{\{x\}, \{x, y\}\}$ for some sets $x$ and $y$. Thus $t = x$ or $t = y$. By \nameref{sub:exercise-3.6}, $t \in \dom{R}$ or $t \in \ran{R}$. @@ -3851,11 +3850,11 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and Let $x \in \dom{\bigcup{\mathscr{A}}}$. By definition of a domain, there exists some $y$ such that - $\pair{x, y} \in \bigcup{\mathscr{A}}$. + $\tuple{x, y} \in \bigcup{\mathscr{A}}$. By definition of the union of sets, - $\exists y, \exists R \in \mathscr{A}, \pair{x, y} \in R$. + $\exists y, \exists R \in \mathscr{A}, \tuple{x, y} \in R$. Equivalently, - $\exists R \in \mathscr{A}, \exists y, \pair{x, y} \in R$. + $\exists R \in \mathscr{A}, \exists y, \tuple{x, y} \in R$. By another application of the definition of a domain, $\exists R \in \mathscr{A}, x \in \dom{R}$. By another application of the definition of the union of sets, @@ -3867,11 +3866,11 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and Let $x \in \ran{\bigcup{\mathscr{A}}}$. By definition of a range, there exists some $t$ such that - $\pair{t, x} \in \bigcup{\mathscr{A}}$. + $\tuple{t, x} \in \bigcup{\mathscr{A}}$. By definition of the union of sets, - $\exists t, \exists R \in \mathscr{A}, \pair{t, x} \in R$. + $\exists t, \exists R \in \mathscr{A}, \tuple{t, x} \in R$. Equivalently, - $\exists R \in \mathscr{A}, \exists t, \pair{t, x} \in R$. + $\exists R \in \mathscr{A}, \exists t, \tuple{t, x} \in R$. By another application of the definition of a range, $\exists R \in \mathscr{A}, x \in \ran{R}$. By another application of the definition of the union of sets, @@ -3912,11 +3911,11 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and Let $x \in \dom{\bigcap{\mathscr{A}}}$. By definition of the \nameref{ref:domain} of a set, - $\exists y, \pair{x, y} \in \bigcap{\mathscr{A}}$. + $\exists y, \tuple{x, y} \in \bigcap{\mathscr{A}}$. By definition of the intersection of sets, - $\exists y, \forall R \in \mathscr{A}, \pair{x, y} \in R$. + $\exists y, \forall R \in \mathscr{A}, \tuple{x, y} \in R$. But this implies that - $\forall R \in \mathscr{A}, \exists y, \pair{x, y} \in R$. + $\forall R \in \mathscr{A}, \exists y, \tuple{x, y} \in R$. By another application of the definition of the \nameref{ref:domain} of a set, $\forall R \in \mathscr{A}, x \in \dom{R}$. By another application of the intersection of sets, @@ -3927,11 +3926,11 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and Let $x \in \ran{\bigcap{\mathscr{A}}}$. By definition of the \nameref{ref:range} of a set, - $\exists t, \pair{t, x} \in \bigcap{\mathscr{A}}$. + $\exists t, \tuple{t, x} \in \bigcap{\mathscr{A}}$. By definition of the intersection of sets, - $\exists t, \forall R \in \mathscr{A}, \pair{t, x} \in R$. + $\exists t, \forall R \in \mathscr{A}, \tuple{t, x} \in R$. But this implies that - $\forall R \in \mathscr{A}, \exists t, \pair{t, x} \in R$. + $\forall R \in \mathscr{A}, \exists t, \tuple{t, x} \in R$. By another application of the definition of the \nameref{ref:range} of a set, $\forall R \in \mathscr{A}, x \in \ran{R}$. By another application of the intersection of sets, @@ -3947,18 +3946,19 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and integer $m$ less than $4$. \begin{answer} - Let $\pair{x_1, x_2, x_3, x_4}$ denote an arbitrary $4$-tuple. + Let $\tuple{x_1, x_2, x_3, x_4}$ denote an arbitrary $4$-tuple. Then \begin{align} - \pair{x_1, x_2, x_3, x_4} - & = \pair{\pair{x_1, x_2, x_3}, x_4} + \tuple{x_1, x_2, x_3, x_4} + & = \tuple{\tuple{x_1, x_2, x_3}, x_4} & \hyperlabel{sub:exercise-7.10-eq1} \\ - & = \pair{\pair{\pair{x_1, x_2}, x_3}, x_4} + & = \tuple{\tuple{\tuple{x_1, x_2}, x_3}, x_4} & \hyperlabel{sub:exercise-7.10-eq2} \end{align} Here \eqref{sub:exercise-7.10-eq1} is an equivalent ordered $2$-tuple and \eqref{sub:exercise-7.10-eq2} is an equivalent ordered $3$-tuple. - Furthermore, $\pair{x_1, x_2, x_3, x_4} = \pair{\pair{x_1, x_2, x_3, x_4}}$, + Furthermore, + $\tuple{x_1, x_2, x_3, x_4} = \tuple{\tuple{x_1, x_2, x_3, x_4}}$, showing it can be represented as an ordered $1$-tuple as well. \end{answer} @@ -3975,13 +3975,13 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \begin{proof} Let $F$ and $G$ be functions such that $\dom{F} = \dom{G}$ and $F(x) = G(x)$ for all $x$ in the common domain. - We prove that $\pair{x, y} \in F$ if and only if $\pair{x, y} \in G$. + We prove that $\tuple{x, y} \in F$ if and only if $\tuple{x, y} \in G$. But this follows immediately: \begin{align*} - \pair{x, y} \in F - & \iff y = F(x) \land \pair{x, F(x)} \in F \\ - & \iff y = G(x) \land \pair{x, G(x)} \in G \\ - & \iff \pair{x, y} \in G. + \tuple{x, y} \in F + & \iff y = F(x) \land \tuple{x, F(x)} \in F \\ + & \iff y = G(x) \land \tuple{x, G(x)} \in G \\ + & \iff \tuple{x, y} \in G. \end{align*} By the \nameref{ref:extensionality-axiom}, $F = G$. \end{proof} @@ -4002,8 +4002,8 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \paragraph{($\Rightarrow$)}% Suppose $f \subseteq g$. - Then for all \nameref{ref:ordered-pair}s $\pair{x, y}$, - $\pair{x, y} \in f$ implies $\pair{x, y} \in g$. + Then for all \nameref{ref:ordered-pair}s $\tuple{x, y}$, + $\tuple{x, y} \in f$ implies $\tuple{x, y} \in g$. Thus every $x \in \dom{f}$ must be a member of $\dom{g}$. Likewise, by definition of a function, $f$ and $g$ are single-valued. Thus $f(x) = y$ and $g(x) = y$. @@ -4014,9 +4014,9 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and Suppose $\dom{f} \subseteq \dom{g}$ and $(\forall x \in \dom{f}) f(x) = g(x)$. - Let $\pair{x, y} \in f$. + Let $\tuple{x, y} \in f$. By hypothesis, $x \in \dom{g}$ and $y = f(x) = g(x)$. - Thus $\pair{x, y} \in g$ as well. + Thus $\tuple{x, y} \in g$ as well. Therefore $f \subseteq g$. \end{proof} @@ -4074,10 +4074,10 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and Suppose $f \cup g$ is a function. Let $x \in (\dom{f}) \cap (\dom{g})$. That is, $x \in \dom{f}$ and $x \in \dom{g}$. - Then there exists only one $y_1$ such that $\pair{x, y_1} \in f$. + Then there exists only one $y_1$ such that $\tuple{x, y_1} \in f$. Likewise there exists only one $y_2$ such that - $\pair{x, y_2} \in g$. - But $\pair{x, y_1} \in f \cup g$ and $\pair{x, y_2} \in f \cup g$. + $\tuple{x, y_2} \in g$. + But $\tuple{x, y_1} \in f \cup g$ and $\tuple{x, y_2} \in f \cup g$. Since $f \cup g$ is single-valued, it follows $y_1 = y_2$. That is, $f(x) = g(x)$. @@ -4090,13 +4090,13 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \begin{enumerate}[(i)] \item Suppose $x \in \dom{f}$ but not in $\dom{g}$. Since $f$ is a function, it follows $f \cup g$ has only one value $y$ - such that $\pair{x, y} \in f \cup g$. + such that $\tuple{x, y} \in f \cup g$. \item Suppose $x \in \dom{g}$ but not in $\dom{f}$. Again, since $g$ is a function, it follows $f \cup g$ has only one - value $y$ such that $\pair{x, y} \in f \cup g$. + value $y$ such that $\tuple{x, y} \in f \cup g$. \item Suppose $x \in \dom{f}$ and $x \in \dom{g}$. By hypothesis, $f(x) = g(x)$ meaning there is only one value $y$ such - that $\pair{x, y} \in f \cup g$. + that $\tuple{x, y} \in f \cup g$. \end{enumerate} The above cases are exhaustive. @@ -4119,26 +4119,26 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and and $g$ in $\mathscr{A}$, either $f \subseteq g$ or $g \subseteq f$. Let $x \in \dom{\bigcup{\mathscr{A}}}$. Then there exists some $y_1$ such that - $\pair{x, y_1} \in \bigcup{\mathscr{A}}$. + $\tuple{x, y_1} \in \bigcup{\mathscr{A}}$. Suppose there also exists some $y_2$ such that - $\pair{x, y_2} \in \bigcup{\mathscr{A}}$. + $\tuple{x, y_2} \in \bigcup{\mathscr{A}}$. By definition of the union of sets, there exists some function - $f \in \mathscr{A}$ such that $\pair{x, y_1} \in f$. + $f \in \mathscr{A}$ such that $\tuple{x, y_1} \in f$. Likewise there exists some function $g \in \mathscr{A}$ such that - $\pair{x, y_2} \in g$. + $\tuple{x, y_2} \in g$. There are two cases to consider: \paragraph{Case 1}% Suppose $f \subseteq g$. - Then $\pair{x, y_1}, \pair{x, y_2} \in g$. + Then $\tuple{x, y_1}, \tuple{x, y_2} \in g$. Since $g$ is a function, i.e. single-valued, $y_1 = y_2$. \paragraph{Case 2}% Suppose $g \subseteq f$. - Then $\pair{x, y_1}, \pair{x, y_2} \in f$. + Then $\tuple{x, y_1}, \tuple{x, y_2} \in f$. Since $f$ is a function, i.e. single-valued, $y_1 = y_2$. \paragraph{Conclusion}% @@ -4181,17 +4181,17 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and By definition of the \nameref{ref:composition} of sets, \begin{equation} \hyperlabel{sub:exercise-3.17-eq1} - F \circ G = \{\pair{u, v} \mid \exists t(uGt \land tFv)\}. + F \circ G = \{\tuple{u, v} \mid \exists t(uGt \land tFv)\}. \end{equation} Consider any $v \in \ran{(F \circ G)}$. By definition of the \nameref{ref:range} of a \nameref{ref:relation}, there - exists some $u_1$ such that $\pair{u_1, v} \in F \circ G$. - Let $u_2$ be a set such that $\pair{u_2, v} \in F \circ G$. + exists some $u_1$ such that $\tuple{u_1, v} \in F \circ G$. + Let $u_2$ be a set such that $\tuple{u_2, v} \in F \circ G$. By \eqref{sub:exercise-3.17-eq1}, there exists a set $t_1$ such that - $\pair{u_1, t_1} \in G$ and $\pair{t_1, v} \in F$. + $\tuple{u_1, t_1} \in G$ and $\tuple{t_1, v} \in F$. Likewise, there exists a set $t_2$ such that - $\pair{u_2, t_2} \in G$ and $\pair{t_2, v} \in F$. + $\tuple{u_2, t_2} \in G$ and $\tuple{t_2, v} \in F$. But $F$ is single-rooted, meaning $t_1 = t_2$. Likewise, because $G$ is single-rooted, $u_1 = u_2$. Thus $F \circ G$ must also be single-rooted. @@ -4208,8 +4208,8 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \hyperlabel{sub:exercise-3.18} Let $R$ be the set - $$\{ \pair{0, 1}, \pair{0, 2}, \pair{0, 3}, - \pair{1, 2}, \pair{1, 3}, \pair{2, 3}\}.$$ + $$\{ \tuple{0, 1}, \tuple{0, 2}, \tuple{0, 3}, + \tuple{1, 2}, \tuple{1, 3}, \tuple{2, 3}\}.$$ Evaluate the following: $R \circ R$, $R \restriction \{1\}$, $R^{-1} \restriction \{1\}$, $\img{R}{\{1\}}$, and $\img{R^{-1}}{\{1\}}$. @@ -4230,9 +4230,9 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \begin{proof} \begin{enumerate}[(i)] - \item $R \circ R = \{ \pair{0, 2}, \pair{0, 3}, \pair{1, 3} \}$. - \item $R \restriction \{1\} = \{ \pair{1, 2}, \pair{1, 3} \}$. - \item $R^{-1} \restriction \{1\} = \{\pair{1, 0}\}$. + \item $R \circ R = \{ \tuple{0, 2}, \tuple{0, 3}, \tuple{1, 3} \}$. + \item $R \restriction \{1\} = \{ \tuple{1, 2}, \tuple{1, 3} \}$. + \item $R^{-1} \restriction \{1\} = \{\tuple{1, 0}\}$. \item $\img{R}{\{1\}} = \{2, 3\}$. \item $\img{R^{-1}}{\{1\}} = \{0\}$. \end{enumerate} @@ -4242,8 +4242,8 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \hyperlabel{sub:exercise-3.19} Let $$A = \{ - \pair{\emptyset, \{\emptyset, \{\emptyset\}\}}, - \pair{\{\emptyset\}, \emptyset} + \tuple{\emptyset, \{\emptyset, \{\emptyset\}\}}, + \tuple{\{\emptyset\}, \emptyset} \}.$$ Evaluate each of the following: $A(\emptyset)$, $\img{A}{\emptyset}$, $\img{A}{\{\emptyset\}}$, $\img{A}{\{\emptyset, \{\emptyset\}\}}$, @@ -4290,14 +4290,14 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \item $\img{A}{\{\emptyset, \{\emptyset\}\}} = \{\{\emptyset, \{\emptyset\}\}, \emptyset\}$. \item $A^{-1} = \{ - \pair{\{\emptyset, \{\emptyset\}\}, \emptyset}, - \pair{\emptyset, \{\emptyset\}} + \tuple{\{\emptyset, \{\emptyset\}\}, \emptyset}, + \tuple{\emptyset, \{\emptyset\}} \}$. \item $A \circ A = - \{\pair{\{\emptyset\}, \{\emptyset, \{\emptyset\}\}}\}$. + \{\tuple{\{\emptyset\}, \{\emptyset, \{\emptyset\}\}}\}$. \item $A \restriction \emptyset = \emptyset$ \item $A \restriction \{\emptyset\} = - \{\pair{\emptyset, \{\emptyset, \{\emptyset\}\}}\}$. + \{\tuple{\emptyset, \{\emptyset, \{\emptyset\}\}}\}$. \item $A \restriction \{\emptyset, \{\emptyset\}\} = A$. \item $\bigcup\bigcup A = \{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\}$. @@ -4318,12 +4318,12 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \nameref{ref:restriction}, intersection, and \nameref{ref:range} of sets, \begin{align*} F \restriction A - & = \{\pair{u, v} \mid uFv \land u \in A\} \\ - & = \{\pair{u, v} \mid + & = \{\tuple{u, v} \mid uFv \land u \in A\} \\ + & = \{\tuple{u, v} \mid uFv \land u \in A \land v \in \ran{F}\} \\ - & = \{\pair{u, v} \mid uFv\} \cap - \{\pair{u, v} \mid u \in A \land v \in \ran{F}\} \\ - & = F \cap \{\pair{u, v} \mid u \in A \land v \in \ran{F}\} \\ + & = \{\tuple{u, v} \mid uFv\} \cap + \{\tuple{u, v} \mid u \in A \land v \in \ran{F}\} \\ + & = F \cap \{\tuple{u, v} \mid u \in A \land v \in \ran{F}\} \\ & = F \cap (A \times \ran{F}). \end{align*} \end{proof} @@ -4341,17 +4341,17 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and By definition of the \nameref{ref:composition} of sets, \begin{align*} (R \circ S) \circ T - & = \{\pair{u, v} \mid + & = \{\tuple{u, v} \mid \exists t(uTt \land t(R \circ S)v)\} \\ - & = \{\pair{u, v} \mid + & = \{\tuple{u, v} \mid \exists t(uTt \land (\exists a(tSa \land aRv))\} \\ - & = \{\pair{u, v} \mid + & = \{\tuple{u, v} \mid \exists t, \exists a, (uTt \land tSa) \land aRv)\} \\ - & = \{\pair{u, v} \mid + & = \{\tuple{u, v} \mid \exists a, \exists t, (uTt \land tSa) \land aRv)\} \\ - & = \{\pair{u, v} \mid + & = \{\tuple{u, v} \mid \exists a, (\exists t(uTt \land tSa)) \land aRv)\} \\ - & = \{\pair{u, v} \mid + & = \{\tuple{u, v} \mid \exists a, u(S \circ T)a \land aRv)\} \\ & = R \circ (S \circ T). \end{align*} @@ -4398,9 +4398,9 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \begin{align*} \img{(F \circ G)}{A} & = \{v \mid (\exists u \in A) u(F \circ G)v\} \\ - & = \{v \mid (\exists u \in A) \pair{u, v} \in F \circ G\} \\ + & = \{v \mid (\exists u \in A) \tuple{u, v} \in F \circ G\} \\ & = \{v \mid (\exists u \in A) - \pair{u, v} \in \{\pair{b, c} \mid + \tuple{u, v} \in \{\tuple{b, c} \mid \exists a(bGa \land aFc)\}\} \\ & = \{v \mid \exists u \in A, \exists a, uGa \land aFv\} \\ & = \{v \mid \exists a, \exists u \in A, uGa \land aFv\} \\ @@ -4415,12 +4415,12 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and By definition of the \nameref{ref:restriction} of a set, \begin{align*} Q \restriction (A \cup B) - & = \{\pair{u, v} \mid uQv \land u \in A \cup B\} \\ - & = \{\pair{u, v} \mid uQv \land (u \in A \lor u \in B)\} \\ - & = \{\pair{u, v} \mid + & = \{\tuple{u, v} \mid uQv \land u \in A \cup B\} \\ + & = \{\tuple{u, v} \mid uQv \land (u \in A \lor u \in B)\} \\ + & = \{\tuple{u, v} \mid (uQv \land u \in A) \lor (uQv \land u \in B)\} \\ - & = \{\pair{u, v} \mid uQv \land u \in A\} \cup - \{\pair{u, v} \mid uQv \land u \in B\} \\ + & = \{\tuple{u, v} \mid uQv \land u \in A\} \cup + \{\tuple{u, v} \mid uQv \land u \in B\} \\ & = (Q \restriction A) \cup (Q \restriction B). \end{align*} @@ -4442,7 +4442,7 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \begin{proof} Let $I_A$ be the identity function on the set $A$. - That is, $I_A = \{\pair{u, u} \mid u \in A\}$. + That is, $I_A = \{\tuple{u, u} \mid u \in A\}$. Let $B$ and $C$ be any sets. We show that (i) $B \circ I_A = B \restriction A$ and (ii) $\img{I_A}{C} = A \cap C$. @@ -4454,30 +4454,30 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \subparagraph{($\subseteq$)}% - Let $\pair{x, y} \in B \circ I_A$. + Let $\tuple{x, y} \in B \circ I_A$. By definition of the \nameref{ref:composition} of sets, there exists some $t$ such that $x(I_A)t$ and $tBy$. By definition of the identity function, $I_A(x) = t$ implies $x = t$. Thus $xBy$. By hypothesis, $x \in \dom{(B \circ I_A)}$. Therefore $x \in \dom{I_A} = A$. - Thus $$\pair{x, y} \in \{\pair{u, v} \mid u \in A \land uBv\} + Thus $$\tuple{x, y} \in \{\tuple{u, v} \mid u \in A \land uBv\} = B \restriction A.$$ \subparagraph{($\supseteq$)}% - Let $\pair{x, y} \in B \restriction A$. + Let $\tuple{x, y} \in B \restriction A$. By definition of the \nameref{ref:restriction} of sets, $x \in A$ and $xBy$. - But $I_A(x) = x$ meaning $\pair{I_A(x), y} \in B$. - In other words, $\pair{x, y} \in B \circ I_A$. + But $I_A(x) = x$ meaning $\tuple{I_A(x), y} \in B$. + In other words, $\tuple{x, y} \in B \circ I_A$. \paragraph{(ii)}% By definition of the \nameref{ref:image} of sets, \begin{align*} \img{I_A}{C} - & = \{v \mid (\exists u \in C) \pair{u, v} \in I_A\} \\ + & = \{v \mid (\exists u \in C) \tuple{u, v} \in I_A\} \\ & = \{v \mid \exists u \in C, u \in A \land u = v\} \\ & = \{v \mid v \in C \land v \in A\} \\ & = C \cap A. @@ -4501,7 +4501,7 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \img{F^{-1}}{A} & = \{x \mid (\exists y \in A) yF^{-1}x\} \\ & = \{x \mid (\exists y \in A) xFy\} \\ - & = \{x \mid (\exists y \in A) \pair{x, y} \in F\} \\ + & = \{x \mid (\exists y \in A) \tuple{x, y} \in F\} \\ & = \{x \mid x \in \dom{F} \land F(x) \in A\} \\ & = \{x \in \dom{F} \mid F(x) \in A\}. \end{align*} @@ -4535,23 +4535,23 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \subparagraph{($\subseteq$)}% - Let $\pair{x, y} \in G \circ G^{-1}$. + Let $\tuple{x, y} \in G \circ G^{-1}$. By definition of the \nameref{ref:composition} of sets, there exists some set $t$ such that $x(G^{-1})t$ and $tGy$. By definition of the \nameref{ref:inverse} of a set, $$x(G^{-1})t \iff tGx.$$ The right hand side of the above biconditional indicates $x \in \ran{G}$. Since $G$ is single-valued, $tGy \land tGx$ implies $x = y$. - Thus $\pair{x, y} \in I_{\ran{G}}$. + Thus $\tuple{x, y} \in I_{\ran{G}}$. \subparagraph{($\supseteq$)}% - Let $\pair{x, x} \in I_{\ran{G}}$ where $x \in \ran{G}$. + Let $\tuple{x, x} \in I_{\ran{G}}$ where $x \in \ran{G}$. By definition of the \nameref{ref:range} of a function, there exists some - $t$ such that $\pair{t, x} \in G$. + $t$ such that $\tuple{t, x} \in G$. By definition of the \nameref{ref:inverse} of a set, it follows - $\pair{x, t} \in G^{-1}$. - Thus $\pair{x, x} \in G \circ G^{-1}$. + $\tuple{x, t} \in G^{-1}$. + Thus $\tuple{x, x} \in G \circ G^{-1}$. \subparagraph{Conclusion}% @@ -4591,7 +4591,7 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \paragraph{($\subseteq$)}% Let $x \in \dom{(F \circ G)}$. - Then there exists a set $y$ such that $\pair{x, y} \in F \circ G$. + Then there exists a set $y$ such that $\tuple{x, y} \in F \circ G$. By definition of the \nameref{ref:composition} of sets, there exists a set $t$ such that $xGt$ and $tFy$. Thus $t \in \dom{F}$. @@ -4613,7 +4613,7 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and such that $uFt$. Thus $xGu \land uFt$. By definition of the \nameref{ref:composition} of sets, - $\pair{x, t} \in F \circ G$. + $\tuple{x, t} \in F \circ G$. Therefore $x \in \dom{(F \circ G)}$. \end{proof} @@ -4803,20 +4803,20 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \paragraph{($\Rightarrow$)}% Suppose $R$ is \nameref{ref:symmetric}. - Let $\pair{x, y} \in R^{-1}$. + Let $\tuple{x, y} \in R^{-1}$. By definition of the \nameref{ref:inverse} of a set, - $\pair{y, x} \in R$. - By symmetry, $\pair{x, y} \in R$. + $\tuple{y, x} \in R$. + By symmetry, $\tuple{x, y} \in R$. Thus $R^{-1} \subseteq R$. \paragraph{($\Leftarrow$)}% Suppose $R^{-1} \subseteq R$. - Let $\pair{x, y} \in R$. + Let $\tuple{x, y} \in R$. By definition of the \nameref{ref:inverse} of a set, - $\pair{y, x} \in R^{-1}$. - Since $R^{-1} \subseteq R$, $\pair{y, x} \in R$. - Therefore $\pair{x, y}$ and $\pair{y, x}$ are both in $R$. + $\tuple{y, x} \in R^{-1}$. + Since $R^{-1} \subseteq R$, $\tuple{y, x} \in R$. + Therefore $\tuple{x, y}$ and $\tuple{y, x}$ are both in $R$. In other words, $R$ is symmetric. \end{proof} @@ -4834,20 +4834,20 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \paragraph{($\Rightarrow$)}% Suppose $R$ is \nameref{ref:transitive}. - Let $\pair{x, y} \in R \circ R$. + Let $\tuple{x, y} \in R \circ R$. By definition of the \nameref{ref:composition} of a set, there exists some $t$ such that $xRt \land tRy$. - That is, $\pair{x, t} \in R$ and $\pair{t, y} \in R$. - Since $R$ is transitive, it follows $\pair{x, y} \in R$. + That is, $\tuple{x, t} \in R$ and $\tuple{t, y} \in R$. + Since $R$ is transitive, it follows $\tuple{x, y} \in R$. \paragraph{($\Leftarrow$)}% Suppose $R \circ R \subseteq R$. - Let $\pair{x, y} \pair{y, z} \in R$. + Let $\tuple{x, y} \tuple{y, z} \in R$. By definition of the \nameref{ref:composition} of a set, - $$R \circ R = \{\pair{u, v} \mid \exists t(uRt \land tRv)\}.$$ - Then $\pair{x, z} \in R \circ R$. - Since $R \circ R \subseteq R$, it follows $\pair{x, z} \in R$. + $$R \circ R = \{\tuple{u, v} \mid \exists t(uRt \land tRv)\}.$$ + Then $\tuple{x, z} \in R \circ R$. + Since $R \circ R \subseteq R$, it follows $\tuple{x, z} \in R$. Thus $R$ is transitive. \end{proof} @@ -4879,21 +4879,21 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \subparagraph{($\subseteq$)}% - Let $\pair{x, y} \in R$. - Since $R$ is symmetric, $\pair{y, x} \in R$. - Since $R$ is transitive, $\pair{x, x} \in R$. - Then there exists a $t$ such that $\pair{x, t} \in R$ and - $\pair{y, t} \in R$, namely $t = x$. + Let $\tuple{x, y} \in R$. + Since $R$ is symmetric, $\tuple{y, x} \in R$. + Since $R$ is transitive, $\tuple{x, x} \in R$. + Then there exists a $t$ such that $\tuple{x, t} \in R$ and + $\tuple{y, t} \in R$, namely $t = x$. By \eqref{sub:exercise-3.33-eq1}, - $\pair{x, y} \in R^{-1} \circ R$. + $\tuple{x, y} \in R^{-1} \circ R$. \subparagraph{($\supseteq$)}% - Let $\pair{x, y} \in R^{-1} \circ R$. + Let $\tuple{x, y} \in R^{-1} \circ R$. By \eqref{sub:exercise-3.33-eq1}, there exists some $t$ such that - $\pair{x, t} \in R$ and $\pair{y, t} \in R$. - But $R$ is symmetric meaning $\pair{t, y} \in R$. - Since $R$ is transitive, it follows $\pair{x, y} \in R$. + $\tuple{x, t} \in R$ and $\tuple{y, t} \in R$. + But $R$ is symmetric meaning $\tuple{t, y} \in R$. + Since $R$ is transitive, it follows $\tuple{x, y} \in R$. \paragraph{($\Leftarrow$)}% @@ -4913,20 +4913,20 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and & \textref{sub:theorem-3e} \nonumber \\ & = R \hyperlabel{sub:exercise-3.33-eq2}. \end{align} - Now let $\pair{x, y} \in R$. - By \eqref{sub:exercise-3.33-eq2} $\pair{x, y} \in R^{-1}$. + Now let $\tuple{x, y} \in R$. + By \eqref{sub:exercise-3.33-eq2} $\tuple{x, y} \in R^{-1}$. By definition of the \nameref{ref:inverse} of a set, - $\pair{y, x} \in R$. + $\tuple{y, x} \in R$. Thus $R$ is symmetric. \subparagraph{(ii)}% - Let $\pair{x, y}, \pair{y, z} \in R$. + Let $\tuple{x, y}, \tuple{y, z} \in R$. By \nameref{spar:exercise-3.33-i}, $R$ is symmetric. - Thus $\pair{z, y} \in R$. + Thus $\tuple{z, y} \in R$. By \eqref{sub:exercise-3.33-eq1}, it follows - $\pair{x, z} \in R^{-1} \circ R$. - Since $R^{-1} \circ R = R$, it follows $\pair{x, z} \in R$. + $\tuple{x, z} \in R^{-1} \circ R$. + Since $R^{-1} \circ R = R$, it follows $\tuple{x, z} \in R$. Thus $R$ is transitive. \end{proof} @@ -4954,12 +4954,12 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and Because $\mathscr{A} \neq \emptyset$, $\bigcap{\mathscr{A}}$ is well-defined. We prove that $\bigcap{\mathscr{A}}$ is a transitive relation. - Let $\pair{x, y}, \pair{y, z} \in \bigcap{\mathscr{A}}$. + Let $\tuple{x, y}, \tuple{y, z} \in \bigcap{\mathscr{A}}$. Then forall $A$ in $\mathscr{A}$, it follows - $\pair{x, y}, \pair{y, z} \in A$. - Since $A$ is transitive, it follows $\pair{x, z} \in A$. + $\tuple{x, y}, \tuple{y, z} \in A$. + Since $A$ is transitive, it follows $\tuple{x, z} \in A$. Since this holds for all $A \in \mathscr{A}$, it follows that - $\pair{x, z} \in A$ as well. + $\tuple{x, z} \in A$ as well. Thus $\bigcap{\mathscr{A}}$ is transitive. \paragraph{(b)}% @@ -4967,14 +4967,14 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and We show that $\bigcup{\mathscr{A}}$ is not necessarily transitive with a counterexample. Suppose $$\mathscr{A} = \{ - \{\pair{1, 2}, \pair{2, 3}, \pair{1, 3}\}, \{\pair{2, 1}\} + \{\tuple{1, 2}, \tuple{2, 3}, \tuple{1, 3}\}, \{\tuple{2, 1}\} \}.$$ Notice that the two members of $\mathscr{A}$ are transitive relations. Now $$\bigcup{\mathscr{A}} = \{ - \pair{1, 2}, \pair{2, 3}, \pair{1, 3}, \pair{2, 1}, + \tuple{1, 2}, \tuple{2, 3}, \tuple{1, 3}, \tuple{2, 1}, \}.$$ - But the above cannot be transitive, for $\pair{1, 2}$ and $\pair{2, 1}$ are - members of the set, but $\pair{1, 1}$ is not. + But the above cannot be transitive, for $\tuple{1, 2}$ and $\tuple{2, 1}$ + are members of the set, but $\tuple{1, 1}$ is not. \end{proof} @@ -5005,7 +5005,7 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and Define $Q$ to be the set \begin{equation} \hyperlabel{sub:exercise-3.36-eq1} - \{\pair{x, y} \in A \times A \mid \pair{f(x), f(y)} \in R\}. + \{\tuple{x, y} \in A \times A \mid \tuple{f(x), f(y)} \in R\}. \end{equation} Show that $Q$ is an equivalence relation on $A$. @@ -5021,27 +5021,27 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and Let $x \in A$. By hypothesis, $f(x) \in B$. Since $R$ is an equivalence relation on $B$, $R$ is reflexive on $B$. - Thus $\pair{f(x), f(x)} \in R$. - But then \eqref{sub:exercise-3.36-eq1} implies $\pair{x, x} \in Q$. + Thus $\tuple{f(x), f(x)} \in R$. + But then \eqref{sub:exercise-3.36-eq1} implies $\tuple{x, x} \in Q$. Thus $Q$ is reflexive on $A$. \paragraph{(ii)}% - Let $\pair{x, y} \in Q$. - By \eqref{sub:exercise-3.36-eq1}, $\pair{f(x), f(y)} \in R$. + Let $\tuple{x, y} \in Q$. + By \eqref{sub:exercise-3.36-eq1}, $\tuple{f(x), f(y)} \in R$. Since $R$ is an equivalence relation on $B$, $R$ is symmetric. - Thus $\pair{f(y), f(x)} \in R$. - But then \eqref{sub:exercise-3.36-eq1} implies $\pair{y, x} \in Q$. + Thus $\tuple{f(y), f(x)} \in R$. + But then \eqref{sub:exercise-3.36-eq1} implies $\tuple{y, x} \in Q$. Thus $Q$ is symmetric. \paragraph{(iii)}% - Let $\pair{x, y}, \pair{y, z} \in Q$. + Let $\tuple{x, y}, \tuple{y, z} \in Q$. By \eqref{sub:exercise-3.36-eq1}, - $\pair{f(x), f(y)}, \pair{f(y), f(z)} \in R$. + $\tuple{f(x), f(y)}, \tuple{f(y), f(z)} \in R$. Since $R$ is an equivalence relation on $B$, $R$ is transitive. - Thus $\pair{f(x), f(z)} \in R$. - But then \eqref{sub:exercise-3.36-eq1} implies $\pair{x, z} \in Q$. + Thus $\tuple{f(x), f(z)} \in R$. + But then \eqref{sub:exercise-3.36-eq1} implies $\tuple{x, z} \in Q$. Thus $Q$ is transitive. \end{proof} @@ -5072,28 +5072,28 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and By definition of a \nameref{ref:partition}, there exists some nonempty set $B \in \Pi$ such that $x \in B$. Thus $(\exists B \in \Pi)(x \in B \land x \in B)$. - By \eqref{sub:exercise-3.37-eq1}, $\pair{x, x} \in R_\Pi$. + By \eqref{sub:exercise-3.37-eq1}, $\tuple{x, x} \in R_\Pi$. Therefore $R_\Pi$ is reflexive on $A$. \paragraph{(ii)}% - Let $\pair{x, y} \in R_\Pi$. + Let $\tuple{x, y} \in R_\Pi$. By \eqref{sub:exercise-3.37-eq1}, there exists some $B \in \Pi$ such that $x \in B \land y \in B$. But then $y \in B \land x \in B$. - Thus $\pair{y, x} \in R_\Pi$. + Thus $\tuple{y, x} \in R_\Pi$. In other words, $R_\Pi$ is symmetric. \paragraph{(iii)}% - Let $\pair{x, y}, \pair{y, z} \in R_\Pi$. + Let $\tuple{x, y}, \tuple{y, z} \in R_\Pi$. By \eqref{sub:exercise-3.37-eq1}, there exists some $B_1 \in \Pi$ such that $x \in B_1 \land y \in B_1$. Likewise there exists some $B_2 \in \Pi$ such that $y \in B_2 \land z \in B_2$. But $\Pi$ is a \nameref{ref:partition} meaning $y \in B_1$ and $y \in B_2$ if $B_1 = B_2$. - Therefore $x \in B_1 \land z \in B_1$ and $\pair{x, z} \in R_\Pi$. + Therefore $x \in B_1 \land z \in B_1$ and $\tuple{x, z} \in R_\Pi$. In other words, $R_\Pi$ is transitive. \end{proof} @@ -5142,7 +5142,7 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \begin{align*} y & \in B = [x]_{R_\Pi} \\ - & = \{t \mid \pair{x, t} \in R_\Pi\} \\ + & = \{t \mid \tuple{x, t} \in R_\Pi\} \\ & = \{t \mid (\exists B_1 \in \Pi)(x \in B_1 \land t \in B_1)\}. & \eqref{sub:exercise-3.38-eq1} \end{align*} @@ -5162,7 +5162,7 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \begin{align*} y & \in \{t \mid (\exists B_1 \in \Pi)(x \in B_1 \land t \in B_1)\} \\ - & = \{t \mid \pair{x, t} \in R_\Pi\} \\ + & = \{t \mid \tuple{x, t} \in R_\Pi\} \\ & = [x]_{R_\Pi} = B. \end{align*} @@ -5183,7 +5183,7 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \begin{align*} B & = \{t \mid (\exists B_1 \in \Pi)(x \in B_1 \land t \in B_1)\} \\ - & = \{ t \mid \pair{x, t} \in R_\Pi \} \\ + & = \{ t \mid \tuple{x, t} \in R_\Pi \} \\ & = [x]_{R_\Pi}. \end{align*} Therefore $B \in A / R_{\Pi}$. @@ -5260,7 +5260,7 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \hyperlabel{sub:exercise-3.41} Let $\mathbb{R}$ be the set of real numbers and define the relation $Q$ on - $\mathbb{R} \times \mathbb{R}$ by $\pair{u, v}Q\pair{x, y}$ iff + $\mathbb{R} \times \mathbb{R}$ by $\tuple{u, v}Q\tuple{x, y}$ iff $u + y = x + v$. \subsubsection{\verified{Exercise 3.41a}}% @@ -5278,27 +5278,27 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \paragraph{(i)}% - Let $\pair{x, y} \in R \times R$. - Since $x + y = x + y$, it immediately follows $\pair{x, y}Q\pair{x, y}$. + Let $\tuple{x, y} \in R \times R$. + Since $x + y = x + y$, it immediately follows $\tuple{x, y}Q\tuple{x, y}$. Thus $Q$ is reflexive on $\mathbb{R}$. \paragraph{(ii)}% - Let $\pair{\pair{u, v}, \pair{x, y}} \in Q$. + Let $\tuple{\tuple{u, v}, \tuple{x, y}} \in Q$. Then $u + y = x + v$. Likewise, $x + v = u + y$. - This immediately implies that $\pair{\pair{x, y}, \pair{u, v}} \in Q$. + This immediately implies that $\tuple{\tuple{x, y}, \tuple{u, v}} \in Q$. Thus $Q$ is symmetric. \paragraph{(iii)}% - Let $\pair{\pair{u, v}, \pair{x, y}} \in Q$ and - $\pair{\pair{x, y}, \pair{a, b}} \in Q$. + Let $\tuple{\tuple{u, v}, \tuple{x, y}} \in Q$ and + $\tuple{\tuple{x, y}, \tuple{a, b}} \in Q$. Then $u + y = x + v$ and $x + b = a + y$. Rearranging terms, we have $u - v = x - y$ and $x - y = a - b$. Thus $u - v = a - b$. Rearranging terms once more yields $u + b = a + v$. - Thus $\pair{\pair{u, v}, \pair{a, b}} \in Q$. + Thus $\tuple{\tuple{u, v}, \tuple{a, b}} \in Q$. Therefore $Q$ is transitive. \end{proof} @@ -5310,21 +5310,21 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \rightarrow (\mathbb{R} \times \mathbb{R}) / Q$ satisfying the equation \begin{equation} \hyperlabel{ssub:exercise-3.41-b-eq1} - G([\pair{x, y}]_Q) = [\pair{x + 2y, y + 2x}]_Q? + G([\tuple{x, y}]_Q) = [\tuple{x + 2y, y + 2x}]_Q? \end{equation} \begin{proof} Let $f \colon \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R} \times \mathbb{R}$ be given by - $f(\pair{x, y}) = \pair{x + 2y, y + 2x}$. + $f(\tuple{x, y}) = \tuple{x + 2y, y + 2x}$. We show (i) that $f$ is \nameref{ref:compatible} with $Q$ and then (ii) there exists such a function satisfying \eqref{ssub:exercise-3.41-b-eq1}. \paragraph{(i)}% \hyperlabel{par:exercise-3.41-b-i} - Let $\pair{u, v}, \pair{x, y} \in \mathbb{R} \times \mathbb{R}$ such that - $\pair{u, v} Q \pair{x, y}$. + Let $\tuple{u, v}, \tuple{x, y} \in \mathbb{R} \times \mathbb{R}$ such + that $\tuple{u, v} Q \tuple{x, y}$. Thus \begin{equation} \hyperlabel{ssub:exercise-3.41-b-eq2} @@ -5332,8 +5332,8 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \end{equation} Next consider \begin{align*} - f(\pair{u, v}) & = \pair{u + 2v, v + 2u}, \\ - f(\pair{x, y}) & = \pair{x + 2y, y + 2x}. + f(\tuple{u, v}) & = \tuple{u + 2v, v + 2u}, \\ + f(\tuple{x, y}) & = \tuple{x + 2y, y + 2x}. \end{align*} Then \begin{align*} @@ -5345,7 +5345,7 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \iff (x + 2y) + (v + 2u) & = (u + 2v) + (y + 2x) \\ \iff (u + 2v) + (y + 2x) & = (x + 2y) + (v + 2u). \end{align*} - This last equality shows $f(\pair{u, v}) \,Q\, f(\pair{x, y})$. + This last equality shows $f(\tuple{u, v}) \,Q\, f(\tuple{x, y})$. Thus $f$ is compatible with $Q$. \paragraph{(ii)}% @@ -5385,11 +5385,11 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \paragraph{(i)}% \hyperlabel{par:exercise-3.43-i} - Let $\pair{x, y}, \pair{y, z} \in R^{-1}$. + Let $\tuple{x, y}, \tuple{y, z} \in R^{-1}$. By definition of the \nameref{ref:inverse} of a set, - $\pair{y, x}$, $\pair{z, y} \in R$. - Since $R$ is transitive, it must be that $\pair{z, x} \in R$. - Then $\pair{x, z} \in R^{-1}$. + $\tuple{y, x}$, $\tuple{z, y} \in R$. + Since $R$ is transitive, it must be that $\tuple{z, x} \in R$. + Then $\tuple{x, z} \in R^{-1}$. Thus $R^{-1}$ is transitive. \paragraph{(ii)}% @@ -5507,7 +5507,7 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and Assume that $<_A$ and $<_B$ are linear orderings on $A$ and $B$, respectively. Define the binary relation $<_L$ on the Cartesian product $A \times B$ by: - $$\pair{a_1, b_1} <_L \pair{a_2, b_2} \quad\text{iff}\quad + $$\tuple{a_1, b_1} <_L \tuple{a_2, b_2} \quad\text{iff}\quad \text{either } a_1 <_A a_2 \text{ or } (a_1 = a_2 \land b_1 <_B b_2).$$ Show that $<_L$ is a linear ordering on $A \times B$. (The relation $<_L$ is called \textit{lexicographic} ordering, being the @@ -5523,8 +5523,8 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \paragraph{(i)}% - Let $\pair{a_1, b_1} <_L \pair{a_2, b_2}$ and - $\pair{a_2, b_2} <_L \pair{a_3, b_3}$. + Let $\tuple{a_1, b_1} <_L \tuple{a_2, b_2}$ and + $\tuple{a_2, b_2} <_L \tuple{a_3, b_3}$. Then either $a_1 <_A a_2$ or $a_1 = a_2 \land b_1 <_B b_2$. Likewise, either $a_2 <_A a_3$ or $a_2 = a_3 \land b_2 <_B b_3$. We consider each combination of cases in turn: @@ -5534,19 +5534,19 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and Suppose $a_1 <_A a_2$ and $a_2 <_A a_3$. Since $<_A$ is a linear ordering, it follows $<_A$ is transitive. Thus $a_1 <_A a_3$. - Therefore $\pair{a_1, b_2} <_L \pair{a_3, b_3}$. + Therefore $\tuple{a_1, b_2} <_L \tuple{a_3, b_3}$. \subparagraph{Case 2}% Suppose $a_1 <_A a_2$, $a_2 = a_3$, and $b_2 <_B b_3$. Then $a_1 < a_3$. - Therefore $\pair{a_1, b_2} <_L \pair{a_3, b_3}$. + Therefore $\tuple{a_1, b_2} <_L \tuple{a_3, b_3}$. \subparagraph{Case 3}% Suppose $a_1 = a_2$, $b_1 <_B b_2$, and $a_2 <_A a_3$. Then $a_1 <_A a_3$. - Therefore $\pair{a_1, b_2} <_L \pair{a_3, b_3}$. + Therefore $\tuple{a_1, b_2} <_L \tuple{a_3, b_3}$. \subparagraph{Case 4}% @@ -5554,7 +5554,7 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and Then $a_1 = a_3$. Since $<_B$ is a linear ordering, it follows $<_B$ is transitive. Thus $b_1 <_B b_3$. - Therefore $\pair{a_1, b_2} <_L \pair{a_3, b_3}$. + Therefore $\tuple{a_1, b_2} <_L \tuple{a_3, b_3}$. \subparagraph{Conclusion}% @@ -5563,7 +5563,7 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \paragraph{(ii)}% - Let $\pair{a_1, b_1}, \pair{a_2, b_2} \in A \times B$. + Let $\tuple{a_1, b_1}, \tuple{a_2, b_2} \in A \times B$. Because $<_A$ and $<_B$ are linear orderings on $A$ and $B$ respectively, it follows $<_A$ and $<_B$ are both trichotomous on their respective sets. @@ -5575,19 +5575,19 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \hyperlabel{spar:exercise-3.45-ii-case-1} Suppose $a_1 <_A a_2$. - Then $\pair{a_1, b_1} <_L \pair{a_2, b_2}$. + Then $\tuple{a_1, b_1} <_L \tuple{a_2, b_2}$. This is trivially the only possible relationship between the ordered pairs. \subparagraph{Case 2}% Suppose $a_1 = a_2$. - If $b_1 <_B b_2$, then $\pair{a_1, b_1} <_L \pair{a_2, b_2}$ is the only - possibility. - If $b_1 = b_2$, then $\pair{a_1, b_1} = \pair{a_2, b_2}$ is the only - possibility. - If $b_2 <_B b_1$, then $\pair{a_2, b_2} <_L \pair{a_1, b_1}$ is the only + If $b_1 <_B b_2$, then $\tuple{a_1, b_1} <_L \tuple{a_2, b_2}$ is the + only possibility. + If $b_1 = b_2$, then $\tuple{a_1, b_1} = \tuple{a_2, b_2}$ is the only possibility. + If $b_2 <_B b_1$, then $\tuple{a_2, b_2} <_L \tuple{a_1, b_1}$ is the + only possibility. \subparagraph{Case 3}% @@ -5597,9 +5597,9 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \subparagraph{Conclusion}% In each of the above cases, we are always left with exactly one of - $$\pair{a_1, b_1} <_L \pair{a_2, b_2}, \quad - \pair{a_1, b_1} = \pair{a_2, b_2}, \quad - \pair{a_2, b_2} <_L \pair{a_1, b_1}.$$ + $$\tuple{a_1, b_1} <_L \tuple{a_2, b_2}, \quad + \tuple{a_1, b_1} = \tuple{a_2, b_2}, \quad + \tuple{a_2, b_2} <_L \tuple{a_1, b_1}.$$ Thus $<_L$ is trichotomous. \end{proof} @@ -5794,7 +5794,7 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \begin{proof} - Note $\sigma$ is defined as $\sigma = \{\pair{n, n^+} \mid n \in \omega\}$. + Note $\sigma$ is defined as $\sigma = \{\tuple{n, n^+} \mid n \in \omega\}$. To prove $\langle \omega, \sigma, 0 \rangle$ is a \nameref{ref:peano-system}, we must show that (i) $0 \not\in \ran{S}$, (ii) $\sigma$ is one-to-one, and (iii) every subset $A$ of $\omega$ @@ -5949,8 +5949,8 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \subparagraph{(c)}% - Note $\{\pair{0, a}\}$ is an acceptable function. - Thus $\pair{0, a} \in h$. + Note $\{\tuple{0, a}\}$ is an acceptable function. + Thus $\tuple{0, a} \in h$. By \nameref{par:recursion-theorem-natural-numbers-i}, $h$ is a function. Therefore $a$ is the only value $h(0)$ takes on. @@ -5959,7 +5959,7 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and Suppose $n^+ \in \dom{h}$. Then there exists some acceptable function $v$ such that $v(n^+) = h(n^+)$. - By definition of acceptable, $\pair{n, v(n)} \in v$. + By definition of acceptable, $\tuple{n, v(n)} \in v$. Since \nameref{par:recursion-theorem-natural-numbers-i} indicates $h$ is a function, $n \in \dom{h}$ and $h(n) = v(n)$. Also by definition of acceptable, $v(n^+) = F(v(n))$. @@ -5976,7 +5976,7 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \subparagraph{(1)}% \hyperlabel{spar:recursion-theorem-natural-numbers-iii-1} - We note that $\{\pair{0, a}\}$ is an acceptable function. + We note that $\{\tuple{0, a}\}$ is an acceptable function. By construction of $h$, $0 \in \dom{h}$. \subparagraph{(2)}% @@ -5985,7 +5985,7 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and Suppose $n \in \dom{h}$. Since $n \in \dom{h}$ there exists an acceptable function $v$ with $n \in \dom{v}$. - Define $$v' = v \cup \{\pair{n^+, F(v(n))}\}.$$ + Define $$v' = v \cup \{\tuple{n^+, F(v(n))}\}.$$ We prove that $v'$ is acceptable: \begin{enumerate}[(a)] @@ -6987,7 +6987,7 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and Hence \nameref{sub:theorem-4b} implies $S = \omega$. Thus for all $m, n \in \omega$, $$m \in n \lor m = n \lor n \in m.$$ We now prove that - $$\in_\omega = \{\pair{m, n} \in \omega \times \omega \mid m \in n\}$$ + $$\in_\omega = \{\tuple{m, n} \in \omega \times \omega \mid m \in n\}$$ is \nameref{ref:irreflexive} and \nameref{ref:connected}. Irreflexivity immediately follows from \nameref{sub:lemma-4l-b}. Connectivity follows immediately from the fact $S = \omega$. @@ -7007,7 +7007,7 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and Relation \begin{equation} \hyperlabel{sub:linear-ordering-natural-numbers-eq1} - \in_\omega = \{\pair{m, n} \in \omega \times \omega \mid m \in n\} + \in_\omega = \{\tuple{m, n} \in \omega \times \omega \mid m \in n\} \end{equation} is a linear ordering on $\omega$. \end{theorem} @@ -7952,10 +7952,10 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \approx B$) if and \begin{proof} By definition, - $$\in_\omega = \{\pair{m, n} \in \omega \times \omega \mid m \in n\}.$$ + $$\in_\omega = \{\tuple{m, n} \in \omega \times \omega \mid m \in n\}.$$ Thus, the \nameref{ref:inverse} of $\in_\omega$ is $$\in_\omega^{-1} = - \{\pair{n, m} \in \omega \times \omega \mid m \in n\}.$$ + \{\tuple{n, m} \in \omega \times \omega \mid m \in n\}.$$ Therefore $$\img{\in_\omega^{-1}}{\{7, 8\}} = \{6, 7\}.$$ \end{proof} diff --git a/preamble.tex b/preamble.tex index 579cd5a..680495b 100644 --- a/preamble.tex +++ b/preamble.tex @@ -92,6 +92,22 @@ % Admonitions % ======================================== +\newcommand\@assumptionbody[1]{ + \begin{equation} + \setlength{\abovedisplayskip}{0pt} + \setlength{\belowdisplayskip}{0pt} + #1 + \end{equation}} + +\NewEnviron{assumption}[1][]{% + \ifstrempty{#1}{% + \begin{tcolorbox}[bottom=8pt] + \@assumptionbody{\BODY} + \end{tcolorbox}}{% + \begin{tcolorbox}[title=#1,bottom=8pt] + \@assumptionbody{\BODY} + \end{tcolorbox}}} + \NewEnviron{note}{% \begin{tcolorbox}[% sharp corners, @@ -156,6 +172,7 @@ \newcommand{\abs}[1]{\left|#1\right|} \newcommand{\ceil}[1]{\left\lceil#1\right\rceil} +\newcommand{\ctuple}[2]{\left< #1, \cdots, #2 \right>} \newcommand{\dom}[1]{\textop{dom}{#1}} \newcommand{\fld}[1]{\textop{fld}{#1}} \newcommand{\floor}[1]{\left\lfloor#1\right\rfloor} @@ -164,9 +181,12 @@ \newcommand{\img}[2]{#1\!\left\llbracket#2\right\rrbracket} \newcommand{\ioc}[2]{\left(#1, #2\right]} \newcommand{\ioo}[2]{\left(#1, #2\right)} +\newcommand{\ltuple}[2]{\left< #1, \ldots, #2 \right>} \newcommand{\powerset}[1]{\mathscr{P}#1} \newcommand{\ran}[1]{\textop{ran}{#1}} \newcommand{\textop}[1]{\mathop{\text{#1}}} +\newcommand{\tuple}[1]{\left< #1 \right>} + \newcommand{\ubar}[1]{\text{\b{$#1$}}} \let\oldemptyset\emptyset