582 lines
15 KiB
TeX
582 lines
15 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{\partial{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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\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{\partial{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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\section{\partial{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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\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{\partial{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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\section{\partial{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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\section{\partial{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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\section{\partial{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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\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{\unverified{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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\end{theorem}
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\unverified{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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TODO
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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{\unverified{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 $A \cap B \cap C$?
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\unverified{Exercise 3.2}}%
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\label{sub:exercise-3.2}
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Give an example of sets $A$ and $B$ for which $\bigcup A = \bigcup B$ but
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$A \neq B$.
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\unverified{Exercise 3.3}}%
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\label{sub:exercise-3.3}
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Show that every member of a set $A$ is a subset of $\bigcup A$.
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(This was stated as an example in this section.)
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\unverified{Exercise 3.4}}%
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\label{sub:exercise-3.4}
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Show that if $A \subseteq B$, then $\bigcup A \subseteq \bigcup B$.
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\unverified{Exercise 3.5}}%
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\label{sub:exercise-3.5}
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Assume that every member of $\mathscr{A}$ is a subset of $B$.
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Show that $\bigcup \mathscr{A} \subseteq B$.
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\unverified{Exercise 3.6a}}%
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\label{sub:exercise-3.6a}
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Show that for any set $A$, $\bigcup \powerset{A} = A$.
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\unverified{Exercise 3.6b}}%
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\label{sub:exercise-3.6b}
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Show that $A \subseteq \powerset{\bigcup A}$.
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Under what conditions does equality hold?
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\unverified{Exercise 3.7a}}%
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\label{sub:exercise-3.7a}
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Show that for any sets $A$ and $B$,
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$$\powerset{A} \cap \powerset{B} = \powerset(A \cap B).$$
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\unverified{Exercise 3.7b}}%
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\label{sub:exercise-3.7b}
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Show that $\powerset{A} \cup \powerset{B} \subseteq \powerset(A \cup B)$.
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Under what conditions does equality hold?
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\unverified{Exercise 3.8}}%
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\label{sub:exercise-3.8}
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Show that there is no set to which every singleton (that is, every set of the
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form $\{x\}$) belongs.
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[\textit{Suggestion}: Show that from such a set, we could construct a set to
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which every set belonged.]
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\unverified{Exercise 3.9}}%
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\label{sub:exercise-3.9}
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Give an example of sets $a$ and $B$ for which $a \in B$ but
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$\powerset{A} \not\in \powerset{B}$.
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\unverified{Exercise 3.10}}%
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\label{sub:exercise-3.10}
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Show that if $a \in B$, then $\powerset{a} \in \powerset{\powerset{\bigcup B}}$.
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[\textit{Suggestion}: If you need help, look in the Appendix.]
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\begin{proof}
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TODO
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\end{proof}
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\end{document}
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