From f560006ece61e326885c08e580d32751636acb6d Mon Sep 17 00:00:00 2001 From: Joshua Potter Date: Sat, 20 May 2023 11:00:07 -0600 Subject: [PATCH] Add definitions/prompts for Enderton "Axioms and Operations." --- Bookshelf/Enderton/Set.tex | 235 ++++++++++++++++++++++++++++++++++--- preamble.tex | 2 +- 2 files changed, 222 insertions(+), 15 deletions(-) diff --git a/Bookshelf/Enderton/Set.tex b/Bookshelf/Enderton/Set.tex index 2beee50..336862c 100644 --- a/Bookshelf/Enderton/Set.tex +++ b/Bookshelf/Enderton/Set.tex @@ -21,14 +21,14 @@ \label{ref:empty-set-axiom} There is a set having no members: - $$\exists\; B, \forall\; x, x \not\in B.$$ + $$\exists B, \forall x, x \not\in B.$$ \section{\defined{Extensionality Axiom}}% \label{ref:extensionality-axiom} If two sets have exactly the same members, then they are equal: - $$\forall\; A, \forall\; B, - \left[\forall\; x, (x \in A \iff x \in B) \Rightarrow A = B\right].$$ + $$\forall A, \forall B, + \left[\forall x, (x \in A \iff x \in B) \Rightarrow A = B\right].$$ \begin{axiom} @@ -36,17 +36,24 @@ If two sets have exactly the same members, then they are equal: \end{axiom} +\section{\partial{Pair Set}}% +\label{ref:pair-set} + +For any sets $u$ and $v$, the \textbf{pair set $\{u, v\}$} is the set whose + only members are $u$ and $v$. + \section{\partial{Pairing Axiom}}% \label{ref:pairing-axiom} For any sets $u$ and $v$, there is a set having as members just $u$ and $v$: - $$\forall\; u, \forall\; v, \exists\; B, \forall\; x, + $$\forall u, \forall v, \exists B, \forall x, (x \in B \iff x = u \text{ or } x = v).$$ -\section{\defined{Powerset}}% -\label{ref:powerset} +\section{\defined{Power Set}}% +\label{ref:power-set} -The \textbf{powerset} of some set $A$ is the set of all subsets of $A$. +For any set $a$, the \textbf{power set $\powerset{a}$} is the set whose members + are exactly the subsets of $a$. \begin{definition} @@ -58,14 +65,28 @@ The \textbf{powerset} of some set $A$ is the set of all subsets of $A$. \label{ref:power-set-axiom} For any set $a$, there is a set whose members are exactly the subsets of $a$: - $$\forall\; a, \exists\; B, \forall\; x, (x \in B \iff x \subseteq a).$$ + $$\forall a, \exists B, \forall x, (x \in B \iff x \subseteq a).$$ + +\section{\partial{Subset Axioms}}% +\label{ref:subset-axioms} + +For each formula $\phi$ not containing $B$, the following is an axiom: + $$\forall t_1, \cdots \forall t_k, \forall c, + \exists B, \forall x, (x \in B \iff x \in c \land \phi).$$ + +\section{\partial{Union Axiom}}% +\label{ref:union-axiom} + +For any set $A$, there exists a set $B$ whose elements are exactly the members + of the members of $A$: + $$\forall A, \exists B, \forall x \left[ x \in B \iff (\exists b \in A) x \in b \right]$$ \section{\partial{Union Axiom, Preliminary Form}}% \label{ref:union-axiom-preliminary-form} For any sets $a$ and $b$, there is a set whose members are those sets belonging either to $a$ or to $b$ (or both): - $$\forall\; a, \forall\; b, \exists\; B, \forall\; x, + $$\forall a, \forall b, \exists B, \forall x, (x \in B \iff x \in a \text{ or } x \in b).$$ \endgroup @@ -209,10 +230,10 @@ Show that if $B \subseteq C$, then $\powerset{B} \subseteq \powerset{C}$. {Enderton.Set.Chapter\_1.exercise\_1\_3} Let $x \in \powerset{B}$. - By definition of the \nameref{ref:powerset}, $x$ is a subset of $B$. + By definition of the \nameref{ref:power-set}, $x$ is a subset of $B$. By hypothesis, $B \subseteq C$. Then $x \subseteq C$. - Again by definition of the \nameref{ref:powerset}, it follows + Again by definition of the \nameref{ref:power-set}, it follows $x \in \powerset{C}$. \end{proof} @@ -230,11 +251,11 @@ Show that $\{\{x\}, \{x, y\}\} \in \powerset{\powerset{B}}.$ Let $x$ and $y$ be members of set $B$. Then $\{x\}$ and $\{x, y\}$ are subsets of $B$. - By definition of the \nameref{ref:powerset}, $\{x\}$ and $\{x, y\}$ are + By definition of the \nameref{ref:power-set}, $\{x\}$ and $\{x, y\}$ are members of $\powerset{B}$. Then $\{\{x\}, \{x, y\}\}$ is a subset of $\powerset{B}$. - By definition of the \nameref{ref:powerset}, $\{\{x\}, \{x, y\}\}$ is a member - of $\powerset{\powerset{B}}$. + By definition of the \nameref{ref:power-set}, $\{\{x\}, \{x, y\}\}$ is a + member of $\powerset{\powerset{B}}$. \end{proof} @@ -371,4 +392,190 @@ List all the members of $V_4$. \end{proof} +\chapter{Axioms and Operations}% +\label{chap:axioms-operations} + +\section{Axioms}% +\label{sec:axioms} + +\subsection{\unverified{Theorem 2A}}% +\label{sub:theorem-2a} + +\begin{theorem}[2A] + + There is no set to which every set belongs. + +\end{theorem} + +\begin{proof} + + TODO + +\end{proof} + +\subsection{\unverified{Theorem 2B}}% +\label{sub:theorem-2b} + +\begin{theorem}[2B] + + For any nonempty set $A$, there exists a unique set $B$ such that for any + $x$, $$x \in B \iff x \text{ belongs to every member of } A.$$ + +\end{theorem} + +\begin{proof} + + TODO + +\end{proof} + +\section{Exercises 3}% +\label{sec:exercises-3} + +\subsection{\unverified{Exercise 3.1}}% +\label{sub:exercise-3.1} + +Assume that $A$ is the set of integers divisible by $4$. +Similarly assume that $B$ and $C$ are the sets of integers divisible by $9$ and + $10$, respectively. +What is in $A \cap B \cap C$? + +\begin{proof} + + TODO + +\end{proof} + +\subsection{\unverified{Exercise 3.2}}% +\label{sub:exercise-3.2} + +Give an example of sets $A$ and $B$ for which $\bigcup A = \bigcup B$ but + $A \neq B$. + +\begin{proof} + + TODO + +\end{proof} + +\subsection{\unverified{Exercise 3.3}}% +\label{sub:exercise-3.3} + +Show that every member of a set $A$ is a subset of $\bigcup A$. +(This was stated as an example in this section.) + +\begin{proof} + + TODO + +\end{proof} + +\subsection{\unverified{Exercise 3.4}}% +\label{sub:exercise-3.4} + +Show that if $A \subseteq B$, then $\bigcup A \subseteq \bigcup B$. + +\begin{proof} + + TODO + +\end{proof} + +\subsection{\unverified{Exercise 3.5}}% +\label{sub:exercise-3.5} + +Assume that every member of $\mathscr{A}$ is a subset of $B$. +Show that $\bigcup \mathscr{A} \subseteq B$. + +\begin{proof} + + TODO + +\end{proof} + +\subsection{\unverified{Exercise 3.6a}}% +\label{sub:exercise-3.6a} + +Show that for any set $A$, $\bigcup \powerset{A} = A$. + +\begin{proof} + + TODO + +\end{proof} + +\subsection{\unverified{Exercise 3.6b}}% +\label{sub:exercise-3.6b} + +Show that $A \subseteq \powerset{\bigcup A}$. +Under what conditions does equality hold? + +\begin{proof} + + TODO + +\end{proof} + +\subsection{\unverified{Exercise 3.7a}}% +\label{sub:exercise-3.7a} + +Show that for any sets $A$ and $B$, + $$\powerset{A} \cap \powerset{B} = \powerset(A \cap B).$$ + +\begin{proof} + + TODO + +\end{proof} + +\subsection{\unverified{Exercise 3.7b}}% +\label{sub:exercise-3.7b} + +Show that $\powerset{A} \cup \powerset{B} \subseteq \powerset(A \cup B)$. +Under what conditions does equality hold? + +\begin{proof} + + TODO + +\end{proof} + +\subsection{\unverified{Exercise 3.8}}% +\label{sub:exercise-3.8} + +Show that there is no set to which every singleton (that is, every set of the + form $\{x\}$) belongs. +[\textit{Suggestion}: Show that from such a set, we could construct a set to + which every set belonged.] + +\begin{proof} + + TODO + +\end{proof} + +\subsection{\unverified{Exercise 3.9}}% +\label{sub:exercise-3.9} + +Give an example of sets $a$ and $B$ for which $a \in B$ but + $\powerset{A} \not\in \powerset{B}$. + +\begin{proof} + + TODO + +\end{proof} + +\subsection{\unverified{Exercise 3.10}}% +\label{sub:exercise-3.10} + +Show that if $a \in B$, then $\powerset{a} \in \powerset{\powerset{\bigcup B}}$. +[\textit{Suggestion}: If you need help, look in the Appendix.] + +\begin{proof} + + TODO + +\end{proof} + \end{document} diff --git a/preamble.tex b/preamble.tex index e18e67b..368cd8e 100644 --- a/preamble.tex +++ b/preamble.tex @@ -131,7 +131,7 @@ \newcommand{\ico}[2]{\left[#1, #2\right)} \newcommand{\ioc}[2]{\left(#1, #2\right]} \newcommand{\ioo}[2]{\left(#1, #2\right)} -\newcommand{\powerset}[1]{\mathscr{P}\;#1} +\newcommand{\powerset}[1]{\mathscr{P}\,#1} \newcommand{\ubar}[1]{\text{\b{$#1$}}} \let\oldemptyset\emptyset