1275 lines
33 KiB
TeX
1275 lines
33 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{Empty Set Axiom}}%
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\label{ref:empty-set-axiom}
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There is a set having no members:
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$$\exists B, \forall x, x \not\in B.$$
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\begin{axiom}
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\lean*{Mathlib/Init/Set}{Set.emptyCollection}
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\end{axiom}
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\section{\defined{Extensionality Axiom}}%
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\label{ref:extensionality-axiom}
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If two sets have exactly the same members, then they are equal:
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$$\forall A, \forall B,
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\left[\forall x, (x \in A \iff x \in B) \Rightarrow A = B\right].$$
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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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\section{\defined{Pair Set}}%
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\label{ref:pair-set}
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For any sets $u$ and $v$, the \textbf{pair set $\{u, v\}$} is the set whose
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only members are $u$ and $v$.
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\begin{definition}
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\statementpadding
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\lean*{Mathlib/Init/Set}{Set.insert}
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\lean*{Mathlib/Init/Set}{Set.singleton}
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\end{definition}
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\section{\defined{Pairing Axiom}}%
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\label{ref:pairing-axiom}
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For any sets $u$ and $v$, there is a set having as members just $u$ and $v$:
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$$\forall u, \forall v, \exists B, \forall x,
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(x \in B \iff x = u \text{ or } x = v).$$
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\begin{axiom}
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\statementpadding
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\lean*{Mathlib/Init/Set}{Set.insert}
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\lean*{Mathlib/Init/Set}{Set.singleton}
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\end{axiom}
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\section{\defined{Power Set}}%
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\label{ref:power-set}
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For any set $a$, the \textbf{power set $\powerset{a}$} is the set whose members
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are exactly the 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{Power Set Axiom}}%
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\label{ref:power-set-axiom}
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For any set $a$, there is a set whose members are exactly the subsets of $a$:
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$$\forall a, \exists B, \forall x, (x \in B \iff x \subseteq a).$$
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\begin{axiom}
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\lean*{Mathlib/Init/Set}{Set.powerset}
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\end{axiom}
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\section{\defined{Subset Axioms}}%
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\label{ref:subset-axioms}
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For each formula $\phi$ not containing $B$, the following is an axiom:
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$$\forall t_1, \cdots \forall t_k, \forall c,
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\exists B, \forall x, (x \in B \iff x \in c \land \phi).$$
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\begin{axiom}
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\lean*{Mathlib/Init/Set}{Set.Subset}
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\end{axiom}
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\section{\defined{Union Axiom}}%
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\label{ref:union-axiom}
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For any set $A$, there exists a set $B$ whose elements are exactly the members
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of the members of $A$:
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$$\forall A, \exists B, \forall x \left[ x \in B \iff (\exists b \in A) x \in b \right]$$
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\begin{axiom}
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\lean*{Mathlib/Data/Set/Lattice}{Set.sUnion}
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\end{axiom}
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\section{\defined{Union Axiom, Preliminary Form}}%
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\label{ref:union-axiom-preliminary-form}
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For any sets $a$ and $b$, there is a set whose members are those sets belonging
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either to $a$ or to $b$ (or both):
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$$\forall a, \forall b, \exists B, \forall x,
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(x \in B \iff x \in a \text{ or } x \in b).$$
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\begin{axiom}
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\lean*{Mathlib/Init/Set}{Set.union}
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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{\verified{Exercise 1.1}}%
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\label{sub:exercise-1.1}
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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{\verified{Exercise 1.1a}}%
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\label{ssub:exercise-1.1a}
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$\{\emptyset\} \_\_\_\_ \{\emptyset, \{\emptyset\}\}$.
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\begin{proof}
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\lean{Bookshelf/Enderton/Set/Chapter\_1}
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{Enderton.Set.Chapter\_1.exercise\_1\_1a}
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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{\verified{Exercise 1.1b}}%
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\label{ssub:exercise-1.11b}
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$\{\emptyset\} \_\_\_\_ \{\emptyset, \{\{\emptyset\}\}\}$.
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\begin{proof}
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\lean{Bookshelf/Enderton/Set/Chapter\_1}
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{Enderton.Set.Chapter\_1.exercise\_1\_1b}
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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{\verified{Exercise 1.1c}}%
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\label{ssub:exercise-1.1c}
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$\{\{\emptyset\}\} \_\_\_\_ \{\emptyset, \{\emptyset\}\}$.
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\begin{proof}
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\lean{Bookshelf/Enderton/Set/Chapter\_1}
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{Enderton.Set.Chapter\_1.exercise\_1\_1c}
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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{\verified{Exercise 1.1d}}%
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\label{ssub:exercise-1.1d}
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$\{\{\emptyset\}\} \_\_\_\_ \{\emptyset, \{\{\emptyset\}\}\}$.
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\begin{proof}
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\lean{Bookshelf/Enderton/Set/Chapter\_1}
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{Enderton.Set.Chapter\_1.exercise\_1\_1d}
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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{\verified{Exercise 1.1e}}%
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\label{ssub:exercise-1.1e}
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$\{\{\emptyset\}\} \_\_ \{\emptyset, \{\emptyset, \{\emptyset\}\}\}$.
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\begin{proof}
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\lean{Bookshelf/Enderton/Set/Chapter\_1}
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{Enderton.Set.Chapter\_1.exercise\_1\_1e}
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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{\verified{Exercise 1.2}}%
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\label{sub:exercise-1.2}
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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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\lean{Bookshelf/Enderton/Set/Chapter\_1}
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{Enderton.Set.Chapter\_1.exercise\_1\_2}
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By the \nameref{ref:extensionality-axiom}, $\emptyset$ 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:extensionality-axiom}, these two sets must be
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different.
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\end{proof}
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\subsection{\verified{Exercise 1.3}}%
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\label{sub:exercise-1.3}
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Show that if $B \subseteq C$, then $\powerset{B} \subseteq \powerset{C}$.
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\begin{proof}
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\lean{Bookshelf/Enderton/Set/Chapter\_1}
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{Enderton.Set.Chapter\_1.exercise\_1\_3}
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Let $x \in \powerset{B}$.
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By definition of the \nameref{ref:power-set}, $x$ 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:power-set}, it follows
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$x \in \powerset{C}$.
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\end{proof}
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\subsection{\verified{Exercise 1.4}}%
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\label{sub:exercise-1.4}
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Assume that $x$ and $y$ are members of a set $B$.
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Show that $\{\{x\}, \{x, y\}\} \in \powerset{\powerset{B}}.$
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\begin{proof}
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\lean{Bookshelf/Enderton/Set/Chapter\_1}
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{Enderton.Set.Chapter\_1.exercise\_1\_4}
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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:power-set}, $\{x\}$ and $\{x, y\}$ are
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members of $\powerset{B}$.
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Then $\{\{x\}, \{x, y\}\}$ is a subset of $\powerset{B}$.
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By definition of the \nameref{ref:power-set}, $\{\{x\}, \{x, y\}\}$ is a
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member of $\powerset{\powerset{B}}$.
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\end{proof}
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\section{Sets - An Informal View}%
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\label{sec:sets-informal-view}
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\subsection{\partial{Exercise 2.1}}%
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\label{sub:exercise-2.1}
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Define the rank of a set $c$ to be the least $\alpha$ such that
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$c \subseteq V_\alpha$.
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Compute the rank of $\{\{\emptyset\}\}$.
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Compute the rank of
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$\{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\}$.
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\begin{proof}
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We first compute the values of $V_n$ for $0 \leq n \leq 3$ under the
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assumption the set of atoms $A$ at the bottom of the hierarchy is empty.
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\begin{align*}
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V_0 & = \emptyset \\
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V_1 & = V_0 \cup \powerset{V_0} \\
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& = \emptyset \cup \{\emptyset\} \\
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& = \{\emptyset\} \\
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V_2 & = V_1 \cup \powerset{V_1} \\
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& = \{\emptyset\} \cup \powerset{\{\emptyset\}} \\
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& = \{\emptyset\} \cup \{\emptyset, \{\emptyset\}\} \\
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& = \{\emptyset, \{\emptyset\}\} \\
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V_3 & = V_2 \cup \powerset{V_2} \\
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& = \{\emptyset, \{\emptyset\}\} \cup
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\powerset{\{\emptyset, \{\emptyset\}\}} \\
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& = \{\emptyset, \{\emptyset\}\} \cup
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\{\emptyset,
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\{\emptyset\},
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\{\{\emptyset\}\},
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\{\emptyset, \{\emptyset\}\}\} \\
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& = \{\emptyset,
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\{\emptyset\},
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\{\{\emptyset\}\},
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\{\emptyset, \{\emptyset\}\}\}
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\end{align*}
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It then immediately follows $\{\{\emptyset\}\}$ has rank $2$ and
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$\{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\}$ has rank $3$.
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\end{proof}
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\subsection{\partial{Exercise 2.2}}%
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\label{sub:exercise-2.2}
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We have stated that $V_{\alpha + 1} = A \cup \powerset{V_\alpha}$.
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Prove this at least for $\alpha < 3$.
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\begin{proof}
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Let $A$ be the set of atoms in our set hierarchy.
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Let $P(n)$ be the predicate, "$V_{n + 1} = A \cup \powerset{V_n}$."
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We prove $P(n)$ holds true for all natural numbers $n \geq 1$ via induction.
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\paragraph{Base Case}%
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Let $n = 1$.
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By definition, $V_1 = V_0 \cup \powerset{V_0}$.
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By definition, $V_0 = A$.
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Therefore $V_1 = A \cup \powerset{V_0}$.
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This proves $P(1)$ holds true.
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\paragraph{Induction Step}%
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Suppose $P(n)$ holds true for some $n \geq 1$.
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Consider $V_{n+1}$.
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By definition, $V_{n+1} = V_n \cup \powerset{V_n}$.
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Therefore, by the induction hypothesis,
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\begin{align}
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V_{n+1}
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& = V_n \cup \powerset{V_n}
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\nonumber \\
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& = (A \cup \powerset{V_{n-1}}) \cup \powerset{V_n}
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\nonumber \\
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& = A \cup (\powerset{V_{n-1}} \cup \powerset{V_n})
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\label{sub:exercise-2.2-eq1}
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\end{align}
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But $V_{n-1}$ is a subset of $V_n$.
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\nameref{sub:exercise-1.3} then implies
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$\powerset{V_{n-1}} \subseteq \powerset{V_n}$.
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This means \eqref{sub:exercise-2.2-eq1} can be simplified to
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$$V_{n+1} = A \cup \powerset{V_n},$$
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proving $P(n+1)$ holds true.
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\paragraph{Conclusion}%
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By mathematical induction, it follows for all $n \geq 1$, $P(n)$ is true.
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\end{proof}
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\subsection{\partial{Exercise 2.3}}%
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\label{sub:exercise-2.3}
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List all the members of $V_3$.
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List all the members of $V_4$.
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(It is to be assumed here that there are no atoms.)
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\begin{proof}
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As seen in the proof of \nameref{sub:exercise-2.1},
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$$V_3 = \{
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\emptyset,
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\{\emptyset\},
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\{\{\emptyset\}\},
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\{\emptyset, \{\emptyset\}\}
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\}.$$
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By \nameref{sub:exercise-2.2}, $V_4 = \powerset{V_3}$ (since it is assumed
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there are no atoms).
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Thus
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\begin{align*}
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& V_4 = \{ \\
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& \qquad \emptyset, \\
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& \qquad \{\emptyset\}, \\
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& \qquad \{\{\emptyset\}\}, \\
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& \qquad \{\{\{\emptyset\}\}\}, \\
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& \qquad \{\{\emptyset, \{\emptyset\}\}\}, \\
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& \qquad \{\emptyset, \{\emptyset\}\}, \\
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& \qquad \{\emptyset, \{\{\emptyset\}\}\}, \\
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& \qquad \{\emptyset, \{\emptyset, \{\emptyset\}\}\}, \\
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& \qquad \{\{\emptyset\}, \{\{\emptyset\}\}\}, \\
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& \qquad \{\{\emptyset\}, \{\emptyset, \{\emptyset\}\}\}, \\
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& \qquad \{\{\{\emptyset\}\}, \{\emptyset, \{\emptyset\}\}\}, \\
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& \qquad \{\emptyset, \{\emptyset\}, \{\{\emptyset\}\}\}, \\
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& \qquad \{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\}, \\
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& \qquad \{\emptyset, \{\{\emptyset\}\}, \{\emptyset, \{\emptyset\}\}\} \\
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& \qquad \{\{\emptyset\}, \{\{\emptyset\}\}, \{\emptyset, \{\emptyset\}\}\}, \\
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& \qquad \{\emptyset, \{\emptyset\}, \{\{\emptyset\}\}, \{\emptyset, \{\emptyset\}\}\} \\
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& \}.
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\end{align*}
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\end{proof}
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\chapter{Axioms and Operations}%
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\label{chap:axioms-operations}
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\section{Axioms}%
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\label{sec:axioms}
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\subsection{\partial{Theorem 2A}}%
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\label{sub:theorem-2a}
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\begin{theorem}[2A]
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There is no set to which every set belongs.
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\note{This was revisited after reading Enderton's proof prior.}
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\end{theorem}
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\begin{proof}
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Let $A$ be an arbitrary set.
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Define $B = \{ x \in A \mid x \not\in x \}$.
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By the \nameref{ref:subset-axioms}, $B$ is a set.
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Then $$B \in B \iff B \in A \land B \not\in B.$$
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If $B \in A$, then $B \in B \iff B \not\in B$, a contradiction.
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Thus $B \not\in A$.
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Since this process holds for any set $A$, there must exist no set to which
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every set belongs.
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\end{proof}
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\subsection{\partial{Theorem 2B}}%
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\label{sub:theorem-2b}
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\begin{theorem}[2B]
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For any nonempty set $A$, there exists a unique set $B$ such that for any
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$x$, $$x \in B \iff x \text{ belongs to every member of } A.$$
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\end{theorem}
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\begin{proof}
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Suppose $A$ is a nonempty set.
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This ensures the statement we are trying to prove does not vacuously hold for
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all sets $x$ (which would yield a contradiction due to
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\nameref{sub:theorem-2b}).
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By the \nameref{ref:union-axiom}, $\bigcup A$ is a set.
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Define $$B = \{ x \in \bigcup A \mid (\forall b \in A), x \in b \}.$$
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By the \nameref{ref:subset-axioms}, $B$ is indeed a set.
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By construction,
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$$\forall x, x \in B \iff x \text{ belongs to every member of } A.$$
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By the \nameref{ref:extensionality-axiom}, $B$ is unique.
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\end{proof}
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\section{Exercises 3}%
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\label{sec:exercises-3}
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\subsection{\verified{Exercise 3.1}}%
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\label{sub:exercise-3.1}
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Assume that $A$ is the set of integers divisible by $4$.
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Similarly assume that $B$ and $C$ are the sets of integers divisible by $9$ and
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$10$, respectively.
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What is in $A \cap B \cap C$?
|
|
|
|
\begin{answer}
|
|
|
|
\lean{Bookshelf/Enderton/Set/Chapter\_1}
|
|
{Enderton.Set.Chapter\_1.exercise\_3\_1}
|
|
|
|
The set of integers divisible by $4$, $9$, and $10$.
|
|
|
|
\end{answer}
|
|
|
|
\subsection{\verified{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{answer}
|
|
|
|
\lean{Bookshelf/Enderton/Set/Chapter\_1}
|
|
{Enderton.Set.Chapter\_1.exercise\_3\_2}
|
|
|
|
Let $A = \{\{1\}, \{2\}\}$ and $B = \{\{1, 2\}\}$.
|
|
|
|
\end{answer}
|
|
|
|
\subsection{\verified{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}
|
|
|
|
\lean{Bookshelf/Enderton/Set/Chapter\_1}
|
|
{Enderton.Set.Chapter\_1.exercise\_3\_3}
|
|
|
|
Let $x \in A$.
|
|
By definition, $$\bigcup A = \{ y \mid (\exists b \in A) y \in b\}.$$
|
|
Then $\{ y \mid y \in x\} \subseteq \bigcup A$.
|
|
But $\{ y \mid y \in x\} = x$.
|
|
Thus $x \subseteq \bigcup A$.
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\verified{Exercise 3.4}}%
|
|
\label{sub:exercise-3.4}
|
|
|
|
Show that if $A \subseteq B$, then $\bigcup A \subseteq \bigcup B$.
|
|
|
|
\begin{proof}
|
|
|
|
\lean{Bookshelf/Enderton/Set/Chapter\_1}
|
|
{Enderton.Set.Chapter\_1.exercise\_3\_4}
|
|
|
|
Let $A$ and $B$ be sets such that $A \subseteq B$.
|
|
Let $x \in \bigcup A$.
|
|
By definition of the union, there exists some $b \in A$ such that $x \in b$.
|
|
By definition of the subset, $b \in B$.
|
|
This immediatley implies $x \in \bigcup B$.
|
|
Since this holds for all $x \in \bigcup A$, it follows
|
|
$\bigcup A \subseteq \bigcup B$.
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\verified{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}
|
|
|
|
\lean{Bookshelf/Enderton/Set/Chapter\_1}
|
|
{Enderton.Set.Chapter\_1.exercise\_3\_5}
|
|
|
|
Let $x \in \bigcup \mathscr{A}$.
|
|
By definition,
|
|
$$\bigcup \mathscr{A} = \{ y \mid (\exists b \in A)y \in b \}.$$
|
|
Then there exists some $b \in A$ such that $x \in b$.
|
|
By hypothesis, $b \subseteq B$.
|
|
Thus $x$ must also be a member of $B$.
|
|
Since this holds for all $x \in \bigcup \mathscr{A}$, it follows
|
|
$\bigcup \mathscr{A} \subseteq B$.
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\verified{Exercise 3.6a}}%
|
|
\label{sub:exercise-3.6a}
|
|
|
|
Show that for any set $A$, $\bigcup \powerset{A} = A$.
|
|
|
|
\begin{proof}
|
|
|
|
\lean{Bookshelf/Enderton/Set/Chapter\_1}
|
|
{Enderton.Set.Chapter\_1.exercise\_3\_6a}
|
|
|
|
We prove that (i) $\bigcup \powerset{A} \subseteq A$ and (ii)
|
|
$A \subseteq \bigcup \powerset{A}$.
|
|
|
|
\paragraph{(i)}%
|
|
\label{par:exercise-3.6a-i}
|
|
|
|
By definition, the \nameref{ref:power-set} of $A$ is the set of all subsets
|
|
of $A$.
|
|
In other words, every member of $\powerset{A}$ is a subset of $A$.
|
|
By \nameref{sub:exercise-3.5}, $\bigcup \powerset{A} \subseteq A$.
|
|
|
|
\paragraph{(ii)}%
|
|
\label{par:exercise-3.6a-ii}
|
|
|
|
Let $x \in A$.
|
|
By definition of the power set of $A$, $\{x\} \in \powerset{A}$.
|
|
By definition of the union,
|
|
$$\bigcup \powerset{A} =
|
|
\{ y \mid (\exists b \in \powerset{A}), y \in b).$$
|
|
Since $x \in \{x\}$ and $\{x\} \in \powerset{A}$, it follows
|
|
$x \in \bigcup \powerset{A}$.
|
|
Thus $A \subseteq \bigcup \powerset{A}$.
|
|
|
|
\paragraph{Conclusion}%
|
|
|
|
By \nameref{par:exercise-3.6a-i} and \nameref{par:exercise-3.6a-ii},
|
|
$\bigcup \powerset{A} = A$.
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\verified{Exercise 3.6b}}%
|
|
\label{sub:exercise-3.6b}
|
|
|
|
Show that $A \subseteq \powerset{\bigcup A}$.
|
|
Under what conditions does equality hold?
|
|
|
|
\begin{proof}
|
|
|
|
\lean{Bookshelf/Enderton/Set/Chapter\_1}
|
|
{Enderton.Set.Chapter\_1.exercise\_3\_6b}
|
|
|
|
Let $x \in A$.
|
|
By \nameref{sub:exercise-3.3}, $x$ is a subset of $\bigcup A$.
|
|
By the definition of the \nameref{ref:power-set},
|
|
$$\powerset{\bigcup A} = \{ y \mid y \subseteq \bigcup A \}.$$
|
|
Therefore $x \in \powerset{\bigcup A}$.
|
|
Since this holds for all $x \in A$, $A \subseteq \powerset{\bigcup A}$.
|
|
|
|
\suitdivider
|
|
|
|
We show equality holds if and only if there exists some set $B$ such that
|
|
$A = \powerset{B}$.
|
|
|
|
\paragraph{($\Rightarrow$)}%
|
|
\label{par:exercise-3.6b-right}
|
|
|
|
Suppose $A = \powerset{\bigcup A}$.
|
|
Then our statement immediately follows by settings $B = \bigcup A$.
|
|
|
|
\paragraph{($\Leftarrow$)}%
|
|
\label{par:exercise-3.6b-left}
|
|
|
|
Suppose there exists some set $B$ such that $A = \powerset{B}$.
|
|
Therefore
|
|
\begin{align*}
|
|
\powerset{\bigcup A}
|
|
& = \powerset{\left(\bigcup {\powerset {B}}\right)} \\
|
|
& = \powerset{B} & \textref{sub:exercise-3.6a} \\
|
|
& = A.
|
|
\end{align*}
|
|
|
|
\paragraph{Conclusion}%
|
|
|
|
By \nameref{par:exercise-3.6b-right} and \nameref{par:exercise-3.6b-left},
|
|
$A = \powerset{\bigcup A}$ if and only if there exists some set $B$ such
|
|
that $A = \powerset{B}$.
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\verified{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}
|
|
|
|
\lean{Bookshelf/Enderton/Set/Chapter\_1}
|
|
{Enderton.Set.Chapter\_1.exercise\_3\_7a}
|
|
|
|
Let $A$ and $B$ be arbitrary sets. We show that
|
|
$\powerset{A} \cap \powerset{B} \subseteq \powerset{(A \cap B)}$ and then
|
|
show that $\powerset{A} \cap \powerset{B} \supseteq \powerset{(A \cap B)}$.
|
|
|
|
\paragraph{($\subseteq$)}%
|
|
|
|
Let $x \in \powerset{A} \cap \powerset{B}$.
|
|
That is, $x \in \powerset{A}$ and $x \in \powerset{B}$.
|
|
By the definition of the \nameref{ref:power-set},
|
|
\begin{align*}
|
|
\powerset{A} & = \{ y \mid y \subseteq A \} \\
|
|
\powerset{B} & = \{ y \mid y \subseteq B \}
|
|
\end{align*}
|
|
Thus $x \subseteq A$ and $x \subseteq B$, meaning $x \subseteq A \cap B$.
|
|
But then $x \in \powerset{(A \cap B)}$, the set of all subsets of
|
|
$A \cap B$.
|
|
Since this holds for all $x \in \powerset{A} \cap \powerset{B}$, it follows
|
|
$$\powerset{A} \cap \powerset{B} \subseteq \powerset{(A \cap B)}.$$
|
|
|
|
\paragraph{($\supseteq$)}%
|
|
|
|
Let $x \in \powerset{(A \cap B)}$.
|
|
By the definition of the \nameref{ref:power-set},
|
|
$$\powerset{(A \cap B)} = \{ y \mid y \subseteq A \cap B \}.$$
|
|
Thus $x \subseteq A \cap B$, meaning $x \subseteq A$ and $x \subseteq B$.
|
|
But this implies $x \in \powerset{A}$, the set of all subsets of $A$.
|
|
Likewise $x \in \powerset{B}$, the set of all subsets of $B$.
|
|
Thus $x \in \powerset{A} \cap \powerset{B}$.
|
|
Since this holds for all $x \in \powerset{(A \cap B)}$, it follows
|
|
$$\powerset{(A \cap B)} \subseteq \powerset{A} \cap \powerset{B}.$$
|
|
|
|
\paragraph{Conclusion}%
|
|
|
|
Since each side of our identity is a subset of the other,
|
|
$$\powerset{(A \cap B)} = \powerset{A} \cap \powerset{B}.$$
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\verified{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}
|
|
|
|
\statementpadding
|
|
|
|
\lean*{Bookshelf/Enderton/Set/Chapter\_1}
|
|
{Enderton.Set.Chapter\_1.exercise\_3\_7b\_i}
|
|
|
|
\lean{Bookshelf/Enderton/Set/Chapter\_1}
|
|
{Enderton.Set.Chapter\_1.exercise\_3\_7b\_ii}
|
|
|
|
Let $x \in \powerset{A} \cup \powerset{B}$.
|
|
By definition, $x \in \powerset{A}$ or $x \in \powerset{B}$ (or both).
|
|
By the definition of the \nameref{ref:power-set},
|
|
\begin{align*}
|
|
\powerset{A} &= \{ y \mid y \subseteq A \} \\
|
|
\powerset{B} &= \{ y \mid y \subseteq B \}.
|
|
\end{align*}
|
|
Thus $x \subseteq A$ or $x \subseteq B$.
|
|
Therefore $x \subseteq A \cup B$.
|
|
But then $x \in \powerset{(A \cup B)}$, the set of all subsets of $A \cup B$.
|
|
|
|
\suitdivider
|
|
|
|
We show equality holds if and only if one of $A$ or $B$ is a subset of the
|
|
other.
|
|
|
|
\paragraph{($\Rightarrow$)}%
|
|
\label{par:exercise-3.7b-right}
|
|
|
|
Suppose
|
|
\begin{equation}
|
|
\label{sub:exercise-3.7b-eq1}
|
|
\powerset{A} \cup \powerset{B} = \powerset{(A \cup B)}.
|
|
\end{equation}
|
|
By the definition of the \nameref{ref:power-set},
|
|
$A \cup B \in \powerset{(A \cup B)}$.
|
|
Then \eqref{sub:exercise-3.7b-eq1} implies
|
|
$A \cup B \in \powerset{A} \cup \powerset{B}$.
|
|
That is, $A \cup B \in \powerset{A}$ or $A \cup B \in \powerset{B}$ (or
|
|
both).
|
|
|
|
For the sake of contradiction, suppose $A \not\subseteq B$ and
|
|
$B \not\subseteq A$.
|
|
Then there exists an element $x \in A$ such that $x \not\in B$ and there
|
|
exists an element $y \in B$ such that $y \not\in A$.
|
|
But then $A \cup B \not\in \powerset{A}$ since $y$ cannot be a member of a
|
|
member of $\powerset{A}$.
|
|
Likewise, $A \cup B \not\in \powerset{B}$ since $x$ cannot be a member of a
|
|
member of $\powerset{B}$.
|
|
Therefore our assumption is incorrect.
|
|
In other words, $A \subseteq B$ or $B \subseteq A$.
|
|
|
|
\paragraph{($\Leftarrow$)}%
|
|
\label{par:exercise-3.7b-left}
|
|
|
|
WLOG, suppose $A \subseteq B$.
|
|
Then, by \nameref{sub:exercise-1.3}, $\powerset{A} \subseteq \powerset{B}$.
|
|
Thus
|
|
\begin{align*}
|
|
\powerset{A} \cup \powerset{B}
|
|
& = \powerset{B} \\
|
|
& = \powerset{A \cup B}.
|
|
\end{align*}
|
|
|
|
\paragraph{Conclusion}%
|
|
|
|
By \nameref{par:exercise-3.7b-right} and \nameref{par:exercise-3.7b-left},
|
|
it follows
|
|
$\powerset{A} \cup \powerset{B} \subseteq \powerset{(A \cup B)}$ if and
|
|
only if $A \subseteq B$ or $B \subseteq A$.
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\partial{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}
|
|
|
|
We proceed by contradiction.
|
|
Suppose there existed a set $A$ consisting of every singleton.
|
|
Then the \nameref{ref:union-axiom} suggests $\bigcup A$ is a set.
|
|
But this set is precisely the class of all sets, which is \textit{not} a set.
|
|
Thus our original assumption was incorrect.
|
|
That is, there is no set to which every singleton belongs.
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\verified{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{answer}
|
|
|
|
\lean{Bookshelf/Enderton/Set/Chapter\_1}
|
|
{Enderton.Set.Chapter\_1.exercise\_3\_9}
|
|
|
|
Let $a = \{1\}$ and $B = \{\{1\}\}$.
|
|
Then
|
|
\begin{align*}
|
|
\powerset{a} & = \{\emptyset, \{1\}\} \\
|
|
\powerset{B} & = \{\emptyset, \{\{1\}\}\}.
|
|
\end{align*}
|
|
It immediately follows that $\powerset{a} \not\in \powerset{B}$.
|
|
|
|
\end{answer}
|
|
|
|
\subsection{\verified{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}
|
|
|
|
\lean{Bookshelf/Enderton/Set/Chapter\_1}
|
|
{Enderton.Set.Chapter\_1.exercise\_3\_10}
|
|
|
|
Suppose $a \in B$.
|
|
By \nameref{sub:exercise-3.3}, $a \subseteq \bigcup B$.
|
|
By \nameref{sub:exercise-1.3}, $\powerset{a} \subseteq \powerset{\bigcup B}$.
|
|
By the definition of the \nameref{ref:power-set},
|
|
$$\powerset{\powerset{\bigcup B}} =
|
|
\{ y \mid y \subseteq \powerset{\bigcup B} \}.$$
|
|
Therefore $\powerset{a} \in \powerset{\powerset{\bigcup B}}$.
|
|
|
|
\end{proof}
|
|
|
|
\section{Algebra of Sets}%
|
|
\label{sec:algebra-sets}
|
|
|
|
\subsection{\verified{Commutative Laws}}%
|
|
\label{sub:commutative-laws}
|
|
|
|
For any sets $A$ and $B$,
|
|
\begin{align*}
|
|
A \cup B = B \cup A \\
|
|
A \cap B = B \cap A
|
|
\end{align*}
|
|
|
|
\begin{proof}
|
|
|
|
\statementpadding
|
|
|
|
\lean*{Mathlib/Data/Set/Basic}{Set.union\_comm}
|
|
|
|
\lean{Mathlib/Data/Set/Basic}{Set.inter\_comm}
|
|
|
|
Let $A$ and $B$ be sets.
|
|
We show (i) $A \cup B = B \cup A$ and then (ii) $A \cap B = B \cap A$.
|
|
|
|
\paragraph{(i)}%
|
|
|
|
By the definition of the union of sets,
|
|
$A \cup B = \{ x \mid x \in A \lor x \in B \}$.
|
|
Since $\lor$ is commutative, it follows
|
|
\begin{align*}
|
|
A \cup B
|
|
& = \{ x \mid x \in A \lor x \in B \} \\
|
|
& = \{ x \mid x \in B \lor x \in A \} \\
|
|
& = B \cup A,
|
|
\end{align*}
|
|
where the last equality follows again from the definition of the union of
|
|
sets.
|
|
|
|
\paragraph{(ii)}%
|
|
|
|
By the definition of the intersection of sets,
|
|
$A \cap B = \{ x \mid x \in A \land x \in B \}$.
|
|
Since $\land$ is commutative, it follows
|
|
\begin{align*}
|
|
A \cap B
|
|
& = \{ x \mid x \in A \land x \in B \} \\
|
|
& = \{ x \mid x \in B \land x \in A \} \\
|
|
& = B \land A,
|
|
\end{align*}
|
|
where the last equality follows again from the definition of the
|
|
intersection of sets.
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\unverified{Associative Laws}}%
|
|
\label{sub:associative-laws}
|
|
|
|
For any sets $A$, $B$ and $C$,
|
|
\begin{align*}
|
|
A \cup (B \cup C) & = (A \cup B) \cup C \\
|
|
(A \cup B) \cup C & = A \cup (B \cup C)
|
|
\end{align*}
|
|
|
|
\begin{proof}
|
|
|
|
TODO
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\unverified{Distributive Laws}}%
|
|
\label{sub:distributive-laws}
|
|
|
|
For any sets $A$, $B$, and $C$,
|
|
\begin{align*}
|
|
A \cap (B \cup C) & = (A \cap B) \cup (A \cap C) \\
|
|
A \cup (B \cap C) & = (A \cup B) \cap (A \cup C)
|
|
\end{align*}
|
|
|
|
\begin{proof}
|
|
|
|
TODO
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\unverified{De Morgan's Laws}}%
|
|
\label{sub:de-morgans-laws}
|
|
|
|
For any sets $A$, $B$, and $C$,
|
|
\begin{align*}
|
|
C - (A \cup B) & = (C - A) \cap (C - B) \\
|
|
C - (A \cap B) & = (C - A) \cup (C - B)
|
|
\end{align*}
|
|
|
|
\begin{proof}
|
|
|
|
TODO
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\unverified{
|
|
Identities Involving \texorpdfstring{$\emptyset$}{the Empty Set}}}%
|
|
\label{sub:identitives-involving-empty-set}
|
|
|
|
For any set $A$,
|
|
\begin{align*}
|
|
A \cup \emptyset & = A \\
|
|
A \cap \emptyset & = \emptyset \\
|
|
A \cap (C - A) & = \emptyset
|
|
\end{align*}
|
|
|
|
\begin{proof}
|
|
|
|
TODO
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\unverified{Monotonicity}}%
|
|
\label{sub:monotonicity}
|
|
|
|
For any sets $A$, $B$, and $C$,
|
|
\begin{align*}
|
|
A \subseteq B & \Rightarrow A \cup C \subseteq B \cup C \\
|
|
A \subseteq B & \Rightarrow A \cap C \subseteq B \cap C \\
|
|
A \subseteq B & \Rightarrow \bigcup A \subseteq \bigcup B
|
|
\end{align*}
|
|
|
|
\begin{proof}
|
|
|
|
TODO
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\unverified{Anti-monotonicity}}%
|
|
\label{sub:anti-monotonicity}
|
|
|
|
For any sets $A$, $B$, and $C$,
|
|
\begin{align*}
|
|
A \subseteq B & \Rightarrow C - B \subseteq C - A \\
|
|
\emptyset \neq A \subseteq B & \Rightarrow \bigcap B \subseteq \bigcap A.
|
|
\end{align*}
|
|
|
|
\begin{proof}
|
|
|
|
TODO
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\unverified{General Distributive Laws}}%
|
|
\label{sub:general-distributive-laws}
|
|
|
|
For any sets $A$ and $\mathscr{B}$,
|
|
\begin{align*}
|
|
A \cup \bigcap \mathscr{B} & =
|
|
\bigcap\; \{ A \cup X \mid X \in \mathscr{B} \}
|
|
\quad\text{for}\quad \mathscr{B} \neq \emptyset \\
|
|
A \cap \bigcup \mathscr{B} & =
|
|
\bigcup\; \{ A \cap X \mid X \in \mathscr{B} \}
|
|
\end{align*}
|
|
|
|
\begin{proof}
|
|
|
|
TODO
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\unverified{General De Morgan's Laws}}%
|
|
\label{sub:general-de-morgans-laws}
|
|
|
|
For any set $C$ and $\mathscr{A} \neq \emptyset$,
|
|
\begin{align*}
|
|
C - \bigcup \mathscr{A} & = \bigcap\; \{ C - X \mid X \in \mathscr{A} \} \\
|
|
C - \bigcap \mathscr{A} & = \bigcup\; \{ C - X \mid X \in \mathscr{A} \}
|
|
\end{align*}
|
|
|
|
\begin{proof}
|
|
|
|
TODO
|
|
|
|
\end{proof}
|
|
|
|
\section{Exercises 4}%
|
|
\label{sec:exercises-4}
|
|
|
|
\subsection{\unverified{Exercise 4.11}}%
|
|
\label{sub:exercise-4.11}
|
|
|
|
Show that for any sets $A$ and $B$,
|
|
$$A = (A \cap B) \cup (A - B) \quad\text{and}\quad
|
|
A \cup (B - A) = A \cup B.$$
|
|
|
|
\begin{proof}
|
|
|
|
TODO
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\unverified{Exercise 4.12}}%
|
|
\label{sub:exercise-4.12}
|
|
|
|
Verify the following identity (one of De Morgan's laws):
|
|
$$C - (A \cap B) = (C - A) \cup (C - B).$$
|
|
|
|
\begin{proof}
|
|
|
|
TODO
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\unverified{Exercise 4.13}}%
|
|
\label{sub:exercise-4.13}
|
|
|
|
Show that if $A \subseteq B$, then $C - B \subseteq C - A$.
|
|
|
|
\begin{proof}
|
|
|
|
TODO
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\unverified{Exercise 4.14}}%
|
|
\label{sub:exercise-4.14}
|
|
|
|
Show by example that for some sets $A$, $B$, and $C$, the set $A - (B - C)$ is
|
|
different from $(A - B) - C$.
|
|
|
|
\begin{proof}
|
|
|
|
TODO
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\unverified{Exercise 4.15}}%
|
|
\label{sub:exercise-4.15}
|
|
|
|
Define the symmetric difference $A + B$ of sets $A$ and $B$ to be the set
|
|
$(A - B) \cup (B - A)$.
|
|
|
|
\subsubsection{\unverified{Exercise 4.15a}}%
|
|
\label{ssub:exercise-4.15a}
|
|
|
|
Show that $A \cap (B + C) = (A \cap B) + (A \cap C)$.
|
|
|
|
\begin{proof}
|
|
|
|
TODO
|
|
|
|
\end{proof}
|
|
|
|
\subsubsection{\unverified{Exercise 4.15b}}%
|
|
\label{ssub:exercise-4.15b}
|
|
|
|
Show that $A + (B + C) = (A + B) + C$.
|
|
|
|
\begin{proof}
|
|
|
|
TODO
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\unverified{Exercise 4.16}}%
|
|
\label{sub:exercise-4.16}
|
|
|
|
Simplify:
|
|
$$[(A \cup B \cup C) \cap (A \cup B)] - [(A \cup (B - C)) \cap A].$$
|
|
|
|
\begin{proof}
|
|
|
|
TODO
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\unverified{Exercise 4.17}}%
|
|
\label{sub:exercise-4.17}
|
|
|
|
Show that the following four conditions are equivalent.
|
|
|
|
\begin{enumerate}[(a)]
|
|
\item $A \subseteq B$,
|
|
\item $A - B = \emptyset$,
|
|
\item $A \cup B = B$,
|
|
\item $A \cap B = A$.
|
|
\end{enumerate}
|
|
|
|
\begin{proof}
|
|
|
|
TODO
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\unverified{Exercise 4.18}}%
|
|
\label{sub:exercise-4.18}
|
|
|
|
Assume that $A$ and $B$ are subsets of $S$.
|
|
List all of the different sets that can be made from these three by use of the
|
|
binary operations $\cup$, $\cap$, and $-$.
|
|
|
|
\begin{proof}
|
|
|
|
TODO
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\unverified{Exercise 4.19}}%
|
|
\label{sub:exercise-4.19}
|
|
|
|
Is $\powerset{(A - B)}$ always equal to $\powerset{A} - \powerset{B}$?
|
|
Is it ever equal to $\powerset{A} - \powerset{B}$?
|
|
|
|
\begin{proof}
|
|
|
|
TODO
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\unverified{Exercise 4.20}}%
|
|
\label{sub:exercise-4.20}
|
|
|
|
Let $A$, $B$, and $C$ be sets such that $A \cup B = A \cup C$ and
|
|
$A \cap B = A \cap C$.
|
|
Show that $B = C$.
|
|
|
|
\begin{proof}
|
|
|
|
TODO
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\unverified{Exercise 4.21}}%
|
|
\label{sub:exercise-4.21}
|
|
|
|
Show that $\bigcup (A \cup B) = \bigcup A \cup \bigcup B$.
|
|
|
|
\begin{proof}
|
|
|
|
TODO
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\unverified{Exercise 4.22}}%
|
|
\label{sub:exercise-4.22}
|
|
|
|
Show that if $A$ and $B$ are nonempty sets, then
|
|
$\bigcap (A \cup B) = \bigcap A \cap \bigcap B$.
|
|
|
|
\begin{proof}
|
|
|
|
TODO
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\unverified{Exercise 4.23}}%
|
|
\label{sub:exercise-4.23}
|
|
|
|
Show that if $\mathscr{B}$ is nonempty, then
|
|
$A \cup \bigcap \mathscr{B} = \bigcap\; \{A \cup X \mid X \in \mathscr{B} \}$.
|
|
|
|
\begin{proof}
|
|
|
|
TODO
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\unverified{Exercise 4.24a}}%
|
|
\label{sub:exercise-4.24a}
|
|
|
|
Show that if $\mathscr{A}$ is nonempty, then
|
|
$\powerset{\bigcap A} = \bigcap\; \{\powerset{X} \mid X \in \mathscr{A} \}$.
|
|
|
|
\begin{proof}
|
|
|
|
TODO
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\unverified{Exercise 4.24b}}%
|
|
\label{sub:exercise-4.24b}
|
|
|
|
Show that
|
|
$$\bigcup\; \{ \powerset{X} \mid X \in \mathscr{A} \} \subseteq
|
|
\powerset{\bigcup A}.$$
|
|
Under what conditions does equality hold?
|
|
|
|
\begin{proof}
|
|
|
|
TODO
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\unverified{Exercise 4.25}}%
|
|
\label{sub:exercise-4.25}
|
|
|
|
Is $A \cup \bigcup \mathscr{B}$ always the same as
|
|
$\bigcup\; \{ A \cup X \mid X \in \mathscr{B} \}$?
|
|
If not, then under what conditions does equality hold?
|
|
|
|
\begin{proof}
|
|
|
|
TODO
|
|
|
|
\end{proof}
|
|
|
|
\end{document}
|