191 lines
4.9 KiB
TeX
191 lines
4.9 KiB
TeX
\documentclass{report}
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\input{../../preamble}
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\makeleancommands{../..}
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\begin{document}
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\header{Elements of Set Theory}{Herbert B. Enderton}
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\tableofcontents
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\begingroup
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\renewcommand\thechapter{R}
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\setcounter{chapter}{0}
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\addtocounter{chapter}{-1}
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\chapter{Reference}%
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\label{chap:reference}
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\section{\defined{Powerset}}%
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\label{ref:powerset}
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The \textbf{powerset} of some set $A$ is the set of all subsets of $A$.
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\begin{definition}
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\lean{Mathlib/Init/Set}{Set.powerset}
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\end{definition}
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\section{\defined{Principle of Extensionality}}%
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\label{ref:principle-extensionality}
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If $A$ and $B$ are sets such that for every object $t$,
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$$t \in A \quad\text{iff}\quad t \in B,$$
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then $A = B$.
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\begin{axiom}
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\lean{Mathlib/Init/Set}{Set.ext}
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\end{axiom}
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\endgroup
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\chapter{Introduction}%
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\label{chap:introduction}
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\section{Baby Set Theory}%
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\label{sec:baby-set-theory}
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\subsection{\partial{Exercise 1}}%
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\label{sub:baby-set-theory-1}
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Which of the following become true when "$\in$" is inserted in place of the
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blank?
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Which become true when "$\subseteq$" is inserted?
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\subsubsection{\partial{Exercise 1a}}%
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\label{ssub:baby-set-theory-1a}
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$\{\emptyset\} \_\_\_\_ \{\emptyset, \{\emptyset\}\}$.
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\begin{proof}
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Because the \textit{object} $\{\emptyset\}$ is a member of the right-hand set,
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the statement is \textbf{true} in the case of "$\in$".
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Because the \textit{members} of $\{\emptyset\}$ are all members of the
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right-hand set, the statement is also \textbf{true} in the case of
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"$\subseteq$".
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\end{proof}
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\subsubsection{\partial{Exercise 1b}}%
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\label{ssub:baby-set-theory-1b}
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$\{\emptyset\} \_\_\_\_ \{\emptyset, \{\{\emptyset\}\}\}$.
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\begin{proof}
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Because the \textit{object} $\{\emptyset\}$ is not a member of the right-hand
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set, the statement is \textbf{false} in the case of "$\in$".
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Because the \textit{members} of $\{\emptyset\}$ are all members of the
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right-hand set, the statement is \textbf{true} in the case of "$\subseteq$".
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\end{proof}
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\subsubsection{\partial{Exercise 1c}}%
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\label{ssub:baby-set-theory-1c}
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$\{\{\emptyset\}\} \_\_\_\_ \{\emptyset, \{\emptyset\}\}$.
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\begin{proof}
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Because the \textit{object} $\{\{\emptyset\}\}$ is not a member of the
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right-hand set, the statement is \textbf{false} in the case of "$\in$".
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Because the \textit{members} of $\{\{\emptyset\}\}$ are all members of the
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right-hand set, the statement is \textbf{true} in the case of "$\subseteq$".
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\end{proof}
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\subsubsection{\partial{Exercise 1d}}%
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\label{ssub:baby-set-theory-1d}
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$\{\{\emptyset\}\} \_\_\_\_ \{\emptyset, \{\{\emptyset\}\}\}$.
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\begin{proof}
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Because the \textit{object} $\{\{\emptyset\}\}$ is a member of the right-hand
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set, the statement is \textbf{true} in the case of "$\in$".
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Because the \textit{members} of $\{\{\emptyset\}\}$ are not all members of the
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right-hand set, the statement is \textbf{false} in the case of
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"$\subseteq$".
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\end{proof}
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\subsubsection{\partial{Exercise 1e}}%
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\label{ssub:baby-set-theory-1e}
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$\{\{\emptyset\}\} \_\_ \{\emptyset, \{\emptyset, \{\emptyset\}\}\}$.
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\begin{proof}
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Because the \textit{object} $\{\{\emptyset\}\}$ is not a member of the
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right-hand set, the statement is \textbf{false} in the case of "$\in$".
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Because the \textit{members} of $\{\{\emptyset\}\}$ are not all members of the
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right-hand set, the statement is \textbf{false} in the case of
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"$\subseteq$".
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\end{proof}
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\subsection{\partial{Exercise 2}}%
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\label{sub:baby-set-theory-2}
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Show that no two of the three sets $\emptyset$, $\{\emptyset\}$, and
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$\{\{\emptyset\}\}$ are equal to each other.
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\begin{proof}
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By the \nameref{ref:principle-extensionality}, $\emptyset$ is only equal to
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$\emptyset$.
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This immediately shows it is not equal to the other two.
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Now consider object $\emptyset$.
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This object is a member of $\{\emptyset\}$ but is not a member of
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$\{\{\emptyset\}\}$.
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Again, by the \nameref{ref:principle-extensionality}, these two sets must be
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different.
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\end{proof}
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\subsection{\partial{Exercise 3}}%
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\label{sub:baby-set-theory-3}
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Show that if $B \subseteq C$, then $\mathscr{P} B \subseteq \mathscr{P} C$.
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\begin{proof}
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Let $x \in \mathscr{P} B$.
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By definition of the \nameref{ref:powerset}, $x$ is a subset of $B$.
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By hypothesis, $B \subseteq C$.
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Then $x \subseteq C$.
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Again by definition of the \nameref{ref:powerset}, it follows
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$x \in \mathscr{P} C$.
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\end{proof}
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\subsection{\partial{Exercise 4}}%
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\label{sub:baby-set-theory-4}
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Assume that $x$ and $y$ are members of a set $B$.
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Show that $\{\{x\}, \{x, y\}\} \in \mathscr{P}\mathscr{P} B.$
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\begin{proof}
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Let $x$ and $y$ be members of set $B$.
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Then $\{x\}$ and $\{x, y\}$ are subsets of $B$.
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By definition of the \nameref{ref:powerset}, $\{x\}$ and $\{x, y\}$ are
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members of $\mathscr{P} B$.
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Then $\{\{x\}, \{x, y\}\}$ is a subset of $\mathscr{P} B$.
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By definition of the \nameref{ref:powerset}, $\{\{x\}, \{x, y\}\}$ is a member
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of $\mathscr{P}\mathscr{P} B$.
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\end{proof}
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\end{document}
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